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andre [41]
3 years ago
5

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?
Physics
1 answer:
kaheart [24]3 years ago
8 0

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

mu=(m+M)v

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s

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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
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15.065ft

Explanation:

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By definition the drag force is expressed as:

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Where

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For second Newton's Law the Force is also defined as,

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Replacing:

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-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

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