Answer:
The distance of separation is decreased
Explanation:
From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.
Answer:
the propagation velocity of the wave is 274.2 m/s
Explanation:
Given;
length of the string, L = 1.5 m
mass of the string, m = 0.002 kg
Tension of the string, T = 100 N
wavelength, λ = 1.5 m
The propagation velocity of the wave is calculated as;
Therefore, the propagation velocity of the wave is 274.2 m/s
Answer:
C
Explanation:
The change in momentum of x has to be the opposite of the change in momentum of Y because the momentum is just transferred from one to another. But I'm still trying to figure it out how to calculate.
Explanation:
a. Net force is mass times acceleration (Newton's second law).
∑F = ma
∑F = (5.0 kg) (2.0 m/s²)
∑F = 10 N
b. The net force is the sum of the individual forces.
10 N = F − 5 N
F = 15 N
c. Friction force here is mgμ.
mgμ = 5 N
(5.0 kg) (10 m/s) μ = 5 N
μ = 0.1
Answer:
D: Increase the distance between the objects.
E: Decrease the mass of one of the objects.
