A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.9 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)

Explanation:

Length of the ladder, L = 10 ft

dx / dt = 0.9 ft/s

x = 6 ft

Let the angle is θ.

According to the diagram

x = L cosθ

Differentiate both sides with respect to t

At,  x = 6 ft

Sin θ = 0.8

So,

0.9 = -  10 x 0.8 x dθ/dt

dθ/dt = - 0.1125 rad/s

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.