Answer:
1210 N/m between the inner and outer edges (repulsive)
252.59 N/m between inner and line charges (attractive)
1210+252.59=1462.59 N/m effective force/metre on the inner edge
Explanation:
f=k*Qq/r^2 was applied, the charges are concentrated at the edges of a ring both inner and the outer were shape edges with uniform charge densities 6.6 micro C/m of distance 4.5-2.7=1.8 cm
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
The person is walking 3.5 miles an hour
We don't know, and we don't have enough information to calculate it.
The weight of the 60kg load is (m g) = 588 Newtons.
IF Hari wanted to<em> lift </em>the load 12m <em>straight up</em>, he would have to do
(Force x distance) = (588 N) x (12 m) = 7,056 Joules of work.
But to drag it, he has to provide enough force to balance out the force of friction, and we don't know how much that is. It depends on the weight of the load, the shape of the load, the smoothness of the part of the load that sits on the ground, and the smoothness of the ground. But the only thing we know is the weight of the load.