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:
a= 17.69 m/s^2
Explanation:
Step one:
given data
A car accelerates uniformly from rest to 23 m/s
u= 0m/s
v= 23m/s
distance= 30m
Step two:
We know that
acceleration= velocity/time
also,
velocity= distance/time
23= 30/t
t= 30/23
t= 1.30 seconds
hence
acceleration= 23/1.30
accelaration= 17.69 m/s^2
Answer:
F = 51.3°
Explanation:
The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

where,
W = Weight of Wagon = 150 N
θ = Angle of Inclinition = 20°
Therefore,

<u>F = 51.3°</u>
There is not enough information given to answer with. The force of gravity at the planet's surface depends on the planet's radius as well as its mass. The planet could have exactly the same mass as Earth has. But if it's radius is only 71% of Earth's radius, then gravity on its surface will be twice as strong as gravity on Earth.