{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW1S - Nguyen Phan Homework 1 Due 4:00 am Inst V...

This preview shows pages 1–4. Sign up to view the full content.

Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points One cubic meter (1.0 m 3 ) of aluminum has a mass of 2700 kg, and a cubic meter of iron has a mass of 7860 kg. Find the radius of a solid aluminum sphere that has the same mass as a solid iron sphere of radius 2 . 2 cm. Correct answer: 3 . 1413 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 2 cm . Basic Concept Density is ρ = m V Solution: Since the masses are the same, ρ Al V Al = ρ Fe V Fe ρ Al 4 3 π r 3 Al = ρ Fe 4 3 π r 3 Fe r Al r Fe 3 = ρ Fe ρ Al r Al = ρ Fe ρ Al 1 3 ( r Fe ) = 7860 kg 2700 kg 1 3 (2 . 2 cm) = 3 . 1413 cm 002 (part 1 of 1) 10 points Consider the following set of equations, where s , s 0 , x and r have units of length, t has units of time, v has units of velocity, g and a have units of acceleration, and k is dimensionless. Which one is dimensionally incorrect ? 1. a = g + kv t + v 2 s 0 2. t = v a + x v 3. s = s 0 + vt + v 2 a 4. t = k r s g + a v correct 5. v 2 = 2 as + ksv t Explanation: For an equation to be dimensionally correct, all its terms must have the same units. (1): t = v a + x v [ t ] = T h v a i + h x v i = LT - 1 LT - 2 + L LT - 1 = T + T = T is consistent. (2): a = g + kv t + v 2 s 0 [ a ] = LT - 2 g + kv t + v 2 s 0 = LT - 2 + LT - 1 T + L 2 T - 2 L = LT - 2 is also consistent. (3): t = k r s g + a v [ t ] = T k r s g + a v = r L LT - 2 + LT - 2 LT - 1 = T + T - 1 is not dimensionally consistent. (4): v 2 = 2 as + ksv t [ v 2 ] = L 2 T - 2 2 as + ksv t = LT - 2 L + LLT - 1 T - 1 = L 2 T - 2

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

View Full Document
Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky 2 is also consistent. (5): s = s 0 + vt + v 2 a [ s ] = L s 0 + vt + v 2 a = L + LT - 1 T + L 2 T - 2 LT - 2 = L is also consistent. So only t = k r s g + a v is dimensionally in- correct . 003 (part 1 of 2) 10 points Stokes law says F = 6 πrηv. F is a force, r the radius and v the velocity. The parameter η has the dimension of 1. [ η ] = M T L correct 2. [ η ] = T 2 L 2 M 3. [ η ] = M T L 2 4. [ η ] = T M 5. [ η ] = T 2 L M 6. [ η ] = M T 7. [ η ] = M T 2 L 2 8. [ η ] = M T 2 L 9. [ η ] = T L M 10. [ η ] = T L 2 M Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 π is dimensionless. The dimen- sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 π r η v ] M L T 2 = L L T [ η ] Solving for [ η ] yields [ η ] = M LT . 004 (part 2 of 2) 10 points Consider a simple pendulum which consists of a string with a bob attached to its end. Its period, ( i.e. , the time interval taken for the bob to complete one cycle of motion), may be written in the form T = k m x g y b z , where k is a dimensionless constant, g the magnitude of the gravitational acceleration, m the mass of the bob, and b the length of the string. The appropriate x , y , and z values are given respectively by 1. ( x, y, z ) = 1 , 1 2 , - 1 2 2. ( x, y, z ) = 0 , - 1 2 , - 1 2 3. ( x, y, z ) = 1 , - 1 2 , 1 2 4. ( x, y, z ) = 0 , - 1 2 , 1 2 correct 5. ( x, y, z ) = 0 , 1 2 , - 1 2 6. ( x, y, z ) = 1 , - 1 2 , - 1 2 7. ( x, y, z ) = (1 , - 1 , 1) 8. ( x, y, z ) = 0 , 1 2 , 1 2 9. ( x, y, z ) = (0 , - 1 , 1)
Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky 3 10. ( x, y, z ) = 1 , 1 2 , 1 2 Explanation: [ T ] = [ k m x g y b z ] T = M x L y T 2 y L z T = M x

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