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AnnyKZ [126]
4 years ago
15

At a particular instant, a stationary observer on the ground sees a package falling with speed v1 at an angle to the vertical. t

o a pilot flying horizontally at constant speed relative to the ground, the package appears to be falling vertically with a speed v2 at that instant. what is the speed of the pilot relative to the ground?
Physics
1 answer:
Lerok [7]4 years ago
7 0
V1 * sin(θ) where θ is the angle v1 makes with the vertical.
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If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that
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Answer:

K_E=mgr_E

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

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\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}

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\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

F=mg=\frac{GM_Em}{r_E^2}

Which means that:

\frac{GM_Em}{r_E}=mgr_E

So finally putting all together we can write:

K_E=\frac{GM_Em}{r_E}=mgr_E

4 0
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