# hw 37 - homework 37 HETTIGER CHRISTOF Due 4:00 am Question...

This preview shows pages 1–2. Sign up to view the full content.

homework 37 – HETTIGER, CHRISTOF – Due: Apr 29 2008, 4:00 am 1 Question 1, chap 18, sect 5. part 1 of 1 10 points A block of mass 8 . 6 kg is completely im- mersed in a liquid of density 1150 kg / m 3 . The block is suspended by a thin wire which experiences a tension of 23 . 3 N. The acceleration of gravity is 9 . 8 m / s 2 . 8 . 6 kg Scale What is the volume of the submerged block? Correct answer: 0 . 00541083 m 3 (tolerance ± 1 %). Explanation: Let : m = 8 . 6 kg , block s mass ρ L = 1150 kg / m 3 , and g = 9 . 8 m / s 2 , Basic Concepts: Archimedes’ Principle: Buoyant Force = Weight of Fluid Displaced Solution: The block is in equilibrium, so summationdisplay F y = 0 = T + ρ L V g - m g , where ρ L V g is the buoyant force on the block. Solving for the volume V gives V = m g - T ρ L g = (8 . 6 kg) (9 . 8 m / s 2 ) - (23 . 3 N) (1150 kg / m 3 ) (9 . 8 m / s 2 ) = 0 . 00541083 m 3 . Question 2, chap 18, sect 5. part 1 of 1 10 points A block of volume 0 . 39 m 3 floats with a fraction 0 . 64 of its volume submerged in a liquid of density 1150 kg / m 3 , as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . V 1 liquid Find the magnitude of the buoyant force on the block. Correct answer: 2812 . 99 N (tolerance ± 1 %). Explanation: Basic Concepts: Archimedes’ Principle: Buoyant Force on Object=Weight of Fluid Displaced by Object. F Buoyant,Part 1 = Weight of displaced liquid = ρ L parenleftbigg ρ 1 ρ L parenrightbigg V 1 g = (1150 kg / m 3 ) (0 . 64) × (0 . 39 m 3 ) (9 . 8 m / s 2 ) = 2812 . 99 N , where ρ 1 ρ L = 0 . 64 . Question 3, chap 18, sect 5. part 1 of 1 10 points A sample of an unknown material weighs 305 . 7 N in air and 222 . 0 N when submerged in an alcohol solution with a density of 0 . 644 × 10 3 kg / m 3 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern