Question: The force between a pair of 0.005 C is 750 N. What is the distance between them?
Answer:
17.32 m
Explanation:
From coulomb's Law,
F = kqq'/r²........................... Equation 1
Where F = Force between the force, q' and q = both charges respectively, k = coulomb's constant, r = distance between both charges.
make r the subject of the equation above
r = √(kqq'/F)..................... Equation 2
From the question,
Given: q = q' = 0.005 C, F = 750 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute these values into equation 2
r = √(9.0×10⁹×0.005×0.005/750)
r = √(300)
r = 17.32 m.
Hence the distance between the pair of charges = 17.32 m
Answer:
There are four main ways of doing that :-
- Velocity
- Acceleration
- Momentum
- Kinetic energy
Hope it helps!
Answer:
405 m
Explanation:
Given:
v₀ = 0 m/s
a = 2.5 m/s²
v = 45 m/s
Find: Δx
v² = v₀² + 2aΔx
(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx
Δx = 405 m
Answer:
The equal and opposite force is equal to the sum of any frictional force to the product of the mass of the wagon and the acceleration of the wagon
Explanation:
Newton's third law states that for every action, or force, there is an equal and opposite reaction
The equal and opposite force = The mass of the Wagon, m × The acceleration of the wagon, a + Any frictional force present, 
Whereby the force with which the horse pulls the wagon = F, mathematically, we have;
The equal and opposite force = -F = m × a + .
