A mover's dolly is used to pull a 115 kg refrigerator up a ramp into a house. The ramp is 2.10 m long and rises 0.850 m. The
mover exerts a force of 496 N up the ramp. How much work is
spent overcoming friction?

The amount of work  spent in overcoming friction is 83.65 J.

From the question,

The amount of work spent in overcoming friction = work done by the mover-potential energy of the refrigerator.

This is expressed mathematically as,

W' = Fd-mgh Equation 1

Where W' = work spent against friction, F = force applied, d = distance, m = mass of the refrigerator, g = acceleration due to gravity, h = height.

From the question,

Given: F = 496 N, d = 2.10 m, m = 115 kg, h = 0.85 m

Constant: g = 9.8 m/s².

Substitute these values into equation 1

W' = 496(2.1)-(115×9.8×0.85)

W' = 1041.6-957.95

W' = 83.65 J.

Hence, the amount of work  spent in overcoming friction is 83.65 J

Learn more about friction here:

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2