Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
C. Seismic energy
This is energy that is released in earthquakes.
Answer: 7291.2 joules
Explanation:
Work is done when force is applied on an object over a distance.
Thus, Workdone = Force X distance
Since Distance moved by box = 12 metres
mass of box = 62kg
Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2
Recall that Force = Mass x acceleration due to gravity
i.e Force = 62kg x 9.8m/s^2
= 607.6 Newton
So, Workdone = Force X Distance
Workdone = 607.6 Newton X 12 metres
Workdone = 7291.2 joules
Thus, 7291.2 joules of work was done.
I’m pretty sure the answer is C. Any change of state or movement requires energy
That's called the "normal" to the surface at that point.
