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Sagot :
The car must be moving at a speed of at least 9.9 m/s so that it doesn't fall off of the track.
There are two forces acting on the car when it is moving on the track: the centripetal force and the force due to gravity. These forces have to be equal to each other in order to ensure that the car remains on the track.
Centripetal force = mv²/r
Force due to gravity = mg
mv²/r = mg
The masses can be canceled out.
It becomes:
v²/r = g
Solve for v.
v² = gr
v = √gr
Now, we can substitute in our values and determine v.
v = √9.8 × 10
v = 9.899 m/s
round to nearest tenth for simplicity
v = 9.90 m/s
So the car has to travel at a velocity of at least 9.90 m/s to remain on the track.
The car must move at a minimum speed of about 9.9 m/s to remain on the looped track without falling off.
To determine the speed at which a car must move to avoid falling off a looped track, we need to ensure the car maintains sufficient centripetal force to counteract gravity.
Given:
Mass of car (m) = 100 kg
Radius of loop (r) = 10 m
At the top of the loop, the gravitational force must equal the required centripetal force:
mg = mv²/r
Solving for v (velocity):
g = v²/r
v² = gr
v = √(gr)
Where g (acceleration due to gravity) = 9.8 m/s²:
v = √(9.8 * 10)
v = √(98)
v = 9.9 m/s
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