MOTION OF A BULLET?

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Offline fortengc

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MOTION OF A BULLET?
« on: 28/02/2011 04:48:03 »
A bullet (mass 15 g) moving at a speed of 425 m/s is brought to a stop in 3.5 cm in a block of wood. Find the force on the bullet, assuming that it is constant

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Offline Geezer

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MOTION OF A BULLET?
« Reply #1 on: 28/02/2011 05:06:31 »
Wait a minute. I'm getting a terrible sense of déjà vu all over again.
There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline fortengc

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MOTION OF A BULLET?
« Reply #2 on: 28/02/2011 05:27:45 »
THIS IS HOW I SOLVED IT.
F=Ma = M v/t
S=(u+v)/2*t implies t= S/(u+v)/2 = 0.035/(425+0)/2=1.647*10^-4 sec
since F= M v/t = 0.015*425/(1.647*10^-4) = 38705.35N = 38.705KN

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Offline Geezer

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There ain'ta no sanity clause, and there ain'ta no centrifugal force æther.

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Offline fortengc

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MOTION OF A BULLET?
« Reply #4 on: 28/02/2011 07:21:35 »
thanks man i had same answer using the other approach of a=v^2/2S can u help me with this other question?

10.   Suppose you begin with 2.0 mg of a pure radioactive substance and 4.0 h later determine that 0.50 mg remain. What is the half-life of the substance?


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Offline lightarrow

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MOTION OF A BULLET?
« Reply #5 on: 28/02/2011 15:48:30 »
thanks man i had same answer using the other approach of a=v^2/2S can u help me with this other question?

10.   Suppose you begin with 2.0 mg of a pure radioactive substance and 4.0 h later determine that 0.50 mg remain. What is the half-life of the substance?


In x hours the initial quantity has halved; since it started with 2 mg and finished with 0.5, how many times has halved? So x =?