If the force acting on a body of mass 4 kg is doubled. by how much will the acceleration change?
1 answer:
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The coefficient of friction between the Tyre and the ground is 0.11
<u>Explanation:</u>
Given:
Radius of the track (r)=125 m.
Speed with which the car travels (v) =42 km/hr
=11.67 m/s
To Find:
Coefficient of friction between the Tyre and the ground.
Formula to be used:
We know that,Frictional force is equal to centripetal force
Frictional force=μmg
therefore 1.08 m=μmg
Cancelling "m" on both sides we get,
μ=1.08/g=1.08/9.8
=0.11
Thus the coefficient of friction between the Tyre and the ground is 0.11
Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:
Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:
This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m