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Citrus2011 [14]
3 years ago
7

What is your mass in kilograms if you weigh 120 pounds

Physics
1 answer:
schepotkina [342]3 years ago
6 0

It depends where you are.

-- If you weigh 120 pounds on the Moon,
then your mass is  329.1 kilograms.

-- If you weigh 120 pounds on Mars,
 then your mass is  143.8 kilograms.

-- If you weigh 120  pounds on the Earth,
then your mass is  54.4 kilograms.

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qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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Korvikt [17]

51.448 g is the required answer!

8 0
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g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th
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Answer:

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Explanation:

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Now, we know that in circular motion,

v² = rg•tanθ

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Thus,

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Answer:

0.749 m

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