A 0.0600-kilogram ball traveling at 60.0 meters

1 answer:

7 0

The momentum of ball is given by:

$\Delta Q=mv \\ \Delta Q=6*10^{-2}*60 \\ \Delta Q=3.6kg*m/s$

Since both have the same momentum, we have:

$\Delta Q=MV \\ 3.6=10^{-2}V \\ \boxed {V=360m/s}$

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

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Given that,

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Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is: