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nexus9112 [7]
3 years ago
12

Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that they

couple and rotate as one unit. The first disk, with rotational inertia 2.11 kg·m2 about its central axis, is set spinning counterclockwise (which may be taken as the positive direction) at 372 rev/min. The second disk, with rotational inertia 8.41 kg·m2 about its central axis, is set spinning counterclockwise at 631 rev/min. They then couple together. (a) What is their angular speed after coupling? If instead the second disk is set spinning clockwise at 631 rev/min, what are their (b) angular velocity (using the correct sign for direction) and (c) direction of rotation after they couple together?

Physics
1 answer:
nadezda [96]3 years ago
4 0

Since no external torque is acting on the system you can use the conservation of angular momentum. I derived the final angular speed below and shown my work on how I did it. It’s now just a matter of plugging in the numbers and using correct placement of negative sign for direction of angular velocity. L in the picture stands for angular momentum. Hope it helps

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I think the answer may be D- mechanical
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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

4 0
3 years ago
Can anyone help with this?
sleet_krkn [62]

Answer:

A. The volume of the object

Explanation:

First of all, it can only PROBABLY be A or D

But why is it only A?

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It's VOLUME here and not MASS here because:

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  • If the Mass would be the answer, then it would be totally not related, because mass isn't related over here, it matters about size.

<h2>Hence, A. The volume of the object </h2>

is your answer!!!!

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