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alisha [4.7K]
3 years ago
12

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

Physics
2 answers:
kirill [66]3 years ago
6 0

Answer:

decreasing the distance of the space shuttle from Earth

increasing the universal gravitational constant acting on the space shuttle*

Explanation:

The magnitude of the gravitational force acting between the Earth and the shuttle is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the mass of the Earth

m is the mass of the shuttle

r is the distance between the shuttle and the Earth's center

By looking at the formula, we see check the effect of each statement given:

- increasing the distance of the space shuttle from Earth  --> FALSE: this means increasing r in the formula, so the gravitational force would decrease

- decreasing the distance of the space shuttle from Earth  --> TRUE: this means decreasing r in the formula, so the gravitational force would increase

- increasing the universal gravitational constant acting on the space shuttle  --> TRUE: this means increasing G in the formula, so the gravitational force would increase. *However, technically is not possible to change the value of G, since it is a universal constant and it has the same value everywhere.

- decreasing the universal gravitational constant acting on the space shuttle: FALSE --> Decreasing G in the formula would decrease the magnitude of the force.

maks197457 [2]3 years ago
5 0
"decreasing the distance of the space shuttle from Earth" 

F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases. 

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The object will feel minimum force of gravity at the
Alika [10]

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A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

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2 years ago
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abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
A ball is projected horizontally from the top of a hill with a velocity of 30m/s if it reaches the ground 5sec later the height
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Answer:

The hill is a total of 150 meters tall

Explanation:

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2 years ago
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