Compare and contrast electrical and magnetic forces
1 answer:
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Answer:
Approximately , assuming that the volume of these two charged objects is negligible.
Explanation:
Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.
Let and
denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question,
and
.
Let denote the distance between these two point charges. In this question,
.
Let denote the Coulomb constant. In standard units,
.
By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:
.
Substitute in the values and evaluate:
.
Answer:
V = 11.83 m/s
Explanation:
Given the following data;
Mass = 2000 kg
Force = 10000N
Distance = 14 m
To find the final velocity of the car;
First of all, we would determine the acceleration of the car;
Acceleration = force/mass
Acceleration = 10000/2000
Acceleration = 5 m/s²
Next, we would use the third equation of motion to find the final velocity;
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
V² = 0² + 2*5*14
V² = 0 + 140
V = √140
V = 11.83 m/s