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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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Answer:

4.16 L

Explanation:

Assuming constant temperature,

At the edge of Typhoon Odessa: P₁ = 1 atm = 1013.3 mbar,

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At the center of Typhoon Odessa: P₂ = (1013.3 - 40.0) mbar = 973.3 mbar

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For a fixed amount of gas at constant temperature (Boyle's law) : P₁V₁ = P₂V₂

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V₂= (4.0) × (1013.3/973.3)

V₂= 4.16 L

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By using an electric field, it is feasible to differentiate between these different forms of radiation.

<h3>What is a radioactive source?</h3>

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

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Define heat capacity of a substance:

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Write the S.I unit of heat capacity:

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Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t
Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.  </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy U_{p} will be:  

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Where m is the mass of the object, g=9.8m/s^{2} the acceleration due gravity and h=500m the height of the object.  

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For m=6kg

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For m=9kg

U_{p}=(9kg)(9.8m/s^{2})(500m)  

U_{p}=44100J=44.1kJ  

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