Asked by John
A large aquarium has viewing ports so people can watch dolphins swim by. What is the net force on a rectangular window, 2.30 m wide and 1.20 m high, if the top of the window is 4.50 m below the surface of the water?
Answers
Answered by
bobpursley
average force on the window is at the 1/2 way point down.
forceon window= (4.5m+1/2 *1.2)*area*densitywater*g
= 5.1m*(2.3*1.2)m^2*1E3kg/m^3*9.8N/kg
Now, that is net force, because above the water one would add atmospheric pressure, but on the inside of the window, one would subtract atmospheric pressure.
forceon window= (4.5m+1/2 *1.2)*area*densitywater*g
= 5.1m*(2.3*1.2)m^2*1E3kg/m^3*9.8N/kg
Now, that is net force, because above the water one would add atmospheric pressure, but on the inside of the window, one would subtract atmospheric pressure.
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