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Troyanec [42]
3 years ago
15

The orbital period of Jupiter is 11.86 years. What is its distance from the sun?

Physics
1 answer:
julsineya [31]3 years ago
6 0
It's 483.8 million miles from the sun..
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Which statement best describes the size of the outer planets? All the outer planets are large. All the outer planets, except Sat
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Answer: a all the outer planets are large

Explanation: because i have the smarts

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3 years ago
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gold has a density of 19.32g/cm^3. aluminum has a density of 2.7g/cm^3. which is heavier: 1g of gold or 1 g of aluminum? explain
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Gold is much heavier because it has a greater density, meaning the atoms are more closely packed. Also, the individual atoms are heavy as well making 1g of gold heavier than 1g of aluminum.
6 0
3 years ago
Which steps can you take to help a person suffering from achohol abuse?
alexandr1967 [171]

Answer:

Call a treatment service. Seek out a healthcare professional. Lastly, report the issue to the local authorities.

Why did I choose these?

You call the treatment services to take care of the person.

You seek out a healthcare professional to get him/her back in shape.

Then you report the issue so that nobody will be endangered by the person that is abusing alcohol. Many accidents occur because of this and that is why you report it.

Explanation:

3 0
4 years ago
Which of these discoveries is generally attributed to kepler?
Ne4ueva [31]
The question is incomplete and no options are given but the answer is;
The orbit of Mars is an ellipse,<span> is generally attributed to Kepler.
Kepler's first law states that; "</span><span>The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.
</span>The planet at that point takes after and follow the ellipse in its orbit, which implies that the distance between Earth and Sun remove is continually changing as the planet goes around its orbit.
4 0
3 years ago
Two point charges of 60.0 C and -12.0 C are separated by a distance of 20.0 cm. A 7.00 C charge is placed midway between these t
iogann1982 [59]

Answer:

4.5\times 10^{14} N

Explanation:

q_1=60 C

q_2=-12 C

q_3=7 C

Distance between q1 and q2,d=20 cm

r=\frac{d}{2}=\frac{20}{2}=10 cm=10\times 10^{-2} m

1m =100 cm

Electric force on charge q3 due to charge q1

F_1=\frac{kq_1q_3}{r^2}=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2} (away from q1)

Electric force on charge q3 due to charge q2

F_2=\frac{kq_2q_3}{r^2}=\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}(attract toward q2)

Net force=F=F_1+F_2

F=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}+\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}

F=\frac{9\times 10^9\times 7}{(10\times 10^{-2})^2}(60+12)

F=4.5\times 10^{14} N

5 0
4 years ago
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