For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
Failed experiments, uncontrolled variables, invalid data, and generalized human error
