Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
The best use of an atomic model to explain the charge of the particles in Thomson's beams is:
<u>An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.</u>
<u>Explanation:</u>
In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.
Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:
-
atoms are spheres of positive charge
- electrons are dotted around inside
Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.
Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y
Explanation:
According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
(1)
Where;
is the Gravitational Constant
is the mass of the Earth
is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
So for satellite X, the orbital period
is:
(2)
Where 
(3)
(4)
For satellite Y, the orbital period
is:
(5)
Where 
(6)
(7)
This means 
Now let's calculate the tangential speed for both satellites:
<u>For Satellite X:</u>
(8)
(9)
<u>For Satellite Y:</u>
(10)
(11)
This means 
Therefore:
Satellite X has a greater period and a slower tangential speed than Satellite Y
Answer:
well...
Explanation:
Acoording to my brain i think that the answer would be C