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mezya [45]
3 years ago
11

Can someone plz do the dimensional analysis for 32ft/6.45 sec to meters/min? Thank You!

Mathematics
1 answer:
Rzqust [24]3 years ago
5 0

Answer:

About 90.73 meters/min.

Step-by-step explanation:

Dimensional analysis is done by multiplying the wanted unit over the current unit in equal measure, or the other way around.

First, we will convert feet to meters, using that 1 ft = .4048 m.

32ft/6.45sec * .3048m/1ft = 32/6.45sec * .3048m = 9.7536m/6.45 sec.

Second, we convert seconds to minutes, using that 60 sec = 1 min.

9.7536m/6.45sec * 60sec/1min = 9.7536*60m/6.45min ≈ 90.7311627m/min.

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NEED HELP ASAP!!
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Answer:

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Step-by-step explanation:

The gravitational pull of the Earth on a person or object is given by Newton's law of gravitation as follows;

F =G\times \dfrac{M \cdot m}{r^{2}}

Where;

G = The universal gravitational constant

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m = The mass of the other object

r = The distance between the centers of the two objects

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The ratio of the gravitational pull on the surface of the Earth, F₁, and the gravitational pull on an astronaut at the international space station, F₂, is therefore given as follows;

\dfrac{F_1}{F_2} = \dfrac{ \dfrac{M \cdot m}{R^{2}}}{\dfrac{M \cdot m}{r^{2}}} = \dfrac{r^2}{R^2}  = \dfrac{(6,800 \ km)^2}{(6,371 \ km)^2} \approx  1.14

∴ F₁ ≈ 1.14 × F₂

F₂ ≈ 0.8778 × F₁

Therefore, the gravitational pull on the astronaut by virtue of the distance from the center of the Earth, F₂ is approximately 88% of the gravitational pull on a person of similar mass on Earth

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Therefore, Cory is correct, the astronauts in the International Space Station, 6,800 km from the Earth's center, are not too far to experience gravity.

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Answer:

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