Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
For conservation of energy we have to:
mgH=mv²/2
Clearing
<span> v=sqrt(2gH)
Then, by definition
</span><span> F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt=
</span> =m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt
the answer is
F=m[sqrt(2gH)]/ Δt
Answer:

Explanation:
Given:
- area of piston on the smaller side of hydraulic lift,

- area of piston on the larger side of hydraulic lift,

- Weight of the engine on the larger side,

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.



is the required effort force.
Fg = G m₁ m₂ / r²
Fg = (6.67×10−¹¹) ( 70) ( 5.972×10²⁴) / (6.3781×10⁶)²
Fg = 685.43 N