Question

A bullet hits a Sand box with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sand box.​

Answer 1

Answer:

Explanation:

V^2=U^2+2as

U=20m/s. V=0m/s as it final comes to rest in sand box

Using above formula First convert 6cm to meters  

-400=2a6/100

a= -3333.34m/s^2

Answer 2

A bullet hits a Sand box with a velocity of 20 m/s and penetrates it up to a distance of 6 cm. Find the deceleration of the bullet in the sand box.

◀️◀️ ANSWER ▶️▶️

here , mass of bullet be m.

KE of bullet (initial) = 1/2× mv^2

KE = 1/2 × m × 20× 20

KE = 200 m

200 m energy required to penetrate sand box up to 6 cm .

here , lets take Force be as F

work done = force × displacement

w = f × 6× 10 ^-2

force = mass × acceleration

200 m = ma × 0.05

= - 3333 m/s^2

that is declaration of bullet in sand box is 3333

I hope this helps you :)