physics

cant find much on how to solve problems like these. any help would be appreciated.

A pressure of 6000 Pa is exerted on a wall of 12 square meters area. What force is exerted on the wall?

What is the pressure at the bottom of a 50 m tall water tower?

What is the pressure at the bottom of a 50 m tall tank of Benzene? Density of Benzene is 0.90 g/cm3

A gold cylinder is suspended in water. Cylinder volume is 1 x 10-6 m3.
Density of gold is 19,300 kg/m3. What is the mass of the cylinder? What is the buoyancy force on the cylinder?

A rectangular block (0.25m x 0.50m x 1.00m) of brass is suspended in water. Density of brass is 8,600 kg/m3. What is the buoyancy force on the block? What is the mass of the block? What is the actual weight of the block? What is the apparent weight of the block when suspended in water?

A block of copper suspended in ethyl alcohol experiences a buoyancy force of 13 N. What is the volume of the block? Density of ethyl alcohol is 810 kg/m3.

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  1. 6000 Pascals is 6000 Newtons per square meter.
    Multiply that by the area of the wall in square meters to get the force in Newtons.

    The density of water is about 1000 kg /m^3
    so the weight of a cubic meter of water is about
    m g = 1000 * 9.81 = 9810 Newtons

    so a column of water 50 m high and one meter square across would weigh
    50 * 9810 N
    That force in ewtons presses down on your one square meter
    so
    we have 50 *9810 Newtons/ square meter
    which is 50 * 9810 Pascals.
    which is about
    490,500 Pascals or about 5 atmospheres due to the water.
    In fact you will have one atmosphere more due to the atmosphere itself but I suspect this means "gage pressure"
    which is the pressure above atmospheric.

    .9 g/cm^3 = 900 kg/m^3
    so multiply the answer we got for the water tower by 0.90

    10^-6 * 19300 = .0193 kg gold

    mass of 10^-6 m^3 water = 10^-3 kg
    buoyant force = weight of water displaced = 10^-3 kg * 9.81 = .00981 N

    The last two are the same idea.

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  2. awesome, thanks

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  3. You are welcome.

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  4. 4a. Mass = 1*10^-6m^3 * 19,300kg/m^3 =
    0.0193 Kg .
    4b. Fb = 1*10^-6m^3 * 1000Kg/m^3 = 0.001 Kg. = 0.0098 N.

    5. V = 0.25 * 0.5 * 1 = 0.125 m^3.
    a. Fb = 0.125m^3 * 1000Kg/m^3 = 125 kg.
    = 1225 N.
    b. M = 0.125m^3 * 8,600kg/m^3 =
    c. = Wb = M*g Newtons.

    6. V*D = 13 N.
    V*810 = 13.
    Solve for V.

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  5. Thank you Henry

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  6. Glad I could help.

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