Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.
I am very sorry I don’t know
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
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Force required to pull one end at a constant speed</h3>
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;
F = ma
where;
- m is mass
- a is acceleration
At a constant speed, the acceleration of the object will be zero.
F = m x 0
F = 0
Thus, the force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
Learn more about constant speed here: brainly.com/question/2681210
Explanation :
Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.
(1) This force is exhibited by all atoms and molecules.
(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.
(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.
(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.
So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".
Explanation:
The acceleration due to gravity g is defined as
and solving for R, we find that
We need the mass M of the planet first and we can do that by noting that the centripetal acceleration 
experienced by the satellite is equal to the gravitational force 
or
The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as
where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as
Solving for <em>M</em>, we get
Putting this expression back into Eqn(1), we get
