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.
cant find much on how to solve problems like these. any help would be appreciated.
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6 answers
awesome, thanks
You are welcome.
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.
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.
Thank you Henry
Glad I could help.