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Salsk061 [2.6K]
3 years ago
11

You throw a bowling ball down an alley with speed v_0v ​0 ​​ , and initially it slides without rolling. After a little while fri

ction kicks in and the ball begins to roll. What is the speed of the bowling ball when it starts rolling without slipping?
Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

Explanation:

As we know that point of contact at ground is taken as reference then there is no external torque about this point on the ball

So we can use angular momentum conservation about this point

L_i = L_f

L_i = mv_0R

L_f = \frac{7}{5}mR^2\omega

so we have

mv_0R = \frac{7}{5}mR^2\omega

R\omega = \frac{5}{7}v_0

So final speed after pure rolling is given as R\omega = \frac{5}{7}v_0

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The Pentium 4 Prescott processor, released in 2004, had a clock rate of 3.6 GHz and voltage of 1.25 V. Assume that, on average,
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Answer:

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% of Static Power = 10

For core i5 Ivy Bridge:

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. Boxes are sitting on a conveyor belt as the conveyor is turned on, moving the boxes toward the right. The belt reaches full sp
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Answer:

The acceleration of the boxes is 1.5 ft/s²

The displacement of the boxes during the speed-up period is 0.1875 ft.

Explanation:

Hi there!

Let´s convert the 45 ft/min into ft/s:

45 ft/min ·  1 min/ 60 s = 0.75 ft/s

It takes the belt 0.5 s to reach this speed. Then, the acceleration of the boxes will be:

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The equation of displacement is the following:

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Since the origin of the frame of reference is located at the point where the boxes begin to move, x0 = 0. Since the boxes were initially at rest, v0 = 0. Then:

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x = 0. 1875 ft

The displacement of the boxes during the speed-up period is 0.1875 ft.

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