A pressure-pump has a horizontal tube of cross-sectional area 10 cm2 for the outflow of water at a speed of 20m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given: density of water = 1000kg/m3]
Question
A pressure-pump has a horizontal tube of cross-sectional area 10 cm2 for the outflow of water at a speed of 20m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given: density of water = 1000kg/m3]
Solution
To solve this problem, we need to use the formula for force exerted by a fluid, which is given by:
F = ρAv²
where:
- F is the force,
- ρ is the density of the fluid,
- A is the cross-sectional area, and
- v is the velocity of the fluid.
Given in the problem, we have:
- ρ = 1000 kg/m³,
- A = 10 cm² = 10 x 10^-4 m² (since 1 cm² = 10^-4 m²), and
- v = 20 m/s.
Substituting these values into the formula, we get:
F = 1000 kg/m³ * 10 x 10^-4 m² * (20 m/s)² = 1000 kg/m³ * 10 x 10^-4 m² * 400 m²/s² = 4,000 N
So, the force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube is 4,000 N.
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