solution_10pdf - yoon (jy4326) – HW10 – markert –...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: yoon (jy4326) – HW10 – markert – (58840) 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 (part 1 of 2) 10.0 points The cross-sectional area of the output piston in a hydraulic device is 17 times the input piston’s area. By how much will the device multiply the input force? Correct answer: 17. Explanation: Let : n = 17 . P in = P out F in A in = F out A out = F out n A in F out = n F in = 17 F in . 002 (part 2 of 2) 10.0 points By what factor will the output piston move compared to the distance the input piston is moved? Correct answer: 0 . 0588235. Explanation: F in d in = F out d out = n F in d out d out = 1 n d in = 1 17 d in = 0 . 0588235 d in . 003 10.0 points The air pressure above the liquid in figure is 1 . 02 atm . The depth of the air bubble in the liquid is 40 . 8 cm and the liquid’s density is 753 kg / m 3 . 40 . 8 cm air air Determine the air pressure in the bubble suspended in the liquid. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 06337 × 10 5 Pa. Explanation: Let : ρ = 753 kg / m 3 , P = 1 . 02 atm , g = 9 . 8 m / s 2 , and h = 40 . 8 cm = 0 . 408 m . P = P + ρ g h = (1 . 02 atm) · 1 . 013 × 10 5 Pa atm + (753 kg / m 3 ) (9 . 8 m / s 2 ) (0 . 408 m) = 1 . 06337 × 10 5 Pa . 004 (part 1 of 2) 10.0 points A heavy liquid with a density 11 g / cm 3 is poured into a U-tube as shown in the left- hand figure below. The left-hand arm of the tube has a cross-sectional area of 15 . 6 cm 2 , and the right-hand arm has a cross-sectional area of 6 . 01 cm 2 . A quantity of 93 . 7 g of a light liquid with a density 1 . 2 g / cm 3 is then poured into the right-hand arm as shown in the right-hand figure below. h 1 h 2 15 . 6 cm 2 6 . 01 cm 2 heavy liquid 11 g / cm 3 L 15 . 6 cm 2 6 . 01 cm 2 light liquid 1 . 2 g / cm 3 yoon (jy4326) – HW10 – markert – (58840) 2 Determine the height L of the light liquid in the column in the right arm of the U-tube, as shown in the second figure above. Correct answer: 12 . 9922 cm. Explanation: Let : m ℓ = 93 . 7 g , A 1 = 15 . 6 cm 2 , A 2 = 6 . 01 cm 2 , ρ ℓ = 1 . 2 g / cm 3 , and ρ h = 11 g / cm 3 . Using the definition of density ρ ℓ = m ℓ V ℓ = m ℓ A 2 L L = m ℓ A 2 ρ ℓ = 93 . 7 g (6 . 01 cm 2 ) (1 . 2 g / cm 3 ) = 12 . 9922 cm . 005 (part 2 of 2) 10.0 points If the density of the heavy liquid is 11 g / cm 3 , by what height h 1 does the heavy liquid rise in the left arm? Correct answer: 0 . 394178 cm. Explanation: After the light liquid has been added to the right side of the tube, a volume A 2 h 2 of heavy liquid is displaced to the left side, raising the heavy liquid on the left side by a height of h 1 with a displaced volume of A 1 h 1 . Since the volume of heavy liquid is not changed, we have A 1 h 1 = A 2 h 2 ....
View Full Document

This note was uploaded on 01/08/2010 for the course CH 53675 taught by Professor Bocknack during the Fall '09 term at University of Texas.

Page1 / 8

solution_10pdf - yoon (jy4326) – HW10 – markert –...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online