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The surface of water in a water tank on the top of a house is 4 m above the tap level. Find the pressure of water at the tap, above atmospheric pressure, when the tap is closed. Is it necessary to specify that the tap is closed? Take `g=10ms^-2`

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To solve the problem, we need to calculate the pressure of water at the tap level when the tap is closed. The pressure at a certain depth in a fluid is given by the formula: \[ P = \rho g h \] where: - \( P \) is the pressure, - \( \rho \) is the density of the fluid (for water, it is approximately \( 1000 \, \text{kg/m}^3 \)), - \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \)), - \( h \) is the height of the water column above the tap (given as \( 4 \, \text{m} \)). ### Step-by-Step Solution: 1. **Identify the variables**: - Density of water, \( \rho = 1000 \, \text{kg/m}^3 \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) - Height of water column, \( h = 4 \, \text{m} \) 2. **Substitute the values into the pressure formula**: \[ P = \rho g h \] \[ P = 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 4 \, \text{m} \] 3. **Calculate the pressure**: \[ P = 1000 \times 10 \times 4 = 40000 \, \text{N/m}^2 \] 4. **Convert the pressure to a more common unit**: \[ P = 40000 \, \text{N/m}^2 = 40000 \, \text{Pa} \text{ (Pascals)} \] 5. **Determine if it is necessary to specify that the tap is closed**: - When the tap is closed, the pressure at the tap remains constant at \( 40000 \, \text{Pa} \). - If the tap were opened, the pressure would decrease as water flows out of the tap. Therefore, it is necessary to specify that the tap is closed to maintain the calculated pressure. ### Final Answer: The pressure of water at the tap, above atmospheric pressure, when the tap is closed is \( 40000 \, \text{Pa} \). Yes, it is necessary to specify that the tap is closed to maintain this pressure.

To solve the problem, we need to calculate the pressure of water at the tap level when the tap is closed. The pressure at a certain depth in a fluid is given by the formula: \[ P = \rho g h \] where: - \( P \) is the pressure, - \( \rho \) is the density of the fluid (for water, it is approximately \( 1000 \, \text{kg/m}^3 \)), - \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \)), ...
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