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Nookie1986 [14]

2 years ago

8

How long would it take for a ball dropped from the top of a 576-foot building to hit the ground? round your answer to two decima

l places?

1 answer:

LUCKY_DIMON [66]2 years ago

4 0

If a ball is if a ball is dropped from a 576ft building it would take about 8 seconds for it to hit the ground.

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A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

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Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

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now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

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