The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.
We Know, K.E. = 1/2 × m × v²
From the expression, we can conclude that Kinetic energy is directly proportional to mass. So, as mass will increase, Kinetic energy will also increase.
In short, Your Correct answer would be Option B
Hope this helps!
Answer: Add an incline or grade to the road track.
Explanation:
Refer to the figure shown below.
When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.
The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.
At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.
When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).
Answer: 6.36
Explanation:
Given
Radius of grindstone, r = 4 m
Initial angular speed of grindstone, w(i) = 8 rad/s
Final angular speed of the grindstone, w(f) = 12 rad/s
Time used, t = 4 s
Angular acceleration of the grinder,
α = Δw / t
α = w(f) - w(i) / t
α = (12 - 8) / 4
α = 4/4 = 1 rad/s²
Number of complete revolution in 4s =
Δθ = w(i).t + 1/2.α.t²
Δθ = 8 * 4 + 1/2 * 1 * 4²
Δθ = 32 + 1/2 * 16
Δθ = 32 + 8
Δθ = 40 rad/s
40 rad/s = 40/2π rpm = 6.36 rpm
Therefore, the grindstone does 6.36 revolutions during the 4 s interval
Answer: The greatest force of gravity on the ball will occur at the point when the ball is near to hit the ground ... hope this helps have a good day :P