Mass number of 43 and 21 electrons what is this atom

Which measurement is a potential difference? A.115J B.115V C.115N D.115C

borishaifa [10]

Answer:

B

Explanation:

Potential difference has a SI Unit of Volt and its symbol is V. Hence answer is B.

A is wrong as it has the unit Joule (J) which is the SI unit for energy.

C is wrong as it has the unit Newton (N) which is the SI unit for force.

D is wrong as it has the unit Coulomb (C) which is the SI unit of charge.

5 0

3 years ago

In some cases fixture wires may be used for

zalisa [80]

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

#SPJ4

3 0

1 year ago

PLEASE HELP!

Anna [14]

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.

5 0

2 years ago

Read 2 more answers

If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo

lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

temperature at t = 0

$T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}$

T(0) = 95°C

temperature after half hour of cooling

$T( t ) = 20 + 75 e^{\dfrac{-t}{50}}$

t = 30 minutes

$T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}$

$T( 30 ) = 20 + 75 \times 0.5488$

T(30) = 61.16° C

average of first half hour will be equal to

$T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt$

$T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30$

$T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30$

$T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]$

$T = \dfrac{1}{30}[600 - 2058.04 + 3750]$

T = 76.39°C

4 0

3 years ago

What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300

Vinvika [58]

The efficiency of an ideal Carnot heat engine can be written as:
$\eta = 1- \frac{T_{cold}}{T_{hot}}$
where
$T_{cold}$ is the temperature of the cold region
$T_{hot}$ is the temperature of the hot region

For the engine in our problem, we have $T_{cold}=300 K$ and $T_{hot}=400 K$ , so the efficiency is
$\eta= 1 - \frac{300 K}{400 K}=0.25$

4 0

2 years ago