Question

An aquarium of length L, width (front to back) W, and depth D is filled to the top with liquid of density rho.Part AFind an expression for the force of the liquid on the bottom of the aquarium.Express your answer in terms of the variables rho, D, L, W, and appropriate constants.F = SubmitMy AnswersGive UpIncorrect; Try Again; 5 attempts remainingThe correct answer does not depend on: pDLWg.Part BFind an expression for the force of the liquid on the front window of the aquarium. Hint: This problem requires an integration.Express your answer in terms of the variables rho, D, L, and appropriate constants.F = SubmitMy AnswersGive UpPart CEvaluate the forces on the front window for a 90-cm-long, 35-cm-wide, 45-cm-deep aquarium filled with water.Express your answer with the appropriate units.F = SubmitMy AnswersGive UpIncorrect; Try Again; 5 attempts remainingPart DEvaluate the forces on the bottom for a 90-cm-long, 35-cm-wide, 45-cm-deep aquarium filled with water.Express your answer with the appropriate units.F =

Answer 1

The force of the liquid on the bottom of the aquarium can be found using the formula F = pghAL, where p is the density of the liquid, g is the acceleration due to gravity, h is the depth of the liquid, A is the area of the bottom of the aquarium, and L is the length of the aquarium. The force on the front window of the aquarium can be found by integrating the pressure over the area of the window. For the given dimensions of a 90-cm-long, 35-cm-wide, 45-cm-deep aquarium filled with water, the force on the front window and the bottom can be calculated as 1357.35 N.

Step-by-step explanation:

The force of the liquid on the bottom of the aquarium can be found using the formula F = pghAL, where p is the density of the liquid, g is the acceleration due to gravity, h is the depth of the liquid, A is the area of the bottom of the aquarium, and L is the length of the aquarium. Since the bottom of the aquarium is a rectangular shape with dimensions W (width) and L (length), the area A = WL. Therefore, the expression for the force on the bottom of the aquarium is F = pghWL.

For the force on the front window of the aquarium, it requires integration. The force on the window can be found by integrating the pressure over the area of the window. The pressure at a depth h is given by P = pgh, where p is the density of the liquid and g is the acceleration due to gravity. The force on the window can be found by integrating this pressure over the area of the window. Since the front window has dimensions W (width) and D (depth), the area of the window is A = WD. Therefore, the expression for the force on the front window of the aquarium is F = ∫pgh dA = pgh∫dA = pghWD.

For the given dimensions of a 90-cm-long, 35-cm-wide, 45-cm-deep aquarium filled with water, we can substitute the values into the formulas to find the forces. The force on the front window is F = pghWD = (1000 kg/m³)(9.8 m/s²)(0.45 m)(0.35 m)(0.90 m) = 1357.35 N. The force on the bottom of the aquarium is F = pghWL = (1000 kg/m³)(9.8 m/s²)(0.45 m)(0.35 m)(0.90 m) = 1357.35 N.

Answer 2

Part A
`\[ F_{\text{bottom}} = 1421.805 \, \text{N} \]`

Part B
`\[ F_{\text{front}} = 1185.989 \, \text{N} \]`

Part C
`\[ F_{\text{front}} \approx 1185.989 \, \text{N} \]`

Part D
`\[ F_{\text{bottom}} \approx 1421.805 \, \text{N} \]`

Part A: Force on the bottom of the aquarium

The force
`\( F_{\text{bottom}} \)` on the bottom of the aquarium is given by:

`\[ F_{\text{bottom}} = \rho g L * W * D \]`

For the provided dimensions and values:

`\[ F_{\text{bottom}} = 1000 * 9.81 * 0.45 * 0.9 * 0.35 \]`

`\[ F_{\text{bottom}} = 1421.805 \, \text{N} \]`

Part B: Force on the front window of the aquarium

The force
`\( F_{\text{front}} \)` on the front window of the aquarium is given by:

`\[ F_{\text{front}} = (1)/(2) \rho g D^2 * W \]`

For the provided dimensions and values:

`\[ F_{\text{front}} = (1)/(2) * 1000 * 9.81 * (0.45)^2 * 0.35 \]`

`\[ F_{\text{front}} = 1185.989 \, \text{N} \]`

Part C: Force on the front window for specific dimensions

Given:

- L = 90 cm

- W = 35 cm

- D = 45 cm

- Water density
`\( \rho = 1000 \)` kg/m³

- Acceleration due to gravity g = 9.81 m/s²

`\[ F_{\text{front}} = (1)/(2) * 1000 * 9.81 * (0.45)^2 * 0.35 \]`

`\[ F_{\text{front}} \approx 1185.989 \, \text{N} \]`

Part D: Force on the bottom for specific dimensions

Given:

- L = 90 cm

- W = 35 cm

- D = 45 cm

- Water density
`\( \rho = 1000 \)` kg/m³

- Acceleration due to gravity g = 9.81 m/s²

`\[ F_{\text{bottom}} = 1000 * 9.81 * 0.45 * 0.9 * 0.35 \]`

`\[ F_{\text{bottom}} \approx 1421.805 \, \text{N} \]`