Answer:
Ek = 1705.28 [J]
Explanation:
In order to solve this problem, we must remember that kinetic energy can be calculated by means of the following equation.
where:
m = mass [kg]
v = velocity [m/s]
Ek = kinetic energy [J] (Units of Joules)
<u>For the person running</u>
<u />![E_{k} =\frac{1}{2}*52.2*(3.5)^{2} \\ E_{k} =319.72[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%2A52.2%2A%283.5%29%5E%7B2%7D%20%5C%5C%20E_%7Bk%7D%20%3D319.72%5BJ%5D)
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<u>For the bullet</u>
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<u />![E_{k} =\frac{1}{2} *0.02*(450)^{2} \\E_{k}=2025 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2A0.02%2A%28450%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D2025%20%5BJ%5D)
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The difference in Kinetic energy is equal to:
Ek = 2025 - 319.72
Ek = 1705.28 [J]
Answer:
False
Explanation:
Physical features of earth's surface like mountains, deserts and oceans are not the barriers in today's modern world. Technological improvements in the transportation and communication has removed these barriers. So they are not barriers anymore.
<span>Poet Kuangchi Chang did not remain in China long enough to be "re-educated." Following the Communist takeover he fled to the United States. His poem "Garden of My Childhood" describes China before the revolution as a peaceful, idyllic garden with a violent horde rapidly approaching. A vine, the wind, and the sea are each personified, and each beckons for him to run. It is not until "eons later," when he is "worlds away," that his "running is all done," and he finds himself at his destination: another garden, just like the one he had left behind.</span>
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
