A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of
1 answer:
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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.
Answer:
dV/dt = 9 cubic inches per second
Explanation:
Let the height of the cylinder is h
Diameter of cylinder = height of the cylinder = h
Radius of cylinder, r = h/2
dh/dt = 3 inches /s
Volume of cylinder is given by
put r = h/2 so,
Differentiate both sides with respect to t.
Substitute the values, h = 2 inches, dh/dt = 3 inches / s
dV/dt = 9 cubic inches per second
Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.