Answer:
Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.
Explanation:
<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
Explanation:
Given that,
Charge 1, 
Charge 2, 
The distance between charges, r = 1.99 m
To find,
The electrostatic force and its nature
Solution,
(a) The electric force between two charges is given by :



(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.
Therefore, this is the required solution.