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Nonamiya [84]
3 years ago
7

1.- Un balín de acero a 20 C tiene un volumen de 0.004 m3. ¿Cuál es la dilatación que sufre cuando su temperatura aumenta a 50 C

?
Physics
1 answer:
Brrunno [24]3 years ago
0 0

Answer:

\Delta V = 1.440\times 10^{-6}\,m^{3}

Explanation:

El coeficiente de expansión volumétrica se calcula a partir de la siguiente ecuación diferencial parcial (The volumetric expansion coefficient is computed by means of the following partial differential equation):

\alpha = \frac{1}{V} \cdot \left(\frac{\partial V}{\partial T} \right)

Se integra la fórmula a continuación (The formula is integrated herein):

\alpha\, dT = \frac{dV}{V}

Supóngase que el coeficiente es constante (Let suppose that coefficient is constant):

\alpha \int\limits^{T_{f}}_{T_{o}}\,dT = \int\limits^{V_{f}}_{V_{o}}\, \frac{dV}{V}

\alpha \cdot (T_{f}-T_{o}) = \ln \frac{V_{f}}{V_{o}}

El volumen final es (The final volume is):

V_{f} = V_{o}\cdot e^{\alpha \cdot (T_{f}-T_{o})}

El coeficiente de expansión volumétrica del acero es 12\times 10^{-6}\,^{\circ}C^{-1} (The volumetric expansion coefficient of steel is 12\times 10^{-6}\,^{\circ}C^{-1}):

V_{f} = (0.004\,m^{3})\cdot e^{(12\times 10^{-6}\,^{\circ}C^{-1})\cdot (50^{\circ}C-20^{\circ}C)}

V_{f} \approx 4.001\times 10^{-3}\,m^{3}

Finalmente, la dilatación experimentada por el balín es (Lastly, the dillatation experimented by the pellet is):

\Delta V = 4.001\times 10^{-3}\,m^{3} - 4.000\times 10^{-3}\,m^{3}

\Delta V = 1.440\times 10^{-6}\,m^{3}

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
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Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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3 years ago
A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having
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Answer:

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

           

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

         F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

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