ch07 - 1. With speed v = 11200 m/s, we find K= 1 2 1 mv =...

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1. With speed v = 11200 m/s, we find Km v ==× = × 1 2 1 2 29 10 11200 18 10 25 2 1 3 (. )( ) . J.
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2. (a) The change in kinetic energy for the meteorite would be () ( ) 2 26 3 1 4 11 41 0k g1 51 0m / s 5 1 0 J 22 fi i i i KK K K m v ∆= − = = × × = −× , or 14 || 5 1 0 J K ∆=× . The negative sign indicates that kinetic energy is lost. (b) The energy loss in units of megatons of TNT would be 14 15 1 megaton TNT 0J 0 .1megaton TNT. 4.2 10 J K §· −∆ = × = ¨¸ × ©¹ (c) The number of bombs N that the meteorite impact would correspond to is found by noting that megaton = 1000 kilotons and setting up the ratio: 0.1 1000kiloton TNT 8. 13kiloton TNT N × ==
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3. (a) From Table 2-1, we have vv a x 2 0 2 2 =+∆ . Thus, () ( ) 2 27 1 5 7 0 22 .4 10 2 3.6 10 0.035 2.9 10 m/s. a x =+ = × + × = × (b) The initial kinetic energy is ( ) 2 7 7 1 3 0 11 1.67 10 kg 2.4 10 m/s 4.8 10 J. i Km v −− == × × = × The final kinetic energy is ( ) 2 7 7 1 3 1.67 10 kg 2.9 10 m/s 6.9 10 J. f v × × = × The change in kinetic energy is K = (6.9 × 10 –13 – 4.8 × 10 –13 ) J = 2.1 × 10 –13 J.
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4. We apply the equation 2 1 00 2 () xt x vt a t =+ + , found in Table 2-1. Since at t = 0 s, x 0 = 0 and 0 12 m/s v = , the equation becomes (in unit of meters) 2 1 2 () 12 t at =+ . With 10 m x = when 1.0 s t = , the acceleration is found to be 2 4.0 m/s a =− . The fact that 0 a < implies that the bead is decelerating. Thus, the position is described by 2 2.0 t t . Differentiating x with respect to t then yields 12 4 .0 dx t dt == . Indeed at t =3.0 s, (3 . 0 ) 0 and the bead stops momentarily. The speed at 10 s t = is (1 0 ) 2 8 m / s , and the corresponding kinetic energy is 22 2 11 (1 .8 1 0 kg)( 28 m/s) 7.1 J. Km v × =
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5. We denote the mass of the father as m and his initial speed v i . The initial kinetic energy of the father is KK i = 1 2 son and his final kinetic energy (when his speed is v f = v i + 1.0 m/s) is f = son . We use these relations along with Eq. 7-1 in our solution. (a) We see from the above that if = 1 2 which (with SI units understood) leads to () 2 2 11 1 1.0 22 2 ii mv m v ªº =+ «» ¬¼ . The mass cancels and we find a second-degree equation for v i : 1 2 1 2 0 2 vv −−= . The positive root (from the quadratic formula) yields v i = 2.4 m/s. (b) From the first relation above i = 1 2 son bg , we have 1 2 1 2 1 2 mv m v i = F H G I K J F H G I K J son 2 and (after canceling m and one factor of 1/2) are led to i son =2 =4.8 ms.
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6. By the work-kinetic energy theorem, () 22 2 2 111 (2.0kg) (6.0m/s) (4.0m/s) 20 J. 222 fi WK m v m v =∆ = = = We note that the directions of & v f and & v i play no role in the calculation.
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7. Eq. 7-8 readily yields (with SI units understood) W = F x x + F y y = 2cos(100º)(3.0) + 2sin(100º)(4.0) = 6.8 J.
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8. Using Eq. 7-8 (and Eq. 3-23), we find the work done by the water on the ice block: () ( ) 3 ˆˆ ˆ ˆ 210i 150 j 15i 12j (210)(15) ( 150) ( 12) 5.0 10 J. WF d =⋅= =+ −=× & &
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9. Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as xv t a t =+ 0 1 2 2 (where x 0 0 = ). We choose to analyze the third and fifth points, obtaining 2 0 2 0 1 0.2m (1.0 s) (1.0 s) 2 1 0.8m (2.0 s) (2.0 s) 2 va Simultaneous solution of the equations leads to 0 0 v = and a = 040 .m s 2 . We now have two ways to finish the problem. One is to compute force from F = ma and then obtain the work from Eq. 7-7. The other is to find K
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This homework help was uploaded on 09/30/2007 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell.

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ch07 - 1. With speed v = 11200 m/s, we find K= 1 2 1 mv =...

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