Answer:
The charge in each ball will be 3 * 10^-12 C
Explanation:
(Assuming the correct charge of the second ball is 8 * 10^-12)
When the balls are brought in contact, all the charges are split evenly among then.
So first we need to find the total charge combined:
(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C
Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:
(9 * 10^-12) / 3 = 3 * 10^-12 C
So the charge in each ball will be 3 * 10^-12 C
D. Mineral and fossil matches from tests done on different continents.
Answer:
3.38 m/s
Explanation:
Mass of child = m₁ = 25
Initial speed of child = u₁ = 5 m/s
Initial speed of cart = u₂ = 0 m/s
Mass of cart = m₂ = 12 kg
Velocity of cart with child on top = v
This is a case of perfectly inelastic collision

Velocity of cart with child on top is 3.38 m/s
The correct answer to the question is : Electric energy
EXPLANATION :
As per the question, we have an electric generator.
Before coming into any conclusion, first we have to understand the function of generator.
The generator is attached to the turbine. When the turbine rotates, the generator also starts rotating with it. Thanks to electromagnetic induction, the electricity is produced in the coil attached to the generator when it rotates.
Hence, from above, it is obvious that kinetic energy is converted into electric energy.
Answer:

Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

The friction forces are computed by


Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]

Simplifying

Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)
