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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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Explanation:

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7 0
3 years ago
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
3 years ago
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