The horizontal door of a substance at a depth of 500m has an area of 0.4m^2 .
Calculate the force exerted by the sea water on the door at this depth.
Wondering what the pressure on the other side of the door is? One has to know in order to calculate net force on the door.
forceondoor/area=pressure due to water+air pressure above.
pressure due to water:
Pressure=densitywater*g*depth
thank you sir. May God bless you
thank you sir. May God bless you.
R.D=density of substance/density of water
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To calculate the force exerted by the sea water on the door at a depth of 500m, you can use the formula for pressure:
Pressure = density * gravity * depth
First, let's find the density of seawater. The density of seawater can vary depending on the salinity, temperature, and depth, but on average it is around 1025 kg/m^3.
Next, we need to consider the gravitational constant. On Earth, the standard value for gravity is approximately 9.8 m/s^2.
Now, let's substitute these values into the formula:
Pressure = 1025 kg/m^3 * 9.8 m/s^2 * 500m
Next, we can calculate the pressure at this depth:
Pressure = 5042500 Pa
Finally, to calculate the force, we can multiply the pressure by the area of the door:
Force = Pressure * Area
Force = 5042500 Pa * 0.4 m^2
Force = 2017000 N
Therefore, the force exerted by the sea water on the door at a depth of 500m is approximately 2017000 Newtons.