To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Given parameters:
Displacement = 8km
Velocity = 3.8km/h
Unknown:
time = ?
Solution:
Velocity is displacement divided by time.
Velocity =
Displacement = velocity x time
Input the parameters:
8 = 3.8 x time
Time =
= 2.1s
The time taken is 2.1s
They traveling at -0.37/ms^
Car A will have highest speed is 83.3m/s .
<h3>What is speed ? </h3>
The rate of change of position of an object in any direction.
The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction
.
speed = distance /time
In case of Car A ,
We have given distance 150Km in 3 min ,
First we have convert the distance km to m
150×1000m
then conversion of min to sec
38×60sec
speed = 15000/180
speed = 83.3m/sec
In case of Car B
we have given 800m in 150 min
lets convert the time into second
150×60
Speed = 800/150×60
speed = 0.88m/ s
In case of Car C
We have given here distance 250 Km and time in 8 hours
convert km to m
25000
and time into sec
88×60×60
speed = 0.86m/ s
Hence ,Car A has highest speed amongst them .
To learn more about speed click here
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Answer:

Explanation:
To develop this exercise we proceed to use the kinetic energy equations,
In the end we replace


Here
meaning the 4 wheels,
So replacing

So,



