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
Answer:
PE = (|accepted value – experimental value| \ accepted value) x 100%
Explanation:
Answer:
send the wagon down a higher hill
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