A boat with a horizontal tow rope pulls a waterskier. She skis off to the side, the rope makes an angle of 15° with the forward direction of motion. If the tension in the rope is 180 N, how much work does the rope two on the scare during the forward displacement of 300 m?

Explanation:

Workdone by the rope is expressed using the formula;

W = Fd sin(theta)

F is the tension in the rope = 180

d is the displacement = 300m

theta is the angle of inclination = 15°

Substitute the given parameters into the formula;

W = 180(300)sin15

W = 54000sin 15

W = 13,976.23

Hence the workdone by the rope is 13,976.23Joules

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14