None of the energy is created nor destroyed
Answer:
critical angle is the angle of incidence in optically denser medium, for which the angle of reflection in the less denser medium is 90 degree
Explanation:
total internal reflection occurs at one particular angle of incidence where the reflected ray lies on the boundary.
this angle is called the critical angle
It depends on what skateboard. It will stop mainly because some skateboards need a hill or curvy platform to function.
Answer:
1977.696 J
Explanation:
Given;
Weight of the box = 28.0 kg
Force applied by the boy = 230 N
angle between the horizontal and the force = 35°
Therefore,
the horizontal component of the force = 230 × cosθ
= 230 × cos 35°
= 188.405 N
Coefficient of kinetic friction, μ = 0.24
Force by friction, f = μN
here,
N = Normal force = Mass × acceleration due to gravity
or
N = 28 × 9.81 = 274.68 N
therefore,
f = 0.24 × 274.68
or
f = 65.9232 N
Now,
work done by the boy, W₁ = 188.405 N × Displacement
= 188.405 N × 30
= 5652.15 J
and,
the
work done by the friction, W₂ = - 65.9232 N × Displacement
= - 65.9232 N × 30 m
= - 1977.696 J
[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
