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Illusion [34]
3 years ago
10

You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happ

ens to their velocities?
a. Both increase at the same rate.
b. The velocity of the first rock increases faster than the velocity of the second.
c. The velocity of the second rock increases faster than the velocity of the first.
d. Both velocities stay constant.
Physics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

option A.

Explanation:

The correct answer is option A.

Two rocks are off a bridge first rock is fallen 4  when the second rock is dropped.

Both the rock is dropped under the effect of acceleration due to gravity so, the rate of change of the velocity for both the rock particle will be the same.

Hence, the first rock will reach ground earlier than the second rock because the rate of change of both the rock is at the same rate.

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B. The sound of the engine will get louder and the pitch higher.

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Answer:

The Total Mechanical Energy

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Explanation:

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A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i
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Answer:

Explanation:

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12 = 1/2 g t²

t = 1.565 s

Horizontal distance of throw

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This distance is to be covered by dog during the time ball falls ie 1.565 s

Speed of dog required = 13.3 / 1.565

= 8.5 m /s

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3 years ago
A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot
spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance <em>s</em><em> </em><em> </em>between the spokes along the rim is

s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}

The 20-cm arrow, traveling at 6 m/s, will travel its length in

t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}

The fastest speed that the wheel can spin without clipping the arrow is

v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}

The angular velocity \omega of the wheel is given by

\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}

In terms of rev/s, we can convert the answer above as follows:

23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

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2.6 seconds (If acceleration due to gravity is taken 10 m/s^2) or 2.65  (If acceleration due to gravity is taken 9.8 m/s^2)
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