Answer:
1.686 m
Explanation:
From coulomb's law,
F = kq1q2/r² ...................................... Equation 1
Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.
making r the subject of the equation,
r = √(kq1q2/F).......................... Equation 2
Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C
Constant: k = 9.0×10⁹ Nm²/C².
Substituting into equation 2
r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)
r = √(14364×10⁻³/5.05)
r = √(14.364/5.05)
r = √2.844
r = 1.686 m
r = 1.686 m.
Thus the distance must be 1.686 m
<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />
According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.
Now, in this case we are on Earth, so the gravity value is 
Note the objects experience the acceleration of gravity regardless of their mass.
Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:
If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at and hit the ground at approximately the same time.
Answer:
1.082 mm
Explanation:
From the question, we can see that we were given The following
Wavelength of the atoms, λ = 502 nm = 502*10^-9 m
Radius of the screen away from the double slit, r = 1.1 m
We know that Y(20) = 10.2 mm = 10.2*10^-3 m
d = (20 * R * λ) / Y(20)
d = (20 * 1.1 * 502*10^-9)/10.2*10^-3
d = 1.1*10^-5 / 10.2*10^-3
d = 1.082 mm
Therefore, we can say that the distance of separation between the two slits is 1.082 mm
Any change in speed or direction of motion is acceleration.
Constant acceleration can mean ...
-- speeding up at a constant rate . . . gaining the same amount
of speed each second.
-- slowing down at a constant rate . . . losing the same amount
of speed each second.
-- changing direction at a constant rate . . . for example, going
around a circular path at a constant speed.
