A car is driving South at 30 Mi/h, when the driver steers bringing the car to head West still moving at 30 Mi/h. Which of the fo
1 answer:
Answer:
D. The linear momentum of the car changed, its kinetic energy did not change.
Explanation:
A. The linear momentum of the car changed, its kinetic energy changed.
B. The linear momentum of the car did not change, its kinetic energy did not change.
C. The linear momentum of the car did not change, its kinetic energy changed.
D. The linear momentum of the car changed, its kinetic energy did not change.
As we know that the direction of motion of the car is changed from South to West so here momentum is not conserved as momentum is a vector quantity
While we know that kinetic energy is given as
so here kinetic energy is a scalar quantity which remains conserved as none of the magnitude changed here
So correct statement is
D. The linear momentum of the car changed, its kinetic energy did not change.
You might be interested in
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).
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).