{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution8

# solution8 - fullwood(tf5349 Homework Fluids Catala(49033...

This preview shows pages 1–3. Sign up to view the full content.

fullwood (tf5349) – Homework Fluids – Catala – (49033) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An engineer weighs a sample of mercury ( ρ = 13 . 6 × 10 3 kg / m 3 ) and finds that the weight of the sample is 4 . 1 N. What is the sample’s volume? The acceler- ation of gravity is 9 . 81 m / s 2 . Correct answer: 3 . 07309 × 10 - 5 m 3 . Explanation: Let : W = 4 . 1 N , ρ = 13 . 6 × 10 3 kg / m 3 , and g = 9 . 81 m / s 2 . m = W g so V = m ρ = W g ρ = 4 . 1 N (9 . 81 m / s 2 )(13 . 6 × 10 3 kg / m 3 ) = 3 . 07309 × 10 - 5 m 3 . 002 10.0 points A pipe contains water at 4 . 9 × 10 5 Pa above atmospheric pressure. If you patch a 4 . 1 mm diameter hole in the pipe with a piece of bubble gum, how much force must the gum be able to withstand? Correct answer: 6 . 46925 N. Explanation: Let : P = 4 . 9 × 10 5 Pa and r = 2 . 05 mm . P = F A F = P A = P ( π r 2 ) = (4 . 9 × 10 5 Pa) π (2 . 05 mm) 2 parenleftbigg 1 m 1000 mm parenrightbigg 2 = 6 . 46925 N . 003 (part 1 of 2) 10.0 points A container is filled with water to a depth of 32 . 8 cm. On top of the water floats a 28 . 7 cm thick layer of oil with a density of 637 kg / m 3 . What is the absolute pressure at the surface of the water? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 1 . 02793 × 10 5 Pa. Explanation: Let : h o = 28 . 7 cm , ρ o = 637 kg / m 3 , g = 9 . 81 m / s 2 , and P 0 = 1 . 01 × 10 5 Pa . P = P 0 + ρ o g h o = 1 . 01 × 10 5 Pa + (637 kg / m 3 ) (9 . 81 m / s 2 ) × (28 . 7 cm) parenleftbigg 1 m 100 cm parenrightbigg = 1 . 02793 × 10 5 Pa . 004 (part 2 of 2) 10.0 points What is the absolute pressure at the bottom of the container? Correct answer: 1 . 06011 × 10 5 Pa. Explanation: Let : h w = 32 . 8 cm and ρ w = 1000 kg / m 3 . P net = P + ρ w g h w = 1 . 02793 × 10 5 Pa

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

View Full Document
fullwood (tf5349) – Homework Fluids – Catala – (49033) 2 + (1000 kg / m 3 ) (9 . 81 m / s 2 ) × (32 . 8 cm) parenleftbigg 1 m 100 cm parenrightbigg = 1 . 06011 × 10 5 Pa . 005 10.0 points The Mariana Trench, in the Pacific Ocean, is about 11 km deep. If atmospheric pressure at sea level is 1 . 01 × 10 5 Pa and the density of sea water is 1025 kg / m 3 , how much pressure would a submarine need to be able to withstand to reach this depth? The acceleration of gravity is 9 . 81 m / s 2 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

solution8 - fullwood(tf5349 Homework Fluids Catala(49033...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online