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
Answer:
D.
R increases
V is constant
I decreases
Explanation:
The resistance of a wire is given by the following formula:
It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, <u>the resistance of circuit increases</u>.
The <u>voltage in the circuit will be constant</u> as the voltage source remains same and it is not changed.
Now, we can use Ohm Law:
V = IR
at constant V:
I ∝ 1/R
it means that current is inversely proportional to resistance. Hence, the increase of resistance causes <u>the current in circuit to decrease.</u>
Therefore, the correct option will be:
<u>D.</u>
<u>R increases
</u>
<u>V is constant
</u>
<u>I decreases</u>
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.
To calculate acceleration, divide your weight by the mass, thus the accleration is:
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