A75 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 620 m/s . part a what is the recoil speed of the hunter?
express your answer to two significant figures and include the appropriate units.

To develop this problem we will apply the concepts related to the conservation of momentum. For this purpose we will define that the initial moment must be preserved and be equal to the final moment. Since the system was at a stationary moment, the moment before the bullet is fired will be 0, while the moment after the bullet is fired will be equivalent to the sum of the products of the mass and the velocity of each object. , this is,

Here,

m = mass of each object

u = Initial velocity of each object

v = Final velocity of each object

Our values are,

Applying our considerations we have that the above formula would become,

Solving for speed 1,

Therefore  the recoil speed of the hunter is -0.3472m/s, said speed being opposite to the movement of the bullet.

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