The correct answer to the question is- C).Electrons are ripped off one material and held tightly by the other material.
EXPLANATION:
Before going to answer this question, first we have to understand the charging by friction.
There are three modes of charging a body, which are known as friction,conduction, and induction.
Friction: Friction is the type of charging a body due to the actual transfer of electrons when two substances of different electron affinities are rubbed with each other. During rubbing, the electrons from less electron affinity substance will be transferred to high affinity substance.
The substance which will accept electrons gets negatively charged, and the other substance will be positively charged.
There will be no transfer of protons from one substance to another as protons are bound inside the nuclei of atoms.
Hence, the correct answer to the question is that electrons are ripped off one material and held tightly by the other material.
its c because The microscopically colliding particles, that include molecules, atoms and electrons, transfer disorganized microscopic kinetic and potential energy, jointly known as internal energy.
<span>So we want to know which object has a gravitational force of the greatest magnitude. Since gravitational force depends on the mass of the object, the object with the greatest mass will have the greatest magnitude of the gravitational force. In this case that is the bowling ball. </span>
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
Oxygen... Hope this Warner helps
