qz10sol_3891f10 - cylinder According to Archimedes’...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
TA: Tomoyuki Nakayama Tuesday, November 16, 2010 PHY 2048: Physic 1, Discussion Section 3891 Quiz 10 (Homework Set #12) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ The figure below right shows a lead ball suspended by thread of negligible mass from an upright cylinder that floats partially submerged in water. The cylinder has a height of 8.00 cm, a face area of 25.0 cm 2 on the top and bottom, and a density of 0.400 g/cm 3 , and 1.50 cm of its height is above the water surface. The water has a density of 1000 kg/m 3 , and the lead has a density of 1.14 × 10 4 kg/m 3 . a) What is the magnitude of the buoyancy exerted on the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cylinder? According to Archimedes’ principle, the buoyancy on the cylinder is equal to the weight of water displaced by the cylinder. Since 1.5 cm is above the surface of water, the rest 6.5 cm is under the surface of water. The buoyancy is B C = ρ W (6.5/8)(Ah)g = 1.59 N b) What is the radius r of the lead ball? Newton’s 2 nd law applied to the whole system yields B B + B C – W B – W C = 0 We express the buoyancies and gravitational forces in terms of volumes and densities and solve the equation for the volume of the ball. ρ W V B g + ρ W (6.5/8)(Ah)g – ρ B V B g – ρ Ahg = 0 ⇒ V B = [(6.5/8) ρ W – ρ C ]Ah/( ρ B – ρ W ) = 7.93 cm 3 The radius of the ball is V B = (4 π /3)r 3 r = 3 √ (3V B /(4 π )) = 1.24 cm...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online