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Homework Solutions 5

# Homework Solutions 5 - homework 01 ASH BEN Due 11:00 pm...

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homework 01 – ASH, BEN – Due: Jan 30 2008, 11:00 pm 1 Question 1, chap 1, sect 1. part 1 of 2 10 points Two points in the xy plane have cartesian coordinates ( x 1 , y 1 ) and ( x 2 , y 2 ), where x 1 = 8 . 6 m, y 1 = 10 m, x 2 = 10 m, and y 2 = 4 m. Determine the distance between these points. Correct answer: 23 . 28 m (tolerance ± 1 %). Explanation: By simple geometry of triangle, the dis- tance between two points is d = radicalBig [ x 2 x 1 ] 2 + [ y 2 y 1 ] 2 = braceleftbigg bracketleftBig ( 10 m) (8 . 6 m) bracketrightBig 2 + bracketleftBig (4 m) ( 10 m) bracketrightBig 2 bracerightbigg 1 2 = 23 . 28 m . Question 2, chap 1, sect 1. part 2 of 2 10 points What is the angle between the line con- necting the two points and x -axis (measured counter-clockwise, within the limits of 180 to +180 )? Correct answer: 143 . 032 (tolerance ± 1 %). Explanation: The angle between the line connecting the two points and x -axis (positive in counter- clockwise direction) is θ = arctan bracketleftbigg y 2 y 1 x 2 x 1 bracketrightbigg = arctan bracketleftbigg (4 m) ( 10 m) ( 10 m) (8 . 6 m) bracketrightbigg = arctan bracketleftbigg (14 m) ( 18 . 6 m) bracketrightbigg = 2 . 49637 rad = 143 . 032 deg . -10 -10 -9 -9 -8 -8 -7 -7 -6 -6 -5 -5 -4 -4 -3 -3 -2 -2 -1 -1 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 ( 10, 4) (8 . 6, 10) 143 . 032 deg Question 3, chap 1, sect 2. part 1 of 1 10 points The standard of time is based on 1. the yearly revolution of the earth about the sun. 2. the daily rotation of the earth. 3. the frequency of light emitted by 86 Kr. 4. None of these correct 5. a precision pendulum clock. Explanation: The second is defined from oscillations of cesium-133 atoms. Question 4, chap 1, sect 4. part 1 of 1 10 points A spherical balloon with radius r inches has volume V = 4 3 π r 3 . Find a function that represents the amount of air required to inflate the balloon from a radius of r inches to a radius of r + 1 inches. 1. 4 3 π ( 3 r + 1 r 2 + 1 )

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homework 01 – ASH, BEN – Due: Jan 30 2008, 11:00 pm 2 2. 4 3 π ( r + 1) 3 3. 4 3 π ( 3 r + 3 r 2 + 1 ) correct 4. 4 3 π ( 1 r + 3 r 2 + 1 ) 5. 4 3 π r 3 Explanation: Δ V = 4 3 π ( r + 1) 3 4 3 π r 3 = 4 3 π ( r 3 + 3 r + 3 r 2 + 1 r 3 ) = 4 3 π ( 3 r + 3 r 2 + 1 ) . Question 5, chap 1, sect 5. part 1 of 1 10 points In one scene in the movie The Godfather II , a solid gold phones is passed around a large table for everyone to see. Suppose the volume of gold in the phone was equal to the volume of 10-centimeter cube of gold (the density of gold is 19,300 kg/m 3 ). Do you think such a phone could be causu- ally passed around a table from hand to hand? What is the weight of the phone? 1. No. It weighs about 45 lbs. correct 2. Yes. It weighs about 9 lbs. 3. No. It weighs about 90 lbs. 4. Yes. It weighs about 4.5 lbs. Explanation: One cubic meter of gold has a mass of about 20,000 kg, which is roughly 45,000 pounds. A ten centimeter cube of gold is about 45 pounds. It would be difficult to pass around this phone.
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Homework Solutions 5 - homework 01 ASH BEN Due 11:00 pm...

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