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Solnce55 [7]
3 years ago
14

For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen

ts are given in the stemplot below. The median point total for this class.

Physics
1 answer:
Talja [164]3 years ago
5 0

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

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Subduction occurs at which of the following tectonic plate boundaries?
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

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\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

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\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

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which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

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