PLS ANSWER ASAP A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

What are the magnitudes of the x- and y- components of the tension?

What is the normal force acting on the skier?

normal force= 480.2 x component=477.4  y component= 110.4

Explanation:

I'm sorry I just did this today in class but i think normal force has to be equal to mg or the mass * gravity of the person and that would be 480.2 newtons. I multiplied 9.8 (gravity) by the mass of 49g

I think for the x and y components you can make a triangle with the angled string and then use sohcahtoa to solve for the tension. I used cos(13) = x/490 to solve for x component and then I used the pythagoran theroem to get the remaining side which would be : a^2 + 477.4^2 = 490^2

hope this can help and that it is correct. good luck

-4 on

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2