HW-17-solutions - pandey(sp34566 HW-17 swinney(57305 This...

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pandey (sp34566) – HW-17 – swinney – (57305) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A skater pushes straight away from a wall. She pushes on the wall with a force whose magnitude is F , so the wall pushes on her with a force F (in the direction of her motion). As she moves away from the wall, her center of mass moves a distance d . Consider the following statements regarding energy. I Δ K trans + Δ E internal = Fd II. Δ K trans + Δ E internal = Fd III. Δ K trans + Δ E internal = 0 IV. Δ K trans = Fd V. Δ K trans = Fd What is the correct form of the energy prin- ciple for the skater as a real system and as a point particle (PP) system? F is applied. Thus, the work done is Fd . Translational kinetic energy is the only type of energy a point particle system can have. IV is correct. 002 10.0points Evaluate the cross product (2ˆ ı + 7ˆ ) × ( ı + 6ˆ ) . 1. 16ˆ ı skater, i.e. by cooling down. The right hand side of the energy principle must therefore be zero for the real system. Furthermore, the left hand side must contain the skater’s change in internal energy for a real system. III is correct. As a point particle system, the skater’s cen- ter of mass moves a distance d while a force
8. Real: III, PP: V 9. Real: II, PP: V 10. Real: IV, PP: III Explanation: As a real system, the wall does no work on the skater since the contact point of the skater’s hand and the wall does not move. Also, the wall does not transfer energy to the Explanation: Because both vectors are in the xy plane, their cross product must be in the ± z di- rection. So we only need to look at the z- component of the cross product. If we label the first vector vector A and the second vector vector B , the z component of the cross product will be A x B y A y B x = 40 . So the resultant vector is 40 ˆ k .
pandey (sp34566) – HW-17 – swinney – (57305) 2 003(part1of2)10.0points An airplane of mass 15375 kg flies level to the ground at an altitude of 14 km with a constant speed of 165 m / s relative to the Earth. What is the magnitude of the airplane’s angular momentum relative to a ground ob- server directly below the airplane in kg · m 2 / s?
Explanation: Since the observer is directly below the air- plane, L = h m v 004(part2of2)10.0points

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