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2 years ago

An adiabatic expansion refers to the fact that:

Physics

1 answer:

Vladimir [108]2 years ago

8 0

The fact that no heat is transferred between a system and its surroundings refers to an adiabatic free expansion.

In gas, the adiabatic compression creates an increase in the temperature while it causes a drop in temperature there is adiabatic expansion against pressure, or a spring.

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Before being allowed to expand in vacuum, the gas is first contained in an insulated container.</span>

The work made by or on the system is zero since there is no external pressure for the gas to enlarge on against. The first law of thermodynamics then insinuates that zero is the net internal energy change of the system since the process does not incorporate any work or heat transfer.

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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

The concept of critical density is just like the idea of escape velocity for a projectile launched from the earth. An object lau

ikadub [295]

Answer:

v = √ 2 G M/ $R_{e}$

Explanation:

To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.

Initial

Em₀ = K + U₀

Final

$Em_{f}$ = $U_{f}$

The kinetic energy is k = ½ m v²

The gravitational potential energy is U = - G m M / r

r is the distance measured from the center of the Earth

How energy is conserved

Em₀ = $Em_{f}$

½ mv² - GmM / $R_{e}$ = -GmM / r

v² = 2 G M (1 / $R_{e}$ – 1 / r)

v = √ 2GM (1 / $R_{e}$ – 1 / r)

The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0

v = √ 2GM / $R_{e}$

5 0

2 years ago

A SMART car can accelerate from rest to a speed of 28 m/s in 20s. What distance does it travel in this time?

sashaice [31]

Speed= distance/ time so distance = speed * time= 28 * 20= 560m .. so the answer is 560m.. l hope it helped :)

6 0

2 years ago

The different shapes of the moon seen from Earth are called

Anastaziya [24]

Answer:

phases of moon

Explanation:

.................

3 0

2 years ago

Read 2 more answers

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it

jek_recluse [69]

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

$\frac{dI}{d \theta}$ = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0

2 years ago