A 120 -kg crate accelerates toward the positive x-direction . If the magnitude of the force due to friction is 74.4 N , what force must pull on the crate to give it a 2.5 m/s ^ s acceleration ?

The applied force to accelerate the crate is 374.4 N

Explanation:

Given;

mass of the crate, m = 120 kg

magnitude of force due to friction, Fk = 74.4N

acceleration of the crate, a = 2.5 m/s²

The net horizontal force on the crate is calculated as;

∑Fx = F - Fk

ma = F - Fk

F = ma + Fk

where;

F is the applied force to accelerate the crate by 2.5 m/s²

F = (120 x 2.5) + (74.4 N)

F = 300 N + 74.4 N

F = 374.4 N

Therefore, the applied force to accelerate the crate is 374.4 N

none, its a machine

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