A 133.5 kg crate, sliding on the floor, is brought to a stop by a 54.5 N force. What is the deceleration of the crate?

0.40m/s^2

Step-by-step explanation:

Given data

Mass= 133.5kg

Force F= 54.5N

The expression for the force acting is

F=ma

substitute

54.5=133.5*a

]Divide both sides by 133.5

a= 54.5/133.5

a= 0.40 m/s^2

Hence the acceleration is 0.40m/s^2

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)