An object is thrown upwards from ground with some velocity. Assume that air friction acts on the body such that the force
of friction is always one-fifths the weight of the body. Let T1 be
the time of ascent and T2 be the time of descent of the body. Let
N = 30 * (T1/T2)^2
. Write the value of N in the bubble sheet. [Note
that the force of air friction is always opposite to the direction of
motion. So during ascent, the friction force on the body is acting
downwards, while during descent, it is acting upwards].
Answers
Answer:
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Answer:
Let the length, breadth and height of the pool and the density of what the liquid in the pool be L,B,H and ρ respectively.
At any depth, h, below the surface of the liquid, the hydrostatic pressure is hρg.
⇒ The pressure at the bottom of the pool is Hρg.
The area of the bottom of the pool is LB.
⇒ The force at the bottom of the pool is LBHρg, which is the same as the mass of of the liquid in the pool times the acceleration due to gravity.
Consider a small strip of of the surface of the pool of thickness dh at a depth h on the wall along the shorter side of the the pool.
The area of this strip is Bdh and the hydrostatic force on this strip is hρg.
⇒ The hydrostatic force on this strip is hρgBdh.
⇒ The hydrostatic force on the entire wall is ∫0HhρgBdh
=ρgB[h22]H0=ρgB(H22).
⇒ The force on each of the smaller side walls of the pool is ρgBH22.
For this particular case, you can substitute the given values of the variables and obtain the required answers.