Physics, asked by balamohan, 1 year ago

a bullet of mass 20 g moving with speed 200 m/s strikes on a piece of wood and comes to rest in 0.02 s. what is the resistance offered by the wood

Answers

Answered by kvnmurty
9
m = 20 grams
initial velocity = u = 200 m/s

t = 0.02 sec.
let the acceleration or deceleration = a m/sec^2

  v = final velocity = u + a t
   0 = 200 + a * 0.02
         a = - 10, 000 m/sec^2

the resistance offered = force against the movement of the bullet
       = F = m a = 20/1000 kg * 10,000 m/sec^2 = 200 Newtons.

Answered by rohankumar
2
Since v=u+at 0=200+(a)0.02 a=-200÷0.02 =-10000 RESISTANCE =-(ACCELERATION)×MASS =-(-10000)×20×10^-3 =(10000×20)÷1000 =200 N
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