What is the height of a water column that is required to exert a pressure of 15 oz/in²?

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Multiple Choice

What is the height of a water column that is required to exert a pressure of 15 oz/in²?

Explanation:
To determine the height of a water column required to exert a pressure of 15 ounces per square inch (oz/in²), we can use the basic formula that relates pressure to the height of a fluid column. The formula is: \[ P = h \times \rho \times g \] where: - P is the pressure (in ounces per square inch), - h is the height (in inches) of the column, - ρ (rho) is the density of the fluid (in this case, water), - g is the acceleration due to gravity, which can be converted appropriately based on the units used. For water, the weight density in ounces can be used. The specific weight of water is approximately 62.4 lb/ft³, which can be converted into ounces (1 lb = 16 oz), yielding roughly 998.4 oz/ft³. Since there are 12 inches in a foot, we can derive that the weight density of water is about 83.2 oz/in³. Using the formula, we can rearrange it to find the height: \[ h = \frac{P}{\rho} \] Substituting our known values: \[ h = \frac{15 \text

To determine the height of a water column required to exert a pressure of 15 ounces per square inch (oz/in²), we can use the basic formula that relates pressure to the height of a fluid column. The formula is:

[ P = h \times \rho \times g ]

where:

  • P is the pressure (in ounces per square inch),

  • h is the height (in inches) of the column,

  • ρ (rho) is the density of the fluid (in this case, water),

  • g is the acceleration due to gravity, which can be converted appropriately based on the units used.

For water, the weight density in ounces can be used. The specific weight of water is approximately 62.4 lb/ft³, which can be converted into ounces (1 lb = 16 oz), yielding roughly 998.4 oz/ft³. Since there are 12 inches in a foot, we can derive that the weight density of water is about 83.2 oz/in³.

Using the formula, we can rearrange it to find the height:

[ h = \frac{P}{\rho} ]

Substituting our known values:

[ h = \frac{15 \text

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