Answer:
The extension of the spring is 0.392 m.
Explanation:
Given;
spring constant, k = 50 N/m
mass attached to the spring, m = 2.0 kg
let the extension of the spring = x
The extension of the spring is calculated by applying Hook's law;
F = kx
mg = kx
Therefore, the extension of the spring is 0.392 m.
Answer:
a = 1,008 10⁻³ m / s²
Explanation:
For this exercise, let's use the kinematic relations of accelerated motion
v² = v₀² - 2 a x
The negative sign is because the acceleration is opposite to the speed, the final speed is zero
0 = v₀² - 2 a x
a = v₀² / 2x
Let's reduce the magnitudes to the SI system
x = 2.4mm (1m / 10³mm) = 2.4 10⁻³m
Let's calculate
a = 2.2²/2 2.4 10⁻³
a = 1,008 10⁻³ m / s²
According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the electrostatic force will become weak by one fourth of the original force.
Twenty is the atomic number of potassium.
