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melisa1 [442]
3 years ago
14

Planet Nine is speculated to be on average 20 times farther away from the Sun than Neptune (on average distance from the Sun). H

ow many miles from the Sun would this place Planet Nine on average? Divide 93 million miles into your result to get how many astronomical units (AU) Planet Nine would be on average away from the Sun.
Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer:

The distance is 55.636 billion miles, or 528.2 AU.

Explanation:

Since the distance from the Sun to Neptune is 2.7818 billion miles, the distance from the Sun to Planet Nine would be 20 times that, which is:

d=(20)(2781800000\ miles)=55636000000\ miles

or 55.636 billion miles.

Since 1 astronomical unit (AU) is 93 million miles, that distance is also:

d=(55636000000\ miles)(\frac{1AU}{93000000\ miles})=598.2\ AU

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. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

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3 years ago
in the diagram the solid eventually becomes a gas what will happen to bring the substance from a solid to a gas
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Since there is no diagram I can see to prove my answer, I must infer that a heat source should be applied to the solid substance and cause the atoms in the solid to spread further apart and become a different state, in which atoms will be able to freely move, gas.
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______________friction is the type of kinetic friction that is difficult to overcome.
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Explanation:

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A drawback to using nuclear fission is that it...
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D. creates radioactive waste.

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8 0
1 year ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
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