A bullet of mass 0.01 kg is fired in to a sand bas of mass 0.49 kg from a tree. The Sand bag with the bullet embedded in to it swings a way at lo m/s find a The momentum after the Collision b, The momentum befor the Collision the velocity of the bullet

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a. (.01+.49)10  

b. same as a.  

c. .01V=momentum in b.

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