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:
Speed = 10 km/hr
Explanation:
Speed = Distance / Time
Since we have to convert it into minutes to hours ( because it is km/hr )
45 minutes = 0.75 hours
Therefore,
Speed = 7.5 / 0.75
= <u>10 km/hr </u>
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<u>Hope it helps!</u>
In order to get the propoerty of work you need to use the following formula
<span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W = 4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
Answer:
the ans is D... good luck