A steel ball of mass 1.07 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle has a radius of 1.38 m. When it is at the bottom of the circle, the ball has a speed of 13.9 m/s. Calculate the magnitude of the tension in the cable when the mass is at the bottom of the circle. (in N)

160.296 N

Explanation:

At the bottom of the circle,

Applying,

Tension  = centrifugal force- weight  of the ball

T = mv²/r-mg Equation 1

Where T = tension in the cable, m = mass of the ball, v = speed of the ball, r = radius of the circle, g = acceleration due to gravity

Given: m = 1.07 kg, v = 13.9 m/s, r = 1.38 m, g = 9.8 m/s²

Substitute into equation 1

T = (1.07(13.9²)/1.38)+[1.07(9.8)]

T = (1.07(193.21)/1.38)+(10.486)

T = (206.7347/1.38)+(10.486)

T = 149.81+10.486

T =160.296 N

Hence the tension in the cable when the mass is at the bottom of the circle = 160.296 N

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form