The problem is based on Bernoulli's theorem, which states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

The equation for Bernoulli's theorem is:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

In this case, since the pipe is horizontal, the height (h) remains constant, so we can ignore the potential energy term (ρgh). The equation simplifies to:

P₁ + ½ρv₁² = P₂ + ½ρv₂²

Given that the initial pressure (P₁) is 4.5×10⁴ N/m², the final pressure (P₂) is 2.0×10⁴ N/m², and the initial velocity (v₁) is 0 (since the water is initially at rest), we can substitute these values into the equation:

4.5×10⁴ + ½ρ(0)² = 2.0×10⁴ + ½ρv₂²

Solving for v₂ (the final velocity), we get:

v₂ = √[(2*(4.5×10⁴ - 2.0×10⁴))/ρ]

Assuming the density of water (ρ) is 1000 kg/m³, we can substitute this value into the equation:

v₂ = √[(2*(4.5×10⁴ - 2.0×10⁴))/1000]

Solving this equation will give you the final velocity of the water.

Question

The problem is based on Bernoulli's theorem, which states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

The equation for Bernoulli's theorem is:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

In this case, since the pipe is horizontal, the height (h) remains constant, so we can ignore the potential energy term (ρgh). The equation simplifies to:

P₁ + ½ρv₁² = P₂ + ½ρv₂²

Given that the initial pressure (P₁) is 4.5×10⁴ N/m², the final pressure (P₂) is 2.0×10⁴ N/m², and the initial velocity (v₁) is 0 (since the water is initially at rest), we can substitute these values into the equation:

4.5×10⁴ + ½ρ(0)² = 2.0×10⁴ + ½ρv₂²

Solving for v₂ (the final velocity), we get:

v₂ = √[(2*(4.5×10⁴ - 2.0×10⁴))/ρ]

Assuming the density of water (ρ) is 1000 kg/m³, we can substitute this value into the equation:

v₂ = √[(2*(4.5×10⁴ - 2.0×10⁴))/1000]

Solving this equation will give you the final velocity of the water.

Knowee AI · Accepted Answer