Answer:Racquet force is twice of Player force
Explanation:
Given
ball arrives at a speed of 
ball returned with speed of 
average Force imparted by racquet on the ball is given by
where 
time of contact of ball with racquet
When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet
From 1 and 2 we get
Hence the magnitude of Force by racquet is twice the Force by player
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
Answer:
Explanation:
Given:
dimension of uniform plate, 
mass of plate, 
Now we find the moment of inertia about the center of mass of the rectangular plate is given as:
where:
length of the plate
breadth of the plate
We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .
Now we find the distance between the center of mass and the corner:
Now using parallel axis theorem:
Answer:
Important enough for someone to ask you to do it.
Explanation:
If someone asked you this question, then it must be important and stuff. This is proven by scientific stuff that no one cares about.
