Answer:
The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, reached is the same for both the cup and the kettle as given by the relation;
Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment
Explanation:
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