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2 years ago

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A system consists of a disk rotating on a frictionless axle and a piece of clay moving toward it, as shown in the figure above.

The outside edge of the disk is moving at a linear speed v, and the clay is moving at speed v/2. The clay sticks to the outside edge of the disk. How does the angular momentum of the system after the clay sticks compare to the angular momentum of the system before the clay sticks, and what is an explanation for the comparison?.

1 answer:

cestrela7 [59]2 years ago

6 0

Explanation:

Forgive me If I am wrong, it's been a while since I've studied Torque.

The formula for the angular momentum is

momentum= I*w.

We can also write I*W as 1/2MR^2 * W so the extra mass coming from the block of clay would most likely cause the angular momentum to increase from the amount it was before.

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umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

<em>Now force:</em>

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

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Data:
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