A hollow copper tube with a diameter of 3.87 cm is sealed at one end and loaded with lead shot to give a total mass of 0.183 kg. When the tube is floated in pure water, what is the depth of its bottom end with respect to the surface of the water?
2007-08-12 06:31:03 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
15.56 cm.
2007-08-12 06:50:40 · answer #1 · answered by John S2005 3 · 0⤊ 0⤋
When it's floating, the upward force of buoyancy exactly equals the downward force of the tube's weight:
buoyancy = mg = (0.183 kg)(9.8 m/sec²)
Also, the buoyancy always equals the weight of the displaced water:
buoyancy = (density of water)·(displaced volume)
= (1 gram/cm^3)·(πr²·depth)
= (1 gram/cm^3)·π(3.87 cm/2)²·depth
Equate the two expressions for buoyancy:
(0.183 kg)(9.8 m/sec²) = (1 gram/cm^3)·π(3.87 cm/2)²·depth
Now just solve for "depth," using simple algebra.
2007-08-12 07:10:50 · answer #2 · answered by RickB 7 · 0⤊ 0⤋
15.56 cm
2007-08-12 07:19:47 · answer #3 · answered by PRADEEP K 1 · 0⤊ 0⤋