1.

A bullet of 25 gm with a speed 200 m/s stops at a distance of 5 cm. Find the resistive force​

Answer»

ong>Answer:

The average resistance force is 10000 N.

Explanation:

Given : A bullet of mass 25G moving with a speed 200 m/s is stopped within 5 cm of the TARGET.

To find : What is average resistance force offered by the target?

Solution :

The mass of the bullet is 25 g.

The initial VELOCITY of the bullet is u=200 m/s.

The FINAL velocity is v=0 m/s.

The distance covered is 5 cm.

Into m, 5 cm=0.05 m.

First we find the acceleration.

Using formula,

2as=v^2-u^22as=v

2

−u

2

Substitute the values,

2a(0.05)=0^2-200^22a(0.05)=0

2

−200

2

0.1a=-400000.1a=−40000

a=-\frac{40000}{0.1}a=−

0.1

40000

a=-400000m/s^2a=−400000m/s

2

CONVERTING mass into kg,

25 g= 0.025 kg.

Now, The average resistance force offered by the target is

F=m\times aF=m×a

F=0.025\times 400000F=0.025×400000

F=10000F=10000

Therefore, The average resistance force is 10000 N.


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