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3 years ago

DUE TODAY PLEASE HELP.....WILL GIVE BRANIST!!!!!!!

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

7 0

girl how someone supposed to help you with that?!

Shtirlitz [24]3 years ago

5 0

Answer:

use water in the ocean getting precipitated and going up then make it fall off a mountain then make it freeze becuase it got stuck on the mountian

thermal energy happend when the water got precipitated

I hope it helps a bit

Sorry if there is mistakes I'm tryna help

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A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 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

Light travelling in one material enters another material in which it travels faster. The

KiRa [710]

Answer:

Explanation:

It light wave will travel at speed of light and go faster in its wavelength

3 0

3 years ago

How could you group pennies,nickels, and dimes together

julsineya [31]

There isn't a direct answer to this, but I can show you!

This is how you group Pennies, Nickels, and Dimes.

1 Penny= 1 cent.

1 Nickel= 5 cents.

1 Dime= 10 cents.

Imagine if there was 5 pennies, 3 dimes, and 8 Nickels.

That might seem like a lot of numbers, but it's pretty simple!

You know how pennies are worth 1 cent? If we had 5 pennies, that would be 5 cents.

1+1+1+1+1=5 cents.

Then, we have 3 dimes. Dimes are each 10 cents.

So, 10x3 (If you know multiplication, if not, you will learn it later) :)

10+10+10= 30 cents.

Now, nickels. We have 8 nickels! Remember, each nickel is 5 cents.

5+5+5+5+5+5+5+5=

40 Nickels.

Now add all of the coins up.

5 cents + 30 cents + 40 cents = 75 cents!

I hope this helped! If you need more help, you can ask me. :)

5 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC Q₃ = 51 nC Q₂ = 30.5 nC

x₁ = - 1.715m x₃ = - 1.085m x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0

3 years ago