Calculate the net torque about point O for the two forces applied as in the figure below. The rod and both forces are in the pla
1 answer:
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Answer:
the diameter of the second pipe is 2.52 in
Explanation:
Given the data in the question;
We know that; the rate of flow is the same;
so
Av1 = Av2
v ∝ √h
=
= √(
)
( π/4.D1² / π/4.D2² ) = √( )
( D1² / D2² ) = √( ) since second is double of first
so
( D1² / D2² ) = √( )
3² / D2² = √2
D2²√2 = 9
D2² = 9/√2
D2² = 9 / 1.4142
D2² = 6.364
D2 = √ 6.364
D2 = 2.52 in
Therefore, the diameter of the second pipe is 2.52 in
Answer:
Explanation:
<u>Dynamics</u>
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
The variable acceleration is calculated by:
The instant velocity is the integral of the acceleration:
Integrating