PHY 303K - Old Midterm #1

PHY 303K - Old Midterm #1 - oldmidterm 01 – FIERRO...

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: oldmidterm 01 – FIERRO, JEFFREY – Due: Feb 11 2008, 11:00 pm 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered Jupiter-like planet has an average radius 11 . 1times that of the Earth and a mass 314 times that of the Earth. Calculate the ratio of new planet’s mass density to the mass density of the Earth. Correct answer: 0 . 229594 (tolerance ± 1 %). Explanation: Let : R np = 11 . 1 R E and M np = 314 M E . The volume of a sphere of radius R is V = 4 3 π R 3 . The (average) density ρ of a body is the ratio of the body’s mass to its volume, ρ = M V . Planets are spherical, so the (average) density of a planet of a given mass M and a given radius R is ρ = M 4 3 π R 3 . Comparing the newly discovered planet to the Earth, we have ρ np ρ E = M np 4 3 π R 3 np M E 4 3 π R 3 E = M np M E parenleftbigg R np R E parenrightbigg 3 = 314 (11 . 1) 3 = . 229594 . Question 2, chap 1, sect 5. part 1 of 1 10 points A piece of wire has a density of 6 . 1 g / cm. What is the mass of 11 . 9 cm of the wire? Correct answer: 72 . 59 g (tolerance ± 1 %). Explanation: Let : ρ = 6 . 1 g / cm and ℓ = 11 . 9 cm . Since g cm · cm = g, we must multiply the den- sity by the length of the wire to obtain the mass. This can also be determined by the definition of linear mass density: ρ = m ℓ , so that m = ρ ℓ = (6 . 1 g / cm) (11 . 9 cm) = 72 . 59 g . Question 3, chap 1, sect 6. part 1 of 1 10 points Identify the equation below which is dimen- sionally incorrect . A, x, y and r have units of length. k here has units of inverse length. v and v have units of velocity. a and g have units of acceleration. ω has units of inverse time. t is time, m is mass, V is volume, ρ is density, and F is force. 1. v = radicalbig v 2 − 2 a x 2. g = F m + ρ m g V 3. y = A cos( k x − ω t ) 4. F = mg bracketleftbigg 1 + v r g bracketrightbigg correct 5. t = − v + radicalBig v 2 + 2 a x a + v 2 a 2 t 6. v = ± ω radicalbig A 2 − x 2 7. v = radicalbig 2 g y + radicalbigg r F m 8. F = mω 2 r oldmidterm 01 – FIERRO, JEFFREY – Due: Feb 11 2008, 11:00 pm 2 Explanation: • Check y = A cos( k x − ω t ): Since k is an inverse length, k x is dimension- less, and similarly, ωt is also dimensionless, so k x − ω t is dimensionless, as an argument of a trigonometry function must always be. The value of a trigonometry function ( e.g. cos( k x − ω t )) is also dimensionless. So the right hand side has the dimension of length. [ y ] = L [ A cos( k x )] = L. This equation is dimensionally correct. • Check F = mω 2 r : The dimension of force is [ F ] = ML T 2 . The dimension of right hand side is bracketleftbig mω 2 r bracketrightbig = M parenleftbigg 1 T parenrightbigg 2 L = M L T 2 ....
View Full Document

This homework help was uploaded on 04/03/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 10

PHY 303K - Old Midterm #1 - oldmidterm 01 – FIERRO...

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