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YW8g492IJo19_1210015909_jwk572

# YW8g492IJo19_1210015909_jwk572 - oldmidterm 01 KIM JI Due...

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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 Jupiter-like planet has an average radius 10 . 6 times 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 = 0 . 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 0 + v 0 t + 1 2 a t 2 Solution: Solving for a , we get a = 2[ x - x 0 - v 0 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

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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? Correct answer: - 19 units 2 (tolerance ± 1 %). Explanation: Basic Concept: vector A · vector B = ( A x ˆ ı + A y ˆ + A z ˆ k ) (1) · ( B x ˆ ı + B y ˆ + B z ˆ k ) , where ˆ ı · ˆ ı = ˆ · ˆ = ˆ k · ˆ k = 1 and ˆ ı · ˆ = ˆ ı · ˆ k = ˆ · ˆ k = ˆ · ˆ ı = ˆ k · ˆ ı = ˆ k · ˆ ı = 0 .
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