A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N, 180.0 on it. If the coefficient of friction is 0.18, calculate the deceleration rate of the wagon as it is caught.

1.736m/s²

Explanation:

According to Newton's second law;

where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

Substitute the given parameters

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

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