Homework 1 Solutions - homework 01 ALIBHAI ZAHID Due 4:00...

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

View Full Document Right Arrow Icon
homework 01 – ALIBHAI, ZAHID – Due: Jan 26 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A piece of pipe has an outer radius, an inner radius, and length as shown in the figure below. 37 cm 4 . 6 cm 2 . 7 cm The density is 7 . 6 g / cm 3 . What is the mass of this pipe? Correct answer: 12 . 253 kg (tolerance ± 1 %). Explanation: Let : r 1 = 4 . 6 cm , r 2 = 2 . 7 cm , = 37 cm , and ρ = 7 . 6 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross- sectional area times the length. Solution: V = ( π r 2 1 - π r 2 2 ) = π [ r 2 1 - r 2 2 ] = π [(4 . 6 cm) 2 - (2 . 7 cm) 2 ] (37 cm) = 1612 . 23 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 - r 2 2 ] = (7 . 6 g / cm 3 ) π [(4 . 6 cm) 2 - (2 . 7 cm) 2 ] (37 cm) = 12253 g = 12 . 253 kg . Question 2 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 . 68 cm. Correct answer: 3 . 82667 cm (tolerance ± 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 68 cm . Density is ρ = m V . Since the masses are the same, ρ Al V Al = ρ Fe V Fe ρ Al parenleftbigg 4 3 π r 3 Al parenrightbigg = ρ Fe parenleftbigg 4 3 π r 3 Fe parenrightbigg parenleftbigg r Al r Fe parenrightbigg 3 = ρ Fe ρ Al r Al = r Fe parenleftbigg ρ Fe ρ Al parenrightbigg 1 3 = (2 . 68 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 3 . 82667 cm . Question 3 part 1 of 1 10 points A cylinder, 16 cm long and 8 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 6 g / cm 3 and 6 . 1 g / cm 3 . 16 cm 8 cm
Image of page 1

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

View Full Document Right Arrow Icon
homework 01 – ALIBHAI, ZAHID – Due: Jan 26 2007, 4:00 am 2 What length of the lighter metal is needed if the total mass is 17741 g? Correct answer: 6 . 24234 cm (tolerance ± 1 %). Explanation: Let : = 16 cm , r = 8 cm , ρ 1 = 4 . 6 g / cm 3 , ρ 2 = 6 . 1 g / cm 3 , and m = 17741 g . Volume of a bar of radius r and length is V = π r 2 and its density is ρ = m V = m π r 2 so that m = ρ π r 2 x - x r Let x be the length of the lighter metal; then - x is the length of the heavier metal. Thus, m = m 1 + m 2 = ρ 1 π r 2 x + ρ 2 π r 2 ( - x ) = ρ 1 π r 2 x + ρ 2 π r 2 - ρ 2 π r 2 x . Therefore m - ρ 2 π r 2 = ρ 1 π r 2 x - ρ 2 π r 2 x and x π r 2 ( ρ 1 - ρ 2 ) = m - ρ 2 π r 2 ℓ . Consequently, x = m - ρ 2 π r 2 π r 2 ( ρ 1 - ρ 2 ) = (17741 g) - (6 . 1 g / cm 3 ) π (8 cm) 2 (16 cm) π (8 cm) 2 (4 . 6 g / cm 3 - 6 . 1 g / cm 3 ) = 6 . 24234 cm . Question 4 part 1 of 1 10 points Convert 46 . 2 to radians. Correct answer: 0 . 806342 rad (tolerance ± 1 %). Explanation: Since π rad = 180 , 46 . 2 parenleftbigg π rad 180 parenrightbigg = 0 . 806342 rad . Question 5 part 1 of 1 10 points Use the fact that the speed of light in a vacuum is about 3 . 00 × 10 8 m/s to determine how many kilometers a pulse from a laser beam travels in exactly five hours. Correct answer: 5 . 4 × 10 9 km (tolerance ± 1 %). Explanation: Basic Concepts: 1 h = 3600 s 1 km = 1 × 10 3 m d = v Δ t Given: Speed of light = 3 . 00 × 10 8 m / s Δ t = 5 h Solution: d = v Δ t = ( 3 × 10 8 m / s ) · (5 h) · 3600 s 1 h · 1 km 1 × 10 3 m = 5 . 4 × 10 9 km
Image of page 2
homework 01 – ALIBHAI, ZAHID – Due: Jan 26 2007, 4:00 am 3 Question 6 part 1 of 1 10 points How many square meters are there in 5 . 6 acres?
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern