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Natasha2012 [34]
3 years ago
12

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa

l pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.
Physics
2 answers:
lianna [129]3 years ago
4 0

B

The earth is larger (approximately 6 times larger) than the moon hence it will have a stronger gravitational pull on the shuttle than the moon has on the shuttle. Because the shuttle is halfway between the moon and earth, the influence of distance in the gravity formula between the two is negated.

Explanation:

Gravitation force depends on distance and mass of objects in the universe. The bigger the mass of an object the higher its gravitational pull while the shorter the distance between the objects the higher the gravitaiton pull between the two. This phenomenon is described by the formulae below;

F = G (M1 – M2) / r2

Where;

F = gravitational force

G = Gravitational constant

M1 = mass of the first object

M2 = mas of the second object

R = distance between the two objects

Learn More:

For more on gravity check out;

brainly.com/question/9934704

brainly.com/question/13418722

#LearnWithBrainly

11111nata11111 [884]3 years ago
3 0
B. The mass of the earth is much greater than that of the moon, and therefore since it is the same distance between the moon and the earth the gravitational pull from the earth is stronger than that of the moon.
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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
Two homogeneous bodies of the same volume
lord [1]

Answer:

No, it is not necessary for them to have same mass.

Explanation:

Let both bodies have a density d1 and d2 respectively.

Since their volumes are equal V1 = V2

we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D

Hence, d1 =  and d2 =    

Taking the ratio of densities,we get

This implies that unless the bodies have same densities, the mass of the two bodies will not be same.

8 0
3 years ago
When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
STALIN [3.7K]

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

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Answer:

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7 0
2 years ago
The ms = 0.5645 and mk=0.1113 for a 30 N box on level ground. What level force is needed to start the box moving
nevsk [136]

Answer:

16.935 N

Explanation:

In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to

F_f = \mu_s (mg)

where

\mu_s = 0.5645 is the coefficient of static friction

(mg) = 30 N is the weight of the box

Therefore, the condition for F must be:

F \geq F_f\\F \geq \mu_k (mg)=(0.5645)(30 N)=16.935 N

So, the applied force must be greater than this value.

3 0
3 years ago
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