A 640 kg automobile slides across an icy street at a speed of 55.8 km/h and collides with a parked car which has a mass of 935 kg. The two cars lock up and slide together. What is the speed of the two cars just after they collide

V = 6.29 m/s

Explanation:

Given that,

The mass of an automobile, m₁ = 640 kg

The mass of  parked car, m₂ = 935 kg

The speed of automobile, u₁ = 55.8 km/h = 15.5 m/s

The speed of the parked car, u₂ = 0

We need to find the speed of the two cars just after they collide . Let V is a common speed. Using the conservation of linear momentum to find it such that,

So, the speed of the two cars is equal to 6.29 m/s.

and 1and 1

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j