A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca
r from sliding down the incline.
1 answer:
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
You might be interested in
Compared to the pucks given, the pair of pucks will rotate at the same rate.
Answer: Option A
<u>Explanation:</u>
The law of conservation of the angular momentum expresses that when no outer torque follows upon an article, no difference in angular momentum will happen. At the point when an item is turning in a shut framework and no outside torques are applied to it, it will have no change in angular momentum.
The conservation of the angular momentum clarifies the angular quickening of an ice skater as she brings her arms and legs near the vertical rotate of revolution. In the event, that the net torque is zero, at that point angular momentum is steady or saved.
By twice the mass yet keeping the speeds unaltered, also twice the angular momentum's to the two-puck framework. Be that as it may, we likewise double the moment of inertia. Since , the turning rate of the two-puck framework must stay unaltered.