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Chapter 7 problems

Chapter 7 problems. Chapter 7 problems (42 (3 votes), 57 (2votes)). Chapter 7 problems (42 (3 votes), 57 (2votes)). Chapter 8 Problems (cont.) (36 (5 votes), 53 (3 votes)). Chapter 8 Problems (cont.).

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Chapter 7 problems

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  1. Chapter 7 problems

  2. Chapter 7 problems(42 (3 votes), 57 (2votes))

  3. Chapter 7 problems(42 (3 votes), 57 (2votes))

  4. Chapter 8 Problems (cont.)(36 (5 votes), 53 (3 votes))

  5. Chapter 8 Problems (cont.)

  6. The book gives a mathematical expression for the coordinates (X,Y,Z) of the center of mass (CoM) of a collection of particles. Someone has suggested that these formulas amount to stating that the CoM is the mass-weighted average position of the collection of particles. Comment briefly on the validity/universality of this statement. (13 Yes; 9 No; 6 didn’t understand or gave confusing answers; 15 no answer) • I don't think it's valid, because formula doesn't exactly give us the mass-weighted average position. • You know, unless I'm understanding the question incorrectly. It is worded somewhat strangely. • If by that statement the person means the relative mass weights the position, I would consider that valid universal as it basically just restates the formula. The word average makes me hesitant because that usually implies dividing by the number of particles which does not happen in this case. (This is the idea of a weighted average!) • This statement appears to be valid. If a m1 is heavier than m2, then the center of mass (the mass weighted average position) will be closer to m1 than m2. This appears to be valid. (Does that help??)

  7. Weighted averagese.g. Neutron lifetime If you wanted to average all of these experimental results, would you just add them up and divide by 8? No not all results are equally precise, so you’d weight each of them by something like 1/(error bar) http://www.iucf.indiana.edu/annrep/2004/n-lifetime.pdf

  8. Chapter 9 Problems

  9. Figure 9.6 The Grand Jete’

  10. Figure 9.4 The Negative mass trick’

  11. Figure 9.8,9 Impulse

  12. A softball (m=0.2kg) approaches a bat at a speed of 25 m/s in a direction that is roughly parallel to the ground. After making contact with the bat for roughly 0.1 sec, the ball reverses direction and is traveling at 30 m/s. What is the average force exerted on the ball by the bat? Explain briefly how you arrived at your answer. (12 forgot that v is a vector (|vf-vi|=55m/s not 5m/s; 9 made other mistakes of various sorts; 6 got it right, 17 didn’t answer). • Favg=-change in m/change in time exchange in velocity. Average force is 20 m/s. (units??) • Favg = -(1(0.20kg)/(0.1s))(30m/s-25m/s); Favg = 10 N • 11N. I multiplied the mass of the bat with the speed of the ball for each direction. One direction product is 5N the other is 6N. I added the two and got 11N. I'm not sure what I am supposed to do with the .1sec. • Since momentum equals mv, the initial momentum is 5.0 kgm/s, and the final momentum is -6.0 kgm/s, so the change in momentum is 11kgm/s, so the average force exhibited on the ball is 11.0 N.!! (remember to watch units!!)

  13. Chapter 9 problems b. What was the direction of travel before the explosion?

  14. Chapter 9 problems

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