Chapter_7_09 - Chapter 7 7.1 On a large meteorite skipped...

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Chapter 7 7.1. On August 10, 1972, a large meteorite skipped across the atmosphere above western United States and Canada, much like a stone skipped across water. The accompanying freball was so bright that it could be seen in the daytime sky. The meteorite’s mass was about 4x10 6 kg; its speed was about 15 km/s. At it entered the atmosphere vertically, it would have hit the earth’s surFace with the same speed. (a) Calculate the meteorite’s loss oF kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple oF the explosive energy oF 1 megaton oF TNT, which is 4.2x10 15 J. (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons oF TNT. To how many “Hiroshima bombs” would the meteorite impact have been equivalent? a. The energy loss would be K = 1 2 mv 2 = 1 2 4 × 10 6 kg (15, 000 m / s ) 2 = 4.5 × 10 14 J b. The equivalent loss in megatons oF TNT # MT = 4.5 × 10 J 4.2 × 10 15 J / = 0.107 = 107 kT c How many Hiroshima Bombs is this? # Hiroshimas = 107 kT 13 kT / Hiroshima = 8.23 We were very lucky. .. 7.2 IF a Saturn V rocket with an Apollo spacecraFt attached has a combined mass oF m = 2.9x10 5 kg and is to reach a speed oF v = 11.2 km/s = 11.2 x10 3 m/s, how much Kinetic Energy will it have K = 1 2 mv 2 = 1.82 × 10 13 J 7.10 A ±oating ice block is pushed through a displacement d = (15 m ) ˆ i (12 m ) ˆ j along a straight embankment by rushing water which exerts a Force F = (210 N ) ˆ i (150 N ) ˆ j on the block. How much work does the Force do on the block during the displacement? W = F d = ((210 N ) ˆ i (150 N ) ˆ j ) ((15 m ) ˆ i m ) ˆ j ) = 4950 J
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7.13 Figure 7-29 shows three forces applied to a greased trunk that moves leftward by 3m over a a frictionless ±oor. The force magnitudes are F 1 =5.00N, F 2 =9.00N, and F 3 =3.00N. During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease? Since there is no motion in the vertical direction, there is no work done by forces and components of forces in that direction. We need only concern ourselves with the horizontal motion F net x = F 2 cos60 F 1 = 9.00 N cos60 5.00 N = 0.5 N The net force is to the left. This is in the same direction as the motion. The work done is therefore positive.
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This note was uploaded on 07/12/2010 for the course PHYS 107 taught by Professor Iashvili during the Spring '10 term at SUNY Buffalo.

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Chapter_7_09 - Chapter 7 7.1 On a large meteorite skipped...

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