A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at
1 answer:
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Part a)
As we know that
here we know that
P1 = 20 bar
V1 = 0.5 m^3
V2 = 2.75 m^3
from above equation
so final state pressure will be 2 bar
Part b)
now in order to find the work done
Answer:
The tension in the rope is 229.37 N.
Explanation:
Given:
Mass of the block is,
Coefficient of static friction is,
Angle of inclination is, °
Draw a free body diagram of the block.
From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.
Forces acting are and normal
. Now, there is no motion in the direction perpendicular to the inclined plane. So,
Consider the direction along the inclined plane.
The forces acting along the plane are and frictional force,
, down the plane and tension,
, up the plane.
Now, as the block is at rest, so net force along the plane is also zero.
Therefore, the tension in the rope is 229.37 N.
