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2546 | Geometry | null | A regular pentagon covers part of another regular polygon, as shown. This regular polygon has $n$ sides, five of which are completely or partially visible. In the diagram, the sum of the measures of the angles marked $a^{\circ}$ and $b^{\circ}$ is $88^{\circ}$. Determine the value of $n$.
(The side lengths of a regular polygon are all equal, as are the measures of its interior angles.)
| [
"9"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2550 | Geometry | null | A circle has centre $O$ and radius 1. Quadrilateral $A B C D$ has all 4 sides tangent to the circle at points $P, Q, S$, and $T$, as shown. Also, $\angle A O B=\angle B O C=\angle C O D=\angle D O A$. If $A O=3$, determine the length of $D S$.
| [
"$\\frac{1}{2 \\sqrt{2}}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2566 | Geometry | null | In the diagram, $A B=21$ and $B C=16$. Also, $\angle A B C=60^{\circ}, \angle C A D=30^{\circ}$, and $\angle A C D=45^{\circ}$. Determine the length of $C D$, to the nearest tenth.
| [
"$\\frac{19 \\sqrt{2}}{\\sqrt{3}+1}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2567 | Geometry | null | In the diagram, the large circle has radius 9 and centre $C(15,0)$. The small circles have radius 4 and centres $A$ and $B$ on the horizontal line $y=12$. Each of the two small circles is tangent to the large circle. It takes a bug 5 seconds to walk at a constant speed from $A$ to $B$ along the line $y=12$. How far does the bug walk in 1 second?
| [
"2"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2573 | Geometry | null | In the diagram, $A B C$ is a right-angled triangle with $P$ and $R$ on $A B$. Also, $Q$ is on $A C$, and $P Q$ is parallel to $B C$. If $R P=2$, $B R=3, B C=4$, and the area of $\triangle Q R C$ is 5 , determine the length of $A P$.
| [
"1"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2579 | Geometry | null | In the diagram, sector $A O B$ is $\frac{1}{6}$ of an entire circle with radius $A O=B O=18$. The sector is cut into two regions with a single straight cut through $A$ and point $P$ on $O B$. The areas of the two regions are equal. Determine the length of $O P$.
| [
"$2 \\sqrt{3} \\pi$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2581 | Algebra | null | In the diagram, a straight, flat road joins $A$ to $B$.
Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds. | [
"27, 77"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2585 | Geometry | null | In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$.
If $A B=718$ and $P Q=250$, determine the length of $B C$. | [
"1375"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2591 | Geometry | null | How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ?
| [
"100"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2593 | Geometry | null | A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$
| [
"$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2594 | Geometry | null | In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre.
| [
"141"
] | false | m | Numerical | null | OE_MM_maths_en_COMP |
|
2595 | Geometry | null | The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ?
| [
"$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$"
] | true | null | Tuple | null | OE_MM_maths_en_COMP |
|
2605 | Geometry | null | In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$.
| [
"$\\frac{\\sqrt{6}}{4}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2716 | Geometry | null | Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$.
| [
"$\\frac{5 \\sqrt{39}}{3}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2719 | Geometry | null | Let $T=T N Y W R$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$.
| [
"8"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2743 | Geometry | null | Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$.
| [
"688"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2769 | Geometry | null | Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace.
| [
"16"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2774 | Geometry | null | Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$.
| [
"7"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2779 | Geometry | null | In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$.
| [
"3"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2806 | Combinatorics | null | This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Find $C(M)$ for the map below.
| [
"6"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2810 | Combinatorics | null | This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.
Determine $W(M)$ for each of the maps in figure a and figure b.
figure a
figure b | [
"12, 8"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2820 | Geometry | null | In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs.
| [
"$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2826 | Geometry | null | Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$.
| [
"$\\frac{9}{5}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2833 | Geometry | null | Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$.
| [
"15"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2837 | Geometry | null | Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$.
| [
"$10 \\sqrt{6}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2868 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Consider the shape below:
Find the 2-signature that corresponds to this shape. | [
"$(12,21,21,21,12)$"
] | false | null | Tuple | null | OE_MM_maths_en_COMP |
|
2869 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
List all 5-labels with 2-signature $(12,12,21,21)$. | [
"12543,13542,14532,23541,24531,34521"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2870 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Find a formula for the number of $(2 n+1)$-labels with the 2 -signature
$$
(\underbrace{12,12, \ldots, 12}_{n}, \underbrace{21,21, \ldots, 21}_{n})
$$ | [
"$\\binom{2n}{n}$"
] | false | null | Expression | null | OE_MM_maths_en_COMP |
|
2871 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Compute the number of 5-labels with 2 -signature $(12,21,12,21)$. | [
"16"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2872 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Determine the number of 9-labels with 2-signature
$$
(12,21,12,21,12,21,12,21) \text {. }
$$
Justify your answer. | [
"7936"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2875 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
For a general $n$, determine the number of distinct possible $p$-signatures. | [
"$p ! \\cdot p^{n-p}$"
] | false | null | Expression | null | OE_MM_maths_en_COMP |
|
2876 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
If a randomly chosen $p$-signature is 575 times more likely of being impossible than possible, determine $p$ and $n$. | [
"7,5"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2879 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Find two 5-labels with unique 2-signatures. | [
"12345, 54321"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2882 | Combinatorics | null | An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
Determine the smallest $p$ for which the 20-label
$$
L=3,11,8,4,17,7,15,19,6,2,14,1,10,16,5,12,20,9,13,18
$$
has a unique $p$-signature. | [
"16"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2886 | Geometry | null | In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I.
| [
"1476"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2889 | Combinatorics | null | The numbers $1,2, \ldots, 8$ are placed in the $3 \times 3$ grid below, leaving exactly one blank square. Such a placement is called okay if in every pair of adjacent squares, either one square is blank or the difference between the two numbers is at most 2 (two squares are considered adjacent if they share a common side). If reflections, rotations, etc. of placements are considered distinct, compute the number of distinct okay placements.
| [
"32"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2906 | Geometry | null | Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units.
| [
"5"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2923 | Geometry | null | Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$.
| [
"49"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2929 | Geometry | null | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1. | [
"$\\frac{5 \\pi}{18}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2930 | Geometry | null | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the number of curvilinear territories remaining at the end of day 3. | [
"54"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2932 | Geometry | null | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the total number of plots sold up to and including day $n$. | [
"$3^{n}+1$"
] | false | null | Expression | null | OE_MM_maths_en_COMP |
|
2933 | Geometry | null | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles. | [
"$2$, $\\frac{2}{15}$"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2940 | Geometry | null | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Find the areas of the circles removed on day 3. | [
"$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
2959 | Geometry | null | Points $A$ and $L$ lie outside circle $\omega$, whose center is $O$, and $\overline{A L}$ contains diameter $\overline{R M}$, as shown below. Circle $\omega$ is tangent to $\overline{L K}$ at $K$. Also, $\overline{A K}$ intersects $\omega$ at $Y$, which is between $A$ and $K$. If $K L=3, M L=2$, and $\mathrm{m} \angle A K L-\mathrm{m} \angle Y M K=90^{\circ}$, compute $[A K M]$ (i.e., the area of $\triangle A K M$ ).
| [
"$\\frac{375}{182}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
2965 | Geometry | null | Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle.
| [
"9"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3050 | Combinatorics | null | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$. | [
"15, 35, 70, 126"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
3051 | Combinatorics | null | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$. | [
"$m=n+k+1$, $j=n+1$"
] | true | null | Expression | null | OE_MM_maths_en_COMP |
|
3058 | Combinatorics | null | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
If $\mathrm{Cl}(n, 1)=a n^{2}+b n+c$, determine the values of $a, b$, and $c$. | [
"$3,-3,1$"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
3060 | Combinatorics | null | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
Compute $\mathrm{Cl}(11,2)$. | [
"1000"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3062 | Combinatorics | null | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
Compute $\mathrm{Cl}(11,3)$. | [
"2025"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3065 | Combinatorics | null | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute Le(17,1). | [
"$\\frac{1}{306}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3066 | Combinatorics | null | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute $\operatorname{Le}(17,2)$. | [
"$\\frac{1}{2448}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3068 | Combinatorics | null | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$. | [
"$\\frac{2011}{2012}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
3070 | Combinatorics | null | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=1}^{\infty} \operatorname{Le}(i, 1)=\operatorname{Le}(n, k)$, determine the values of $n$ and $k$. | [
"$0,0$"
] | true | null | Numerical | null | OE_MM_maths_en_COMP |
|
3071 | Combinatorics | null | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, compute expressions for $n$ and $k$ in terms of $m$. | [
"$m-1,m-1$"
] | true | null | Expression | null | OE_MM_maths_en_COMP |
|
3093 | Geometry | null | $\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles.
| [
"$72 \\sqrt{2}$"
] | false | null | Numerical | null | OE_MM_maths_en_COMP |
|
800 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A quarantined physics student decides to perform an experiment to land a small box of mass $m=60 \mathrm{~g}$ onto the center of a target a distance $\Delta d$ away. The student puts the box on a top of a frictionless ramp with height $h_{2}=0.5 \mathrm{~m}$ that is angled $\theta=30^{\circ}$ to the horizontal on a table that is $h_{1}=4 \mathrm{~m}$ above the floor. If the student pushes the spring with spring constant $k=6.5 \mathrm{~N} / \mathrm{m}$ down by $\Delta x=0.3 \mathrm{~m}$ compared to its rest length and lands the box exactly on the target, what is $\Delta d$ ? Answer in meters. You may assume friction is negligible.
| [
"2.47"
] | false | m | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
801 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A wooden bus of mass $M=20,000 \mathrm{~kg}$ ( $M$ represents the mass excluding the wheels) is on a ramp with angle $30^{\circ}$. Each of the four wheels is composed of a ring of mass $\frac{M}{2}$ and radius $R=1 \mathrm{~m}$ and 6 evenly spaced spokes of mass $\frac{M}{6}$ and length $R$. All components of the truck have a uniform density. Find the acceleration of the bus down the ramp assuming that it rolls without slipping.
Answer in $\mathrm{m} / \mathrm{s}^{2}$.
| [
"3.32"
] | false | m | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
804 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A ball is situated at the midpoint of the bottom of a rectangular ditch with width $1 \mathrm{~m}$. It is shot at a velocity $v=5 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ relative to the horizontal. How many times does the ball collide with the walls of the ditch until it hits the bottom of the ditch again? Assume all collisions to be elastic and that the ball never flies out of the ditch.
| [
"2"
] | false | null | Numerical | 2e-1 | OE_MM_physics_en_COMP |
|
805 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball (of negligible size) of mass $M=5 \mathrm{~kg}$.
Let $h$ be the minimum height that $m$ was dropped such that $M$ would be able to move all the way around the loop. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$.
Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If $m$ was dropped from a height $3 R$, find the minimum value of $\frac{m}{M}$ such that the combined mass can fully move all the way around the loop. Let this minimum value be $k$. Compute $\alpha=\frac{k^{2}}{h^{2}}$. (Note that this question is only asking for $\alpha$ but you need to find $h$ to find $\alpha$ ). Assume the balls are point masses (neglect rotational effects). | [
"$2.37$"
] | false | $\mathrm{~m}^{2}$ | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
809 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | The following diagram depicts a single wire that is bent into the shape below. The circuit is placed in a magnetic field pointing out of the page, uniformly increasing at the rate $\frac{d B}{d t}=2.34 \mathrm{~T} / \mathrm{s}$. Calculate the magnitude of induced electromotive force in the wire, in terms of the following labelled areas $\left(\mathrm{m}^{2}\right)$. Note that $B$ is non-inclusive of $C$ and that $A=4.23, B=2.74$, and $C=0.34$.
| [
"$1.9$"
] | false | $\mathrm{Tm}^{2} \mathrm{~s}^{-1}$ | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
815 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | For his art project, Weishaupt cut out $N=20$ wooden equilateral triangular blocks with a side length of $\ell=10 \mathrm{~cm}$ and a thickness of $t=2 \mathrm{~cm}$, each with the same mass and uniform density. He wishes to stack one on top of the other overhanging the edge of his table. In centimeters, what is the maximum overhang? Round to the nearest centimeter. A side view is shown below. Assume that all triangles are parallel to each other.
Note: This diagram is not to scale.
| [
"21"
] | false | cm | Numerical | 5e-1 | OE_MM_physics_en_COMP |
|
818 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | The graph provided plots the $y$-component of the velocity against the $x$-component of the velocity of a kiddie roller coaster at an amusement park for a certain duration of time. The ride takes place entirely in a two dimensional plane.
Some students made a remark that at one time, the acceleration was perpendicular to the velocity. Using this graph, what is the minimum x-velocity the ride could be travelling at for this to be true? Round to the nearest integer and answer in meters per second. The diagram is drawn to scale, and you may print this page out and make measurements.
| [
"1"
] | false | m/s | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
820 | Modern Physics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | While exploring outer space, Darth Vader comes upon a purely reflective spherical planet with radius $R_{p}=$ $40,000 \mathrm{~m}$ and mass $M_{p}=8.128 \times 10^{24} \mathrm{~kg}$. Around the planet is a strange moon of orbital radius $R_{s}=$ $6,400,000 \mathrm{~m}\left(R_{s} \gg R_{p}\right)$ and mass $M_{s}=9.346 \times 10^{19} \mathrm{~kg}\left(M_{s} \ll M_{p}\right)$. The moon can be modelled as a blackbody and absorbs light perfectly. Darth Vader is in the same plane that the planet orbits in. Startled, Darth Vader shoots a laser with constant intensity and power $P_{0}=2 \times 10^{32} \mathrm{~W}$ at the reflective planet and hits the planet a distance of $\frac{R_{p}}{2}$ away from the line from him to the center of the planet. Upon hitting the reflective planet, the light from the laser is plane polarized. The angle of the planet's polarizer is always the same as the angle of reflection. After reflectance, the laser lands a direct hit on the insulator planet. Darth Vader locks the laser in on the planet until it moves right in front of him, when he turns the laser off. Determine the energy absorbed by the satellite. Assume the reflective planet remains stationary and that the reflective planet is a perfect polarizer of light.
| [
"$1.33 \\cdot 10^{35}$"
] | false | J | Numerical | 1e34 | OE_MM_physics_en_COMP |
|
822 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Mario is racing with Wario on Moo Moo Meadows when a goomba, ready to avenge all of his friends' deaths, came and hijacked Mario's kart. A graph representing the motion of Mario at any instant is shown below. The velocity acquired by Mario is shown on the x-axis, and the net power of his movement is shown on the $\mathrm{y}$-axis. When Mario's velocity is $6 \mathrm{~m} / \mathrm{s}$, he eats a mushroom which gives him a super boost.
You may need to make measurements. Feel free to print this picture out as the diagram is drawn to scale. Find the total distance from Mario runs from when his velocity is $0 \mathrm{~m} / \mathrm{s}$ to when his velocity just reaches $9 \mathrm{~m} / \mathrm{s}$ given that Mario's mass is $m=89 \mathrm{~kg}$. Answer in meters and round to one significant digit. | [
"40"
] | false | m | Numerical | 5e0 | OE_MM_physics_en_COMP |
|
824 | Thermodynamics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | An engineer has access to a tetrahedron building block with side length $\ell=10 \mathrm{~cm}$. The body is made of a thermal insulator but the edges are wrapped with a thin copper wiring with cross sectional area $S=2 \mathrm{~cm}^{2}$. The thermal conductivity of copper is $385.0 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. He stacks these tetrahedrons (all facing the same direction) to form a large lattice such that the copper wires are all in contact. In the diagram, only the front row of a small section is coloured. Assume that the lattice formed is infinitely large.
At some location in the tetrahedral building block, the temperature difference between two adjacent points is $1^{\circ} \mathrm{C}$. What is the heat flow across these two points? Answer in Watts.
Note: Two adjacent points refer to two adjacent points on the tetrahedron.
| [
"4.62"
] | false | N | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
825 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Three unit circles, each with radius 1 meter, lie in the same plane such that the center of each circle is one intersection point between the two other circles, as shown below. Mass is uniformly distributed among all area enclosed by at least one circle. The mass of the region enclosed by the triangle shown above is $1 \mathrm{~kg}$. Let $x$ be the moment of inertia of the area enclosed by all three circles (intersection, not union) about the axis perpendicular to the page and through the center of mass of the triangle. Then, $x$ can be expressed as $\frac{a \pi-b \sqrt{c}}{d \sqrt{e}}$ $\mathrm{kg} \mathrm{m}^{2}$, where $a, b, c, d, e$ are integers such that $\operatorname{gcd}(a, b, d)=1$ and both $c$ and $e$ are squarefree. Compute $a+b+c+d+e$.
| [
"39"
] | false | null | Numerical | 1e0 | OE_MM_physics_en_COMP |
|
827 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Two infinitely long current carrying wires carry constant current $i_{1}=2 \mathrm{~A}$ and $i_{2}=3 \mathrm{~A}$ as shown in the diagram. The equations of the wire curvatures are $y^{2}-8 x-6 y+25=0$ and $x=0$. Find the magnitude of force (in Newtons) acting on one of the wires due to the other.
Note: The current-carrying wires are rigidly fixed. The units for distances on the graph should be taken in metres. | [
"$7.5398 \\cdot 10^{-6}$"
] | false | N | Numerical | 1e-7 | OE_MM_physics_en_COMP |
|
829 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Two electrons are in a uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}}$ where $E_{0}=10^{-11} \mathrm{~N} / \mathrm{C}$. One electron is at the origin, and another is $10 \mathrm{~m}$ above the first electron. The electron at the origin is moving at $u=10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ from the line connecting the electrons at $t=0$, while the other electron is at rest at $t=0$. Find the minimum distance between the electrons. You may neglect relativistic effects.
| [
"6.84"
] | false | m | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
830 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Consider a long uniform conducting cylinder. First, we divide the cylinder into thirds and remove the middle third. Then, we perform the same steps on the remaining two cylinders. Again, we perform the same steps on the remaining four cylinders and continuing until there are 2048 cylinders.
We then connect the terminals of the cylinder to a battery and measure the effective capacitance to be $C_{1}$. If we continue to remove cylinders, the capacitance will reach an asymptotic value of $C_{0}$. What is $C_{1} / C_{0}$ ?
You may assume each cylindrical disk to be wide enough to be considered as an infinite plate, such that the radius $R$ of the cylinders is much larger than the $d$ between any successive cylinders.
Note: The diagram is not to scale. | [
"1.017"
] | false | null | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
831 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A square based pyramid (that is symmetrical) is standing on top of a cube with side length $\ell=10 \mathrm{~cm}$ such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of $\mu=0.3$. The cube is then given an initial speed $v$ in some direction parallel to the floor. What is the maximum possible value of $v$ such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second.
| [
"1.07"
] | false | m/s | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
832 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | During quarantine, the FBI has been monitoring a young physicists suspicious activities. After compiling weeks worth of evidence, the FBI finally has had enough and searches his room.
The room's door is opened with a high angular velocity about its hinge. Over a very short period of time, its angular velocity increases to $\omega=8.56 \mathrm{rad} / \mathrm{s}$ due to the force applied at the end opposite from the hinge. For simplicity, treat the door as a uniform thin rod of length $L=1.00 \mathrm{~m}$ and mass $M=9.50 \mathrm{~kg}$. The hinge (pivot) is located at one end of the rod. Ignore gravity. At what distance from the hinge of the door is the door most likely to break? Assume that the door will break at where the bending moment is largest. (Answer in metres.)
| [
"$\\frac{\\sqrt{3}}{3}$"
] | false | m | Numerical | 1e-2 | OE_MM_physics_en_COMP |
|
833 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A solid half-disc of mass $m=1 \mathrm{~kg}$ in the shape of a semi-circle of radius $R=1 \mathrm{~m}$ is kept at rest on a smooth horizontal table. QiLin starts applying a constant force of magnitude $F=10 \mathrm{~N}$ at point A as shown, parallel to its straight edge. What is the initial linear acceleration of point B? (Answer in $\mathrm{m} / \mathrm{s}^{2}$ )
Note: the diagram above is a top down view. | [
"$15.9395$"
] | false | $\mathrm{m} / \mathrm{s}^{2}$ | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
834 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A regular tetrahedron of mass $m=1 \mathrm{~g}$ and unknown side length is balancing on top of a hemisphere of mass $M=100 \mathrm{~kg}$ and radius $R=100 \mathrm{~m}$. The hemisphere is placed on a flat surface such that it is at its lowest potential. For a certain value of the length of the regular tetrahedron, the oscillations become unstable. What is this side length of the tetrahedron?
| [
"$200\\sqrt{6}$"
] | false | m | Numerical | 5e0 | OE_MM_physics_en_COMP |
|
838 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball of mass $M=5 \mathrm{~kg}$.
The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different alternative. Now let the circular loop have a uniform coefficient of friction $\mu=0.6$, while the rest of the path is still frictionless. Assume that the balls can once again collide with a restitution coefficient of $e=\frac{1}{2}$. Considering the balls to be point masses, find the minimum value of $h$ such that the ball of mass $M$ would be able to move all the way around the loop. Both balls can be considered as point masses. | [
"$72.902$"
] | false | m | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
839 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the "ground" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless.
Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree.
Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$.
| [
"$60.2$"
] | false | $^{\circ}$ | Numerical | 2e-1 | OE_MM_physics_en_COMP |
|
841 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A bicycle wheel of mass $M=2.8 \mathrm{~kg}$ and radius $R=0.3 \mathrm{~m}$ is spinning with angular velocity $\omega=5 \mathrm{rad} / \mathrm{s}$ around its axis in outer space, and its center is motionless. Assume that it has all of its mass uniformly concentrated on the rim. A long, massless axle is attached to its center, extending out along its axis. A ball of mass $m=1.0 \mathrm{~kg}$ moves at velocity $v=2 \mathrm{~m} / \mathrm{s}$ parallel to the plane of the wheel and hits the axle at a distance $h=0.5 \mathrm{~m}$ from the center of the wheel. Assume that the collision is elastic and instantaneous, and that the ball's trajectory (before and after the collision) lies on a straight line.
Find the time it takes for the axle to return to its original orientation. Answer in seconds and round to three significant figures. | [
"0.568"
] | false | s | Numerical | 5e-2 | OE_MM_physics_en_COMP |
|
843 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A small toy car rolls down three ramps with the same height and horizontal length, but different shapes, starting from rest. The car stays in contact with the ramp at all times and no energy is lost. Order the ramps from the fastest to slowest time it takes for the toy car to drop the full $1 \mathrm{~m}$. For example, if ramp 1 is the fastest and ramp 3 is the slowest, then enter 123 as your answer choice.
Figure 1
Figure 2
Figure 3
| [
"213"
] | false | null | Numerical | 0 | OE_MM_physics_en_COMP |
|
845 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | In a typical derby race, cars start at the top of a ramp, accelerate downwards, and race on a flat track, and are always set-up in the configuration shown below.
A common technique is to change the location of the center of mass of the car to gain an advantage. Alice ensures the center of mass of her car is at the rear and Bob puts the center of mass of his car at the very front. Otherwise, their cars are exactly the same. Each car's time is defined as the time from when the car is placed on the top of the ramp to when the front of the car reaches the end of the flat track. At the competition, Alice's car beat Bob's. What is the ratio of Bob's car's time and Alice's car's time?
Assume that the wheels are small and light compared to the car body, neglect air resistance, and the height of the cars are small compared to the height of the ramp. In addition, neglect all energy losses during the race and the time it takes to turn onto the horizontal surface from the ramp. Express your answer as a decimal greater than 1 . | [
"1.03"
] | false | null | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
846 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A simple crane is shown in the below diagram, consisted of light rods with length $1 \mathrm{~m}$ and $\sqrt{2} \mathrm{~m}$. The end of the crane is supporting a $5 \mathrm{kN}$ object. Point $B$ is known as a "pin." It is attached to the main body and can exert both a vertical and horizontal force. Point $A$ is known as a "roller" and can only exert vertical forces. Rods can only be in pure compression or pure tension.
In $\mathrm{kN}$, what is the force experienced by the rod $C D$ ? Express a positive number if the member is in tension and a negative number if it is in compression. | [
"10"
] | false | kN | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
848 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A small block of mass $m$ and charge $Q$ is placed at rest on an inclined plane with a slope $\alpha=40^{\circ}$. The coefficient of friction between them is $\mu=0.3$. A homogenous magnetic field of magnitude $B_{0}$ is applied perpendicular to the slope. The speed of the block after a very long time is given by $v=\beta \frac{m g}{Q B_{0}}$. Determine $\beta$. Do not neglect the effects of gravity.
| [
"0.6"
] | false | null | Numerical | 5e-2 | OE_MM_physics_en_COMP |
|
850 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A sprinkler fountain is in the shape of a semi-sphere that spews out water from all angles at a uniform speed $v$ such that without the presence of wind, the wetted region around the fountain forms a circle in the $X Y$ plane with the fountain centered on it.
Now suppose there is a constant wind blowing in a direction parallel to the ground such that the force acting on each water molecule is proportional to their weight. The wetted region forms the shape below where the fountain is placed at $(0,0)$. Determine the exit speed of water $v$ in meters per second. Round to two significant digits. All dimensions are in meters.
| [
"2.5"
] | false | m/s | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
852 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Poncho is a very good player of the legendary carnival game known as Pico-Pico. Its setup consists of a steel ball, represented by a point mass, of negligible radius and a frictionless vertical track. The goal of Pico-Pico is to flick the ball from the beginning of the track (point $A$ ) such that it is able to traverse through the track while never leaving the track, successfully reaching the end (point $B)$. The most famous track design is one of parabolic shape; specifically, the giant track is of the shape $h(x)=5-2 x^{2}$ in meters. The starting and ending points of the tracks are where the two points where the track intersects $y=0$. If $\left(v_{a}, v_{b}\right]$ is the range of the ball's initial velocity $v_{0}$ that satisfies the winning condition of Pico-Pico, help Poncho find $v_{b}-v_{a}$. This part is depicted below:
| [
"0.1231"
] | false | m/s | Numerical | 1e-2 | OE_MM_physics_en_COMP |
|
854 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Anyone who's had an apple may know that pieces of an apple stick together, when picking up one piece a second piece may also come with the first piece. The same idea is tried on a golden apple. Consider two uniform hemispheres with radius $r=4 \mathrm{~cm}$ made of gold of density $\rho_{g}=19300 \mathrm{~kg} \mathrm{~m}^{-3}$. The top half is nailed to a support and the space between is filled with water.
Given that the surface tension of water is $\gamma=0.072 \mathrm{~N} \mathrm{~m}^{-1}$ and that the contact angle between gold and water is $\theta=10^{\circ}$, what is the maximum distance between the two hemispheres so that the bottom half doesn't fall? Answer in millimeters. | [
"0.0281"
] | false | mm | Numerical | 2e-3 | OE_MM_physics_en_COMP |
|
855 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | In the following two problems we will look at shooting a basketball. Model the basketball as an elastic hollow sphere with radius 0.1 meters. Model the net and basket as shown below, dimensions marked. Neglect friction between the backboard and basketball, and assume all collisions are perfectly elastic.
For this problem, you launch the basketball from the point that is 2 meters above the ground and 4 meters from the backboard as shown. You attempt to make a shot by hitting the basketball off the backboard as depicted above. What is the minimum initial speed required for the ball to make this shot?
Note: For this problem, you may assume that the size of the ball is negligible. | [
"7.19"
] | false | m/s | Numerical | 1e-2 | OE_MM_physics_en_COMP |
|
857 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Consider a toilet paper roll with some length of it hanging off as shown. The toilet paper roll rests on a cylindrical pole of radius $r=1 \mathrm{~cm}$ and the coefficient of static friction between the role and the pole is $\mu=0.3$.
The length of the paper hanging off has length $\ell=30 \mathrm{~cm}$ and the inner radius of the roll is $R_{i}=2 \mathrm{~cm}$. The toilet paper has thickness $s=0.1 \mathrm{~mm}$ and mass per unit length $\lambda=5 \mathrm{~g} / \mathrm{m}$. What is the minimum outer radius $R_{o}$ such that the toilet paper roll remains static? Answer in centimeters. | [
"2.061"
] | false | cm | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
858 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | In the circuit shown below, a capacitor $C=4 \mathrm{~F}$, inductor $L=5 \mathrm{H}$, and resistors $R_{1}=3 \Omega$ and $R_{2}=2 \Omega$ are placed in a diamond shape and are then fed an alternating current with peak voltage $V_{0}=1 \mathrm{~V}$ of unknown frequency. Determine the magnitude of the maximum instantaneous output voltage shown in the diagram.
| [
"0.65"
] | false | V | Numerical | 5e-2 | OE_MM_physics_en_COMP |
|
859 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A uniform bar of length $l$ and mass $m$ is connected to a very long thread of negligible mass suspended from a ceiling. It is then rotated such that it is vertically upside down and then released. Initially, the rod is in unstable equilibrium. As it falls down, the minimum tension acting on the thread over the rod's entire motion is given by $\alpha m g$. Determine $\alpha$.
| [
"0.165"
] | false | mg | Numerical | 5e-3 | OE_MM_physics_en_COMP |
|
862 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Colliding Conducting Slab A thin conducting square slab with side length $s=5 \mathrm{~cm}$, initial charge $q=0.1 \mu \mathrm{C}$, and mass $m=100 \mathrm{~g}$ is given a kick and sent bouncing between two infinite conducting plates separated by a distance $d=0.5 \mathrm{~cm} \ll s$ and with surface charge density $\pm \sigma= \pm 50 \mu \mathrm{C} / \mathrm{m}^{2}$. After a long time it is observed exactly in the middle of the two plates to be traveling with velocity of magnitude $v=3 \mathrm{~m} / \mathrm{s}$ and direction $\theta=30^{\circ}$ with respect to the horizontal line parallel to the plates. How many collisions occur after it has traveled a distance $L=15 \mathrm{~m}$ horizontally from when it was last observed? Assume that all collisions are elastic, and neglect induced charges. Note that the setup is horizontal so gravity does not need to be accounted for.
| [
"25273"
] | false | null | Numerical | 1e2 | OE_MM_physics_en_COMP |
|
864 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Spinning Cylinder Adithya has a solid cylinder of mass $M=10 \mathrm{~kg}$, radius $R=0.08 \mathrm{~m}$, and height $H=0.20 \mathrm{~m}$. He is running a test in a chamber on Earth over a distance of $d=200 \mathrm{~m}$ as shown below. Assume that the physical length of the chamber is much greater than $d$ (i.e. the chamber extends far to the left and right of the testing area). The chamber is filled with an ideal fluid with uniform density $\rho=700 \mathrm{~kg} / \mathrm{m}^{3}$. Adithya's cylinder is launched with linear velocity $v=10 \mathrm{~m} / \mathrm{s}$ and spins counterclockwise with angular velocity $\omega$. Adithya notices that the cylinder continues on a horizontal path until the end of the chamber. Find the angular velocity $\omega$. Do not neglect forces due to fluid pressure differences. Note that the diagram presents a side view of the chamber (i.e. gravity is oriented downwards with respect to the diagram).
Assume the following about the setup and the ideal fluid:
- fluid flow is steady in the frame of the center of mass of the cylinder
- the ideal fluid is incompressible, irrotational, and has zero viscosity
- the angular velocity of the cylinder is approximately constant during its subsequent motion
Hint: For a uniform cylinder of radius $R$ rotating counterclockwise at angular velocity $\omega$ situated in an ideal fluid with flow velocity $u$ to the right far away from the cylinder, the velocity potential $\Phi$ is given by
$$
\Phi(r, \theta)=u r \cos \theta+u \frac{R^{2}}{r} \cos \theta+\frac{\Gamma \theta}{2 \pi}
$$
where $(r, \theta)$ is the polar coordinate system with origin at the center of the cylinder. $\Gamma$ is the circulation and is equal to $2 \pi R^{2} \omega$. The fluid velocity is given by
$$
\mathbf{v}=\nabla \Phi=\frac{\partial \Phi}{\partial r} \hat{\mathbf{r}}+\frac{1}{r} \frac{\partial \Phi}{\partial \theta} \hat{\theta}
$$ | [
"$1.25$"
] | false | $\mathrm{s}^{-1}$ | Numerical | 1e-2 | OE_MM_physics_en_COMP |
|
867 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Consider a rectangular loop made of superconducting material with length $\ell=200 \mathrm{~cm}$ and width $w=2 \mathrm{~cm}$. The radius of this particular wire is $r=0.5 \mathrm{~mm}$. This superconducting rectangular loop initially has a current $I_{1}=5 \mathrm{~A}$ in the counterclockwise direction as shown in the figure below. This rectangular loop is situated a distance $d=1 \mathrm{~cm}$ above an infinitely long wire that initially contains no current. Suppose that the current in the infinitely long wire is increased to some current $I_{2}$ such that there is an attractive force $F$ between the rectangular loop and the long wire. Find the maximum possible value of $F$. Write your answer in newtons.
Hint: You may neglect the magnetic field produced by the vertical segments in the rectangular loop.
| [
"$1.12 \\times 10^{-3}$"
] | false | N | Numerical | 1e-4 | OE_MM_physics_en_COMP |
|
874 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Consider the following simple model of a bow and arrow. An ideal elastic string has a spring constant $k=10 \mathrm{~N} / \mathrm{m}$ and relaxed length $L=1 \mathrm{~m}$ which is attached to the ends of an inflexible fixed steel rod of the same length $L$ as shown below. A small ball of mass $m=2 \mathrm{~kg}$ and the thread are pulled by its midpoint away from the rod until each individual part of the thread have the same length of the rod, as shown below. What is the speed of the ball in meters per seconds right after it stops accelerating? Assume the whole setup is carried out in zero gravity.
| [
"$2.23$"
] | false | m/s | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
875 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A truck (denoted by $S$ ) is driving at a speed $v=2 \mathrm{~m} / \mathrm{s}$ in the opposite direction of a car driving at a speed $u=3 \mathrm{~m} / \mathrm{s}$, which is equipped with a rear-view mirror. Both $v$ and $u$ are measured from an observer on the ground. Relative to this observer, what is the speed (in $\mathrm{m} / \mathrm{s}$ ) of the truck's image $S^{\prime}$ through the car's mirror? Car's mirror is a plane mirror.
| [
"$8$"
] | false | m/s | Numerical | 1e-1 | OE_MM_physics_en_COMP |
|
879 | Optics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | These days, there are so many stylish rectangular home-designs (see figure A). It is possible from the outline of those houses in their picture to estimate with good precision where the camera was. Consider an outline in one photograph of a rectangular house which has height $H=3$ meters (see figure B for square-grid coordinates). Assume that the camera size is negligible, how high above the ground (in meters) was the camera at the moment this picture was taken? | [
"$0.9$"
] | false | m | Numerical | 1e-2 | OE_MM_physics_en_COMP |
|
880 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | Consider a thin rigid wire-frame MNPP'N'M' in which MNN'M' and NPP'N' are two squares of side $L$ with resistance per unit-length $\lambda$ and their planes are perpendicular. The frame is rotated with a constant angular velocity $\omega$ around an axis passing through $\mathrm{NN}$ ' and put in a region with constant magnetic field $B$ pointing perpendicular to $\mathrm{NN}^{\prime}$. What is the total heat released on the frame per revolution (in Joules)? Use $L=1 \mathrm{~m}, \lambda=1 \Omega / \mathrm{m}, \omega=2 \pi \mathrm{rad} / \mathrm{s}$ and $B=1 \mathrm{~T}$.
| [
"$6.58$"
] | false | J | Numerical | 2e-2 | OE_MM_physics_en_COMP |
|
882 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A scale of uniform mass $M=3 \mathrm{~kg}$ of length $L=4 \mathrm{~m}$ is kept on a rough table (infinite friction) with $l=1 \mathrm{~m}$ hanging out of the table as shown in the figure below. A small ball of mass $m=1 \mathrm{~kg}$ is released from rest from a height of $h=5 \mathrm{~m}$ above the end of the scale. Find the maximum angle (in degrees) that the scale rotates by in the subsequent motion if ball sticks to the scale after collision. Take gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.
| [
"$18.21$"
] | false | $^{\circ}$ | Numerical | 5e-1 | OE_MM_physics_en_COMP |
|
886 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | The following information applies for the next two problems. A circuit has a power source of $\mathcal{E}=5.82 \mathrm{~V}$ connected to three elements in series: an inductor with $L=12.5 \mathrm{mH}$, a capacitor with $C=48.5 \mu \mathrm{F}$, and a diode with threshold voltage $V_{0}=0.65 \mathrm{~V}$. (Of course, the polarity of the diode is aligned with that of the power source.) You close the switch, and after some time, the voltage across the capacitor becomes constant. (Note: An ideal diode with threshold voltage $V_{0}$ is one whose IV characteristic is given by $I=0$ for $V<V_{0}$ and $V=V_{0}$ for $I>0$.)
How much time (in seconds) has elapsed before the voltage across the capacitor becomes constant? | [
"$2.446 \\times 10^{-3}$"
] | false | s | Numerical | 1e-4 | OE_MM_physics_en_COMP |
|
887 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | At Hanoi-Amsterdam High School in Vietnam, every subject has its own flag (see Figure A, taken by Tung X. Tran). While the flags differ in color, they share the same central figure. Consider a planar conducting frame of that figure rotating at a constant angular velocity in a uniform magnetic field (see Figure B). The frame is made of thin rigid wires with same uniform curvature and same resistance per unit length. What fraction of the total heat released is released by the outermost wires? | [
"$0.864$"
] | false | J | Numerical | 5e-2 | OE_MM_physics_en_COMP |
|
889 | Mechanics | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | A tesseract is a 4 dimensional example of cube. It can be drawn in 3 dimensions by drawing two cubes and connecting their vertices together as shown in the picture below:
Now for the 3D equivalent. The lines connecting the vertices are replaced with ideal springs of constant $k=10 \mathrm{~N} / \mathrm{m}$ (in blue in the figure). Now, suppose the setup is placed in zero-gravity and the outer cube is fixed in place with a sidelength of $b=2 \mathrm{~m}$. The geometric center of the inner cube is placed in the geometric center of the outer cube, and the inner cube has a side-length $a=1 \mathrm{~m}$ and mass $m=1.5 \mathrm{~kg}$. The inner cube is slightly displaced from equilibrium. Consider the period of oscillations
- $T_{1}$ : when the springs have a relaxed length of 0 ;
- $T_{2}$ : when the springs are initially relaxed before the inner cube is displaced.
What is $T_{1}+T_{2}$ ? | [
"2.35"
] | false | s | Numerical | 5e-2 | OE_MM_physics_en_COMP |
|
892 | Electromagnetism | - Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,
$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$
- Solar Mass
$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$
- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,
$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$
- Planck's constant,
$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$
- Permittivity of free space,
$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$
- Coulomb's law constant,
$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$
- Permeability of free space,
$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$
- Magnetic constant,
$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$
- 1 atmospheric pressure,
$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$
- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$ | An open electrical circuit contains a wire loop in the shape of a semi-circle, that contains a resistor of resistance $R=0.2 \Omega$. The circuit is completed by a conducting pendulum in the form of a uniform rod with length $\ell=0.1 \mathrm{~m}$ and mass $m=0.05 \mathrm{~kg}$, has no resistance, and stays in contact with the other wires at all times. All electrical components are oriented in the $y z$ plane, and gravity acts in the $z$ direction. A constant magnetic field of strength $B=2 \mathrm{~T}$ is applied in the $+x$ direction.
Ignoring self inductance and assuming that $\alpha \ll 1$, the general equation of motion is in the form of $\theta(t)=A(t) \cos (\omega t+\varphi)$, where $A(t) \geq 0$. Find $\omega^{2}$. | [
"$145$"
] | false | $\mathrm{~s}^{-1}$ | Numerical | 0 | OE_MM_physics_en_COMP |
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