The charge present determines a force to be attractive or repulsive.
The charges acquired by two bodies determines the Force as Attractive Or Repulsive.
Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.
Examples of the experiments and observations:
- On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.
This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.
- In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.
The more the distance between the charges, the less is the Electric Force.
The lesser the distance between the charges, the more is the Electric Force.
If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.
Hence, the charge present determines a force to be attractive or repulsive.
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Answer:
F₁ = 4 F₀
Explanation:
The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:
F₀ = mv₀²/r --------------- equation (1)
where,
F₀ = Force on string at t₀
m = mass of ball
v₀ = speed of ball at t₀
r = radius of circular path
Now, at time t₁:
v₁ = 2v₀
F₁ = mv₁²/r
F₁ = m(2v₀)²/r
F₁ = 4 mv₀²/r
using equation (1):
<u>F₁ = 4 F₀</u>
<span>In general the larger the pore space (the higher the porosity) the easier it is</span>
Answer: +2 e
Explanation: Net charge of a system is sum of charges on all the constituents of the system.
The system contains two spheres. Charge on one sphere is:
q₁ = +6 e
Charge on the second sphere is:
q₂ = -4 e
Net charge of the system is:
q = q₁ + q₂
q = +6 e + (-4 e) = + 2 e
Thus, the net charge of system is +2 e.
Answer:
-5 m/s^2
Explanation:
10/2=5 and is is a stop so it is negitive
