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kirill [66]
3 years ago
5

Jupiter is 318 times more massive than Earth. Based upon mass alone (i.e, if you were to place Jupiter at the same distance from

the Sun as Earth), what would be the gravitational force between Jupiter and the Sun in units of "Earth-pulls?"
Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

318 Earth-pull

Explanation:

Using Newton's law of gravitational force

F = GMm / r²

F is directly proportional to mass

let earth  = m

then m .......... Earth-pull

        318 m = 318 m × Earth-pull / m = 318 Earth-pull

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When searching for your word processing file to finish writing your report, you should look for a file with which extension?
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Answer:

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Explanation:

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3 years ago
Karla Ayala pulls a sled on an icy road (dangerous!). Because of Karla's pull, the tension force is 151 N, and the rope makes a
skelet666 [1.2K]

Answer:

W = 1418.9 J = 1.418 KJ

Explanation:

In order to find the work done by the pull force applied by Karla, we need to can use the formula of work done. This formula tells us that work done on a body is the product of the distance covered by the object with the component of force applied in the direction of that displacement:

W = F.d

W = Fd Cosθ

where,

W = Work Done = ?

F = Force = 151 N

d = distance covered = 10 m

θ = Angle with horizontal = 20°

Therefore,

W = (151 N)(10 m) Cos 20°

<u>W = 1418.9 J = 1.418 KJ</u>

6 0
3 years ago
Which of the following statement(s) is
alexgriva [62]

Answer:

Explanation:

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3 years ago
Help!! If an atom has 7 protons and 8 neutrons, and is not in an excited state, how many electrons would Bohr say the atom has?
quester [9]
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8 0
3 years ago
Read 2 more answers
A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp
Sergeu [11.5K]

Answer:

U=1.29\times 10^{-7}\ J

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm  ,K₁= 4.7

d₂=2.1 mm  , K₂=2.6

We know that capacitance given as

C_1=\dfrac{K_1\varepsilon _oA}{d_1}

C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_1=1.26\times 10^{-10}\ F

C_2=\dfrac{K_2\varepsilon _oA}{d_2}

C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}

C_2=0.701\times 10^{-10}\ F

Net capacitance

C=\dfrac{C_1C_2}{C_1+C_2}

C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F

C=4.5\times 10^{-11}\ F

We know that stored energy given as

U=\dfrac{CV^2}{2}

V= 76 V

U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J

U=1.29\times 10^{-7}\ J

3 0
3 years ago
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