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3 years ago

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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Using a skillet on the stove to cook scrambled eggs is an example of which heat transfer?

MatroZZZ [7]

Answer: conduction.

Explanation:

1) A skillet or pan will be in direct contact with the food.

2) Being the skillet a solid matter, it transfers heat by conduction.

3) There are three heat transfer mechanisms: convection, conduction, radiation.

i) Convection: only happens in fluids: liquids and gases.

It is a slow mechanism and is due to the motion of the particles of fluid.

This is not the case, since the skillet is solid and the eggs are in direct contact with it. If the eggs were separated from the skillet convection might happen but it would be very slow and the eggs would take so long time to be cooked.

ii) Conduction: happens by direct contact among neighboring particles, due to their vibrations. The vibrations are only transmitted to neighboring particles, which is only possibble when they are in direct contact, so it is the classical mechanism in solids. So, the heat transfer using the skillet to cook scrambles eggs is an example of conduction.

iii) Radiation: is the transfer of heat by infrared waves. It is an important mechanism when the substances are separated by vaccum or gases and the difference of temperature is high. Yet, this is not the importante mechanism here. You feel heat radiation when you put your hand close to the skillet.

4 0

3 years ago

Read 2 more answers

Where is the planet moving faster?

valentinak56 [21]

Answer:

I looked it up but got perihelion so I don't know if that will help at all but try um.... I don't know try L

Explanation:

7 0

2 years ago

What amount of force is needed to propel and object of 27 kg to an acceleration of 11,550 m/s^2? (1 point)

aliya0001 [1]

Answer:

311,850 N

Explanation:

We can solve the problem by using Newton's second law:

$F=ma$

where

F is the net force applied on an object

m is the mass of the object

a is its acceleration

For the object in this problem,

m = 27 kg

$a=11550 m/s^2$

Substituting, we find the force required:

$F=(27)(11550)=311,850 N$

3 0

2 years ago

An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface char

jarptica [38.1K]

Yes that is the correct answer

8 0

3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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3 years ago