The mine car and its contents have a total mass of 6.0 Mg and a center of gravity at G. If the coefficient of static friction be
tween the wheels and the tracks is μs = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked. Does the car move?
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Answer:
The total time it is in the air for the ball is 1.6326 s
Given:
Initial velocity = 8
To find:
the total time it is in the air = ?
Formula used:
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
Solution:
A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.
The time taken by the ball to reach the maximum height is given by,
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
t =
t = 0.8163 s
Thus, time taken by the ball to reach the ground again = time taken to reach maximum height
So, Total time required for ball to reach ground = 2t = 2 × 0.8163
Total time required for ball to reach ground = 1.6326 s
The total time it is in the air for the ball is 1.6326 s