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Notes Ask Your Teache 0/2 points l Previous Answers HRW 10 7.P00 1.ssm. A proton (mass m = 1.67 x 10-27 kg) is being accelerated along a straight line at 3.00 x 1015 m/s2 in a machine. The proton has an initial speed of 2.00 x 107 m/s and travels 4.3 cm. (a) What is its speed? (Give your answer to at least four significant figures.) 6580 X m/s (b) What is the increase in its kinetic energy? The constant-acceleration kinematic equations apply. After you find the final speed, use the definition of kinetic energy Submit Answer Save Progress Practice Another Version e Ask Your Teach 0/3 points I Previous Answers HRW10 7.P003. On August 10, 1972, a large meteorite skipped across the atmosphere above western United States and western Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky and was brighter than the usual meteorite trail. The meteorite\'s mass was about 4 x 106 kg; its speed was about 15 km/s. Had it entered the atmosphere vertically, it would have hit the Earth\'s surface with about the same speed. °Notes In Notes Ask Your Teach (a) Calculate the meteorite\'s loss of kinetic energy (in joules) that would have been associated with the vertical impact. -500000000000X (b) Express the energy as a multiple of the explosive energy of 1 megaton of TNT, which is 4.2 x 1015 J. -0.1 x megaton TNT (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?

Solution

1) Final velocity V = 2.5651 x 10 ⁷ m/s ( By using formula v²- u² = 2.a.s )

2) Increase in kinetic energy = 4.308 x 10^-13 J ( By formula 1/2 x m x (v²- u²))

3) Loss of kinetic energy = 45 x 10 ⁷ J ( By formula 1/2. m .v² )

4) In Megaton TNT = 1.071 x 10^-7

5) 8.2 x 10^-6 bombs will equate the energy of hiroshima bomb.