You pull downward with a force of 26.0 N on a rope that passes over a disk-shaped pulley of mass 1.31 kg and radius 0.0791 m. The other end of the rope is attached to a 0.601-kg mass. Calculate the tension in the rope on both sides of the pulley. Enter tension for the part of the rope that you are pulling on first. Then enter the tension for the part of the rope with the mass.

Tension for the part of the rope that was pulled on first, T1 = 26N

Tension of the other side of the rope, T2 = 9.4N

Explanation:

The first rope creates a tension, T1

Tension T1 = 26N

pulley of mass, M = 1.31 kg

radius, r = 0.0791 m

Mass of rope, m = 0.601 kg

Tension from downward pull on the other side of the rope gives an equation:

T2-m*g =m*a

a = (T2-m*g)/m ...equation 1

Where m = mass attached to the rope

Tension from pulley system also gives an equation:

(T1-T2) = 1/2 *M*a

Where M = mass of the pulley

a= 2(T1-T2)/M ...equation 2

Equating equation 1 and equation 2

a = a

(T2-m*g)/m = 2(T1-T2)/M

Let g= 9.81m/s^2

[T2 -(0.601×9.81)]/0.601 = 2(26-T2)/1.31

(T2 - 5.89581)/0.601 = 1.5267(26-T2)

T2 - 5.89581 = 0.9175(26-T2)

T2 - 5.89581 = 23.855 - 0.9175T2

T2 + 0.9175T2 = 23.855 - 5.89581

1.9175T2 = 17.95919

T2 = 17.95919/1.9175

T2 = 9.3659N

Tension from the other side of the pull, T2 is approximately 9.4N

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

answera compound;