A piston above a liquid in a closed container has an area of 1m2. The piston carries a load of 350 kg. What will be the external
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The last equation gives you the tension in the string on the right:
Answer:
9.8 × 10⁴Pa
Explanation:
Given:
Velocity V₁ = 12m/s
Pressure P₁ = 3 × 10⁴ Pa
From continuity equation we have
ρA₁V₁ = ρA₂V₂
A₁V₁ = A₂V₂
making V₂ the subject of the equation;
the pipe is widened to twice its original radius,
r₂ = 2r₁
then the cross-sectional area A₂ = 4A₁
⇒
This implies that the water speed will drop by a factor of because of the increase the pipe cross-sectional area.
The Bernoulli Equation;
Energy per unit volume before = Energy per unit volume after
p₁ + ρV₁² + ρgh₁ = p₂ +
ρV₂² + ρgh₂
Total pressure is constant and = P =
ρV₂²ρV²
p₁ + ρV₁² = p₂ +
ρV₂²
Making p₂ the subject of the equation above;
p₂ = p₁ + ρV₁² -
ρV₂²
But so,
p₂ = p₁ + ρV₁² -
ρ
p₂ = 3.0 x 10⁴ + ( × 1000 × 12²) - (
× 1000 × 12²/4² )
P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³
P₂ = 9.79 × 10⁴Pa
P₂ = 9.8 × 10⁴Pa