An aircraft travelled from a point X to a point Y 100m away in the N45°E direction. It then travelled westward to a point Z which is N45°W from X. The distance of the aircraft from X is A 50km
B 100/√2 km
C 100√2 km
D 100km

amplitude

hope this

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