Answer:
B
Explanation:
Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.
A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.
C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.
D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.
You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.
<h3>What are some uses of fixture wires?</h3>
Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG
In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.
To know more about fixture wires, refer
brainly.com/question/26098282
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Answer:
Explanation:
At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume
ΔV = 9.2 - 5.6 = 3.6 L
3.6 L = 3.6 x 10⁻³ m³
ΔV = 3.6 x 10⁻³ m³
P = 3.7 x 10³ Pa
So work done
= 3.7 x 10³ x 3.6 x 10⁻³ J
= 13.32 J .
( c ) is the answer , because work is done by the gas so it will be positive.
Answer:
T = 76.39°C
Explanation:
given,
coffee cup temperature = 95°C
Room temperature= 20°C
expression

temperature at t = 0

T(0) = 95°C
temperature after half hour of cooling

t = 30 minutes


T(30) = 61.16° C
average of first half hour will be equal to

![T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820t%20-%20%5Cdfrac%7B75%20e%5E%7B%5Cdfrac%7B-t%7D%7B50%7D%7D%7D%7B%5Cdfrac%7B1%7D%7B50%7D%7D%29%5D_0%5E30)
![T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820t%20-%203750e%5E%7B%5Cdfrac%7B-t%7D%7B50%7D%7D%5D_0%5E30)
![T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820%5Ctimes%2030%20-%203750%20e%5E%7B%5Cdfrac%7B-30%7D%7B50%7D%7D%20%2B%203750%5D)
![T = \dfrac{1}{30}[600 - 2058.04 + 3750]](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B600%20-%202058.04%20%2B%203750%5D)
T = 76.39°C
The efficiency of an ideal Carnot heat engine can be written as:

where

is the temperature of the cold region

is the temperature of the hot region
For the engine in our problem, we have

and

, so the efficiency is