A15 ft ladder is sliding down a wall. the base of the ladder is sliding away from the wall at 1.5 ft/sec. how fast is the top of the ladder sliding down the wall at the instant the base of the ladder is 9 ft from the wall?

1.125 ft/sec

Explanation:

Given

length of ladder =15 ft

Base of ladder is 9 m away

Angle which ladder makes with horizontal is

Let at any instant base of ladder is x m away and top of ladder is ym away from ground

therefore

differentiating w.r.t we get

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