Answer:
Short circuit
Explanation:
The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.
The wire across the bulb is connected from one terminal to another without any resistance in between them.
So, the correct option is (d) " short circuit ".
Answer: The silver pellet will release heat
Explanation:
Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.
Hahahahhahaha A ROSTER NEVER LAYS THE EGGS IT IS THE HEN THAT LAYS THE EGGS.....ITS LIKE MEN GETTING PREGNANT LOL
To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore


Here,
m = mass
= Velocity (Final and initial)
First case) When the particle goes from 10m/s to 20m/s



Second case) When the particle goes from 20m/s to 30m/s



As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.