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anyanavicka [17]
3 years ago
12

(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e

quation between the two if 32ºF corresponds to 0ºC and 212ºF corresponds to 100ºC.
Physics
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

C=-\dfrac{10}{17}(F-32)

Explanation:

Given that

32° F corresponds to 0 °C.  ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)

Lets C in y- direction and F in x- direction,so we can say that

C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)

C-0=\dfrac{100-0}{32-212}(F-32)

C=-\dfrac{10}{17}(F-32)

So the linear relationship is

C=-\dfrac{10}{17}(F-32)

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What is the voltage across each resistor?
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Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

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3 years ago
A cube with sides of length 2cm has a mass of 7.36g . calculate the density of the cube
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Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3

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What quantity measures the number of complete cycles an oscillation makes per second? A. period B. amplitude C. frequency D. for
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Hello, there Jcparris

Your answer is going to be C. Frequency

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A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
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Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

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Now,

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=  \frac{3000}{1000}

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