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Aleonysh [2.5K]
4 years ago
10

Write a linear equation that expresses the relationship between the temperature in degrees Celsius (C) and degrees Fahrenheit (F

). Use the fact that water freezes at 0°C (32°F) and boils at 100°C (212°F).
Physics
1 answer:
UkoKoshka [18]4 years ago
3 0

Answer:

\displaystyle F=\frac{9}{5}C+32

\displaystyle C=\frac{5}{9}(F-32)

Explanation:

<u>Temperature Units Conversion </u>

The conversion formula between Celsius and Fahrenheit temperature scales is well-known. But we'll use the provided data to derive the formula. Let's model the relationship between Fahrenheit (F) and Celsius (C) as a linear function like

F=mC+b

Where m and b must be computed according to the pair of conditions given. The values for each temperature scale are (C,F)=(0,32) and (100,212). Replacing the first value

32=m\times 0+b

It means that  

b=32

By using the second point

212=m\times 100+32

Solving for m

\displaystyle m=\frac{212-32}{100}=\frac{180}{100}

Simplifying

\displaystyle m=\frac{9}{5}

So, the conversion formula is

\displaystyle F=\frac{9}{5}C+32

Which is the widely known formula for temperature conversion

Solving for C, we get the inverse relation

\boxed{\displaystyle C=\frac{5}{9}(F-32)}

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Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol
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Answer:

R \approx 2.418\times 10^{9}\,km

Explanation:

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Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

T = \frac{2\pi}{\omega}

Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

\omega = \sqrt{\frac{a_{r}}{R} }

Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}

Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

R \approx 2.418\times 10^{9}\,km

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