A rock with a weight of 52N is found to have an apparent weight of 36N when completely submerged in water.The density of water is 1000kg m ^-3.(g=9.8N kg ^-1).Calculate the buoyant force acting on the rock when it is submerged in water??
2006-12-17 22:01:14 · 3 answers · asked by Rasathi 1 in Science & Mathematics ➔ Physics
Fb = Fg(real) - Fg(appearent)
Fb = 52 - 36
Fb = 16 N
We can also from the givens, get
mass of the rock
volume of the rock
density of the rock
mass of the displaced water
2006-12-17 22:20:30 · answer #1 · answered by ray2_moot 2 · 0⤊ 0⤋
it is just substracting 36 from 52.because the force of buoyancy is equal to the volume of water diplaced and this indicates the difference between the actual weight and apparent weight that is the weight in water in this case.so 52-36=16 answer will be 16N
2006-12-18 06:17:29 · answer #2 · answered by secab b 1 · 0⤊ 0⤋
Unless I am mistaken, the bouyant force would be equal to the difference between 52N and 36N
52-36 = 16N
2006-12-18 06:19:12 · answer #3 · answered by z_o_r_r_o 6 · 0⤊ 0⤋