Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where is the force of attraction or repulsion
is Coulumb's constant=
and are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,
Question6:
Given: q1=, r=5 m, F=
therefore by Coulumb's law,
⇒
Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Sorry!
This cannot be answered. We don't have weight, height, etc.
Electricity is a form of energy. Electricity is the flow of electrons. All matter is made up of atoms, and an atom has a center, called a nucleus. The negative charge of an electron is equal to the positive charge of a proton, and the number of electrons in an atom is usually equal to the number of protons. Hope this helped. ;)
Answer:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2