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Unformatted text preview: oldmidterm 01 – KIM, JI – Due: Dec 18 2007, noon 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered Jupiterlike planet has an average radius 10 . 6times 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 . 26364 (tolerance ± 1 %). Explanation: Let : R np = 10 . 6 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 (10 . 6) 3 = . 26364 . Question 2, chap 2, sect 5. part 1 of 1 10 points A body moving with uniform acceleration has a velocity of 6 . 01 cm / s when its x coordi nate is 3 . 69 cm. If its x coordinate 3 . 05 s later is 8 cm, what is the magnitude of its acceleration? Correct answer: 6 . 45429 cm / s 2 (tolerance ± 1 %). Explanation: Basic Concept: Kinematic Equations x = x + v t + 1 2 a t 2 Solution: Solving for a , we get a = 2[ x x v t ] t 2 = 6 . 45429 cm / s 2 2 . Question 3, chap 2, sect 6. part 1 of 1 10 points An object is thrown downward with an ini tial speed of 9 m / s from a height of 45 m above the ground. At the same instant, a sec ond object is propelled vertically from ground level with a speed of 59 m / s . The acceleration of gravity is 9 . 8 m / s 2 . At what height above the ground will the two objects pass each other? Correct answer: 36 . 8982 m (tolerance ± 1 %). Explanation: Let : v up = 59 m / s , v down = 9 m / s , and h = 45 m . Basic Concepts: y down = h v down t 1 2 g t 2 y up = v up t 1 2 g t 2 . Solution: The two objects pass one another when y down = y up . h v down t 1 2 g t 2 = v up t 1 2 g t 2 oldmidterm 01 – KIM, JI – Due: Dec 18 2007, noon 2 h = v down t + v up t = ⇒ t = h v down + v up = 0 . 661765 s . So, H = v up t 1 2 g t 2 = (59 m / s) (0 . 661765 s) 1 2 (9 . 8 m / s 2 ) (0 . 661765 s) 2 = 36 . 8982 m . Question 4, chap 3, sect 4. part 1 of 1 10 points Consider vectors vector A and vector B with coordinate components shown in the illustration below. A B A perspective drawing: Each coordinate has a length of ± 5 units. To indicate the coordinates of each vector, a line is projected to the hor izontal plane then two lines are pro jected to the horizontal coordinates. As well, a line is directly projected to the vertical coordinate. What is the scalar product vector A · vector B for these two vectors?...
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Mass

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