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Brrunno [24]
3 years ago
10

How does the electric force between two charged particles change if one

Physics
1 answer:
ExtremeBDS [4]3 years ago
7 0

Answer:

brainly.com/question/11848211

^ Similar/same question as yours!

The pressure between two costs is proportional to the product of the charges.

If solely one of the expenses is decreased by way of a element of 3, then the pressure is decreased with the aid of a thing of 3.

If each prices are reduced by a thing of 3, then the pressure is reduced via a issue of 9.

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Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred
madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=\dfrac{T_2}{T_1}\times P_1

P_2=\dfrac{600}{300}\times 2

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

m=\dfrac{PV}{RT}

m=\dfrac{200\times 3}{0.287\times 300}\ kg

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}

S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}

S₂ - S₁ = 3.42 KJ/K

 

5 0
3 years ago
The amount of light that enters the pupil is controlled by the:<br> retina.<br> lens.<br> inis.
nignag [31]

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

5 0
3 years ago
Define any two characteristics of sound​
Oliga [24]

Answer:

five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed

5 0
3 years ago
What can happen to the atomic particles when u rub two objects together ​
MAVERICK [17]
Look it up on google it has the answer
3 0
3 years ago
1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper
rewona [7]

Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W =  R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n =  34,53 mol

Finally, in (2):

W =  (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

4 0
3 years ago
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