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3 years ago

Every choice has a positive or negative consequence. a. True b. False

Physics

2 answers:

pickupchik [31]3 years ago

7 0

This is A, true, because no matter what you do, it will have a reaction that happens.

vitfil [10]3 years ago

3 0

The answer is A. True

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PLEASE HURRYYY

babymother [125]

your answer is make up artist

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4 years ago

Leo la conclusión e identificó dos situaciones que causan que una lengua desaparezca buscó otra situación que también incide al

Novosadov [1.4K]

Answer:

yes smh

Explanation:

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3 years ago

I had gotten this one wrong please help thank you

Nata [24]

Check the 1st 2nd 3rd and 4th boxes

4 0

4 years ago

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

Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an lrc-series circuit is l d 2q dt 2

Aloiza [94]

DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150

5 0

3 years ago