Answer:
True
Explanation:
If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell
According to Gauss law
∅ = EA =-Q/∈₀
Where ∅ is the electric flux through the gaussian surface and E is the electric field strength
If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero
The force needed to overcome sliding friction is more than the force needed to overcome rolling friction or static or even fluid
Answer:
b is the answer tennis ball
Volume=mass/density
volume=455.6/19.3
volume=23.6 mL
