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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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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

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Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
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