Answer:
Explanation:
Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.
For faster car on the road,
v = 2v
..........(1)
For the slower car on the road,
............(2)
Equation (1) becomes,
So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.
Before we go through the questions, we need to calculate and determine some values first.
r = 11.5 m
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N
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1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.
<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>
<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>
<span>Centripetal force = 7119.55 N </span>
<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>
<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>
<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>
√<span>(gr) </span>
√<span>(9.8 x 11.5) = 10.62 m/s</span>
If a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2 and then recombine at Point B
<h3>
Parallel circuit</h3>
A parallel circuit is a circuit with separate branches with a common endpoint. In a parallel circuit, the voltage across each branch is the same but the currents vary. The total current is the sum of the currents flowing through each component.
If a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2 and then recombine at Point B.
Find out more on Parallel circuit at: brainly.com/question/80537
