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Lynna [10]
3 years ago
9

An 8.0g bullet, moving at 400 m/s, goes through a stationary block of wood in 4.0 x 10^-4s, emerging at a speed of 100 m/s. (a)

what average force did the wood exert on the bullet? (b) how thick is the wood?
Physics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

Explanation:

Initial velocity (u) of the Bullet = 400 m/sec

Final velocity (v) of the Bullet = 100 m/sec

Bullet passed through the block in (t) = 0.0004 sec  

Using 1st Equation of motion :

400 m/s = 100 m/s - a (0.0004)

<em>Deceleration of Bullet = 750,000 m/sec^2 </em>

<em />

(a) F (force exerted by the wooden block on the bullet) = F (force exerted by the bullet on the wooden block)

F = m * a = 0.008 * 750,000<em> = </em>6000 N

(b) Using 3rd Equation of motion :

v^{2} = u^{2} - 2aS

10000 = 160000 - 2 * 750,000 * S

Thickness of wood (S) = 0.1 m

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<h3>What is the rate?</h3>

Rate refers to the pace at which something happens. We know that the easiest way to obtain the rate is by the use of proportion. Now we have been told that the dog is able to run 500 miles in 12 seconds.

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