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:
28852 J
Explanation:
When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.
The work (W) is:
W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.
Knowing that cos(26°) = 0.899, and F = m*g
W = 51.9*9.8*63.1*0.899
W = 28852 J
Answer:
13800 N
Explanation:
Impulse is the product of average force and time expressed as I=Ft where I is the impulse which results into change in momentum, F is the average force and t is the time of impact. Making F the subject of formula then
Substituting I with 13.8 N.s and time, t witg 0.001 s then the average force is calculated as
Therefore, the average force is equivalent to 13800 N
Answer:
1) T = 4.5 s
2) T = 4.5 s
3) v = 9.9 m/s
Explanation:
We can use the equation
T = 2π√(L/g)
1) T = 2π√(5m/9.81 m/s²) = 4.5 s
2) T = 2π√(L/g)
T = 2π√(5m/9.81 m/s²) = 4.5 s
3) v = √(2gR)
v = √(2(9.81 m/s²)(5m))
v = 9.9 m/s
