At A, coaster is only associated with potential energy.
At B, coaster is associated with kinetic as well as potential energy.
Since the track is frictionless, no energy will be lost when coaster reaches from point A to point B. Therefore, according to conservation of energy, total energy at A should be equal to total energy at B.
Total energy at A = mgh = mg(12)
Total energy at B = mgh+ mv²/2 = mg(2) + mv²/2
∴12mg = 2mg + mv²/2
∴(12g-2g)×2 = v²
∴v² = 20g
∴v = 14m/s.
Again conserving energy at points B and C.
Total energy at B = 2mg + m(14)²/2
Total energy at C = 4mg + mv²/2
∴2mg + m(14²)/2 = 4mg + mv²/2
Solving this you get,
v = 12.52 m/s.
Therefore, speed of roller coaster at point C is 12.52 m/s.
Answer:
a) 
, b) 
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
Where 
is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
Since 
, then the angular velocity is equal to zero. Therefore:
Theater=sin^-1(30/80)
theater=22
F=400gsin22=150N
F=ma
150=400a
a=0.375ms^-2
v^2-u^2=2as
v^2-0=2x0.375x80
v=7.75ms^-1
It is know as smoke because if you cook food smoke will go up in the air and that makes vapor and also water from the ground it suck up
<span>No, the distance from the last stop to the school and the time it takes to travel that distance are required.</span>
