Midterm 1

# Midterm 1 - midterm 01 – BAUTISTA, ALDO – Due: Sep 21...

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Unformatted text preview: midterm 01 – BAUTISTA, ALDO – Due: Sep 21 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:1, V3:2, V4:5, V5:1. Question 1 part 1 of 1 10 points A cylinder, 18 cm long and 5 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 8 g / cm 3 and 6 . 2 g / cm 3 . 1 8 c m 5 cm What length of the lighter metal is needed if the total mass is 7597 g? 1. 9 . 01252 cm 2. 9 . 35069 cm 3. 9 . 64025 cm 4. 9 . 93908 cm 5. 10 . 2782 cm 6. 10 . 6229 cm correct 7. 10 . 9801 cm 8. 11 . 3402 cm 9. 11 . 7004 cm 10. 12 . 0628 cm Explanation: Let : ‘ = 18 cm , r = 5 cm , ρ 1 = 4 . 8 g / cm 3 , ρ 2 = 6 . 2 g / cm 3 , and m = 7597 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 ‘ . Consequently, x = m- ρ 2 π r 2 ‘ π r 2 ( ρ 1- ρ 2 ) = (7597 g)- (6 . 2 g / cm 3 ) π (5 cm) 2 (18 cm) π (5 cm) 2 (4 . 8 g / cm 3- 6 . 2 g / cm 3 ) = 10 . 6229 cm . Question 2 part 1 of 1 10 points The height of a helicopter above the ground is given by h = c t 3 , where c = 1 m / s 3 , h is in meters, and t is in seconds. The helicopter takes off at t = 0 s. After 4 s it releases a small mailbag. The acceleration of gravity is 9 . 8 m / s 2 . How long after its release does the mailbag reach the ground? 1. 7 . 49452 s midterm 01 – BAUTISTA, ALDO – Due: Sep 21 2006, 4:00 am 2 2. 8 . 0528 s 3. 8 . 61023 s 4. 9 . 16695 s 5. 9 . 72307 s 6. 10 . 2787 s 7. 10 . 9849 s correct 8. 11 . 3887 s 9. 11 . 9753 s 10. 12 . 9641 s Explanation: Given : t = 4 s . Under free fall, h ( t ) = y ( t ) = y + v t + 1 2 a t 2 . The initial height of the mailbag is the height of the helicopter 4 s after takeoff h = h ( t ) = (1 m / s 3 ) (4 s) 3 = 64 m , and it starts its free fall motion from this point. Its initial velocity is equal to the veloc- ity of the helicopter at that time v = dh dt = 3 c t 2 = 3 (1 m / s 3 ) (4 s) 2 = 48 m / s . Thus the equation of motion governing the mailbag is y ( t ) = 0 = h + v t- 1 2 g t 2 . In quadratic form, 1 2 g t 2- v t- h = 0 . From the quadratic formula, t = v ± q v 2 + 2 g h g . Since D = v 2 + 2 g h = (48 m / s) 2 + 2 ( 9 . 8 m / s 2 ) (64 m) = 3558 . 4 m 2 / s 2 , the time for the mailbag to reach the ground is t = v ± √ D g = 48 m / s ± p 3558 . 4 m 2 / s 2 9 . 8 m / s 2 . The helicopter takes off at t = 0 s. The negative solution is rejected, so t = 48 m / s + p 3558 . 4 m 2 / s 2 9 . 8 m / s 2 = 10 . 9849 s ....
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## This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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Midterm 1 - midterm 01 – BAUTISTA, ALDO – Due: Sep 21...

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