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ivolga24 [154]
3 years ago
15

Help now

Physics
1 answer:
sergiy2304 [10]3 years ago
5 0

Answer:

The two neighboring planets with the greatest distance between them is <em><u>Neptune and Uranus</u></em>. This happens when they are at the opposite sides of the Sun at approximately. They are 4500 million miles away from each other.

If by adjacent you mean next to each other, in terms of the order of planets in our Solar System, then the closest would be <u><em>Earth and Venus</em></u>; which are separated by about 38,000,000 kilometres when in opposition.

I am not sure about the general patterns..

<u><em>Venus has the longest day</em></u> of any planet in our solar system. It completes one rotation every 243 Earth days.

A year on <u><em>Neptune</em></u> is the longest of any planet, lasting the equivalent of 164.8 years (or 60,182 Earth days). It is obvious because it is also the farthest with the longest orbit.

Outer planets are less dense than the inner planets because they posses a big volume but contains not much mass. Try to imagine 2 rocks, one is big, but is hollow or have a lot of holes; and the other is small, but solid with no holes.

Happy learning!!

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If you push a cork into the neck of a bottle filled with air, what happens to the pressure inside the bottle?
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The pressure inside the bottle will build. When you push the cork inside of the bottle the air inside has less space to work with, so it bunches up, making the pressure build up as well.

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3 years ago
Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle
Vinil7 [7]

Answer:

1/9 of that just outside the smaller sphere

Explanation:

The electric field strength produced by a charged sphere outside the sphere itself is equal to that produced by a single point charge:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Calling R the radius of the first sphere, the electric field just outide the surface of the first sphere is

E_0=k\frac{Q}{R^2}

The second sphere has a radius which is 3 times that of the smaller sphere:

R'=3R

So, the electric field just outside the second sphere is

E'=k\frac{Q}{R'^2}=k\frac{Q}{(3R)^2}=\frac{1}{9}(k\frac{Q}{R^2})=\frac{E_0}{9}

So, the correct answer is 1/9.

7 0
3 years ago
A pitcher throws a 0.143-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just
o-na [289]

Answer:

200N appox

Explanation:

Step one:

Given data

mass of ball= 0.143kg

initial velocity of ball u= 42m/s

final velocity of the ball= 49m/s

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Step two

From the relation Ft=mΔv

we can find the average force as

F=mΔv/t

substitute

F=0.143*(49-42)/0.005

F=0.143*7/0.005

F=1.001/0.005

F=200.2

F= 200N appox

The average force is  200N

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3 years ago
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