How do you find the total force exerted on the five exposed faces of a cube that is on a side lying on the bottom of a swimming pool that is 20ft by 15ft by 10ft deep filled with water?

Answer 1

I guess you can use the hydrostatic pressure at a depth h:

#p=p_0+rhogh#

Where:

#p# is your required pressure at depth #h# (#=10ft=3.05m#), #g=9.81m/(s^2)# is acceleration of gravity and #p_0# is the pressure at the surface of water (#1 atm=1.013*10^5Pa#). The density of water should be: #rho=1000 (kg)/m^3# So you should get: #p=1.013*10^5+(1000*9.81*3.05)=1.312*10^5Pa~=1.3atm# and: #pressure=(force)/(area)#
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Answer 2

To find the total force exerted on the five exposed faces of a cube submerged in water, you can use the concept of pressure and the formula for pressure at a certain depth in a fluid.

The pressure exerted by a fluid at a certain depth is given by the formula:

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

where:

  • ( P ) is the pressure,
  • ( \rho ) is the density of the fluid (in this case, water),
  • ( g ) is the acceleration due to gravity, and
  • ( h ) is the depth of the fluid.

Given that the dimensions of the swimming pool are 20ft by 15ft by 10ft, and the cube's side is lying on the bottom of the pool, the depth of the water is 10ft.

Now, to find the total force exerted on the five exposed faces of the cube, you need to calculate the pressure exerted on each face and then multiply it by the area of the face.

Since the cube is submerged, each face will experience the same pressure. Thus, you only need to calculate the pressure once.

The pressure exerted by the water can be calculated using the formula mentioned earlier, where ( \rho ) is the density of water (( 62.4 , \text{lbs/ft}^3 )) and ( g ) is the acceleration due to gravity (( 32 , \text{ft/s}^2 )). The depth (( h )) is 10ft.

After calculating the pressure, you can then multiply it by the area of one face of the cube (since all faces of a cube have the same area) and then multiply by 5 to get the total force exerted on the five faces.

[ \text{Total force} = \text{Pressure} \times \text{Area of one face} \times 5 ]

[ \text{Area of one face} = (\text{Side length of cube})^2 ]

[ \text{Pressure} = \rho \cdot g \cdot h ]

Substitute the values and calculate to find the total force exerted on the five exposed faces of the cube.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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