Answer:
Explanation:
Given
1 mole of perfect, monoatomic gas
initial Temperature


Work done in iso-thermal process
=initial pressure
=Final Pressure

Since it is a iso-thermal process therefore q=w
Therefore q=39.64 J
(b)if the gas expands by the same amount again isotherm-ally and irreversibly
work done is





<u>Answer:</u>
The power and voltage are related through Power Rule
.
<u>
Explanation :
</u>
Power Rule states that the current I that flows the element in the circuit with a loss in voltage V, then the amount of power dissipated by that element in the circuit is the multiplicative product of voltage and the current.
Mathematically, power law is

The rate of work done or the rate of energy consumption or production is termed as power.Unit of power is denoted as W(watts).
Potential energy between any two points on a circuit is called as Voltage and is measured in volts (V)
.
So, the speed of the ball after 2 seconds after free fall is <u>20 m/s</u>.
<h3>Introduction</h3>
Hi ! I'm Deva from Brainly Indonesia. In this material, we can call this event "Free Fall Motion". There are two conditions for free fall motion, namely falling (from top to bottom) and free (without initial velocity). Because the question only asks for the final velocity of the ball, in fact, we may use the formula for the relationship between acceleration and change in velocity and time. In general, this relationship can be expressed in the following equation :

With the following conditions :
- a = acceleration (m/s²)
= speed after some time (m/s)
= initial speed (m/s)- t = interval of time (s)
<h3>Problem Solving</h3>
We know that :
- a = acceleration = 9,8 m/s² >> because the acceleration of a falling object is following the acceleration of gravity (g).
= initial speed = 0 m/s >> the keyword is free fall- t = interval of time = 2 s
What was asked :
= speed after some time = ... m/s
Step by step :




So, the speed of the ball after 2 seconds after free fall is 20 m/s.
Answer:True
Explanation:Because the deflection in galvanometer is calculated by the following formula;