To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal
force and reduce vehicle traction it is important to
1 answer:
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).
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Answer:
The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Explanation:
Given that,
Mass of object = 5 kg
Speed = 3 m/s
Mass of stationary object = 3 kg
Moving object deflected = 30°
Stationary object deflected = 31°
We need to calculate the velocity of each ball after collision
Using conservation of momentum
Along x-axis
Put the value into the fomrula
....(I)
Along y -axis
Put the value into the formula
...(II)
From equation (I) and (II)
Put the value of v₁ in equation (I)
Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Answer:D
Explanation:
Given
mass of object
Distance traveled
velocity acquired
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy
where is friction work if any
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .