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anonymous
 one year ago
A 0.145 kg baseball is thrown horizontally at 15 m/s from a platform, so that it starts at 5.2 m above the ground. How much has its kinetic energy increased when it reaches the ground?
16 J , 24 J , 0 J , or 7.4 J
? :/
anonymous
 one year ago
A 0.145 kg baseball is thrown horizontally at 15 m/s from a platform, so that it starts at 5.2 m above the ground. How much has its kinetic energy increased when it reaches the ground? 16 J , 24 J , 0 J , or 7.4 J ? :/

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the initial kinetic energy is: \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {15^2} = ...Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we get 16.3125? so our solution is 16 J?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that result is right! Nevertheless it is not our final answer!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay! what is netx?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to consider that situation: dw:1433103218745:dw our ball, in free falling, has gained a vertical velocity

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, we have to compute that vertical speed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! how do we do taht?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1such vertical speed is given by the so called formula of Torricelli, which is: \[v = \sqrt {2gh} = \sqrt {2 \times 9.81 \times 5.2} = ...m/\sec \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.010.10006 ? what happens now? :O

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! Now we have this situation: dw:1433103512624:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely the final velocity has the subsequent magnitude: \[{v_f} = \sqrt {{{10}^2} + {{15}^2}} = ...m/\sec \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! So the final kinetic energy is: \[KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {18^2} = ...Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay! we get 23.49 so about 24 J is our solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since that the requested energy change is: \[\Delta KE = 23.49  16 = ...Joules\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay so we get 7.49 so about 7.4 J is our answer? :O

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! that's right!
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