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MAVERICK [17]
3 years ago
6

ILL ADD BRAINLIEST

Physics
1 answer:
Marina86 [1]3 years ago
6 0

This is kinda confusing. I wish u just to a screenshot of the problem but here goes...

Forest at highest latitudes- Hardwood trees/deer, squirrel, foxes

Praries/temperate climate- Mostly small mammals/scrubs/steppes

High humidity/rainfall near equator- Abundant thick vegatation/manny species

No trees/ polar bears/ mosses- 25cm rain/few animals

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Two 10-N weights are hanging from two Springs A and B. Spring A has a greater spring constant and therefore will ____. a. have a
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During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

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Answer:

there isnt a pic

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