Answer
Radius of the wheel r = 2.1 m
Moment of inertia I = 2500 Kg m²
Tangential force applied F = 18 N
Time interval t = 16 s
Initial angular speed ω1 = 0
Final angular speed ω2 = ?
Let α be the angular acceleration.
Torque applied τ = Iα
F r = Iα
Angular acceleration α = F r/I
= 
= 0.015 rad/s²
(a)From rotational kinematic relation
Final angular speed ω₂ = ω₁ + αt
= 0 + (0.015 rad/s^2 * 16 s)
= 0.24 rad/s
(b) Work done W = 0.5 Iω₂² - (1/2)Iω₁²
= 0.5*( 2500 Kg m²)(0.24 rad/s)^2 - 0
= 72 J
(c) Average power supplied by the child P = W/t = 
= 4.5 watt
Answer:
the velocity is changing therefore the acceleration is changing too.
Explanation:
From the formula, V=displacement/time
displacement/distance =velocity*time
distance=1200*2.5
distance=3*10³km
Ptotal=Ptotal —> m1v1+m2v2=m1v1’+m2v2’ —> (1kg)(2m/s)+(1kg)(0m/s)=(1kg)(-1m/s)+(1kg)(v2’) —> v2’=3m/s
answer: v=3m/s
Answer:
Explanation:
If v be the velocity just after the rebound
Kinetic energy will be converted into potential energy
1/2 m v² = mgh
v² = 2gh
v = √ 2gh
= √ 2 x 9.8 x .96
= 4.33 m / s