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Answer:
option B
Explanation:
given,
Force exerted by the hydraulic jack piston = F₁ = 250 N
diameter of piston, d₁ = 0.02 m
r₁ = 0.01 m
diameter of second piston, d₂ = 0.15 m
r₂ = 0.075 m
mass of the jack to lift = ?
now,




F₂ = 14062.5 N
F = m g


m = 1435 Kg
hence, the correct answer is option B
Answer:
P_(pump) = 98,000 Pa
Explanation:
We are given;
h2 = 30m
h1 = 20m
Density; ρ = 1000 kg/m³
First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,
Thus, it can be expressed as;
P_(tank)+ P_(pump) = P_(nozzle)
Now, the pressure would be given by;
P = ρgh
So,
ρgh_1 + P_(pump) = ρgh_2
Thus,
P_(pump) = ρg(h_2 - h_1)
Plugging in the relevant values to obtain;
P_(pump) = 1000•9.8(30 - 20)
P_(pump) = 98,000 Pa