Abullet of mass 0.093 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 2.5 kg that is sitting at rest on a nearly frictionless surface. (a) what is the speed of the block after the bullet embeds itself in the block?

Explanation:

It is given that,

Mass of the bullet, m₁ = 0.093 kg

Initial speed of bullet, u₁ = 100 m/s

Mass of block, m₂ = 2.5 kg

Initial speed of block, u₂ = 0

We need to find the speed of the block after the bullet embeds itself in the block. Let it is given by V. On applying the conservation of linear momentum as :

V = 3.58 m/s

So, the speed of the bullet is 3.58 m/s. Hence, this is the required solution.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j