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zlopas [31]
2 years ago
12

What is the voltage of this circuit.

Physics
1 answer:
liberstina [14]2 years ago
6 0

Answer:

110.4 v

Explanation:

Find the equivalent resistance of the two parallel circuits

R1 and R2 in parallel   =   R1 * R2 / (R1+ R2) =   <u>1 ohm </u>

 similarly R4 and R5 =  <u>4.77 ohms</u>

Now you can add the three resistances into one  (R3=10)

    1 ohm + 10 ohm + 4.77 ohm = 15.77 ohms

Now

V = IR

V =  7 amps  * 15.77 ohm = 110.4 v

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Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

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3 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
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Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

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The start brightness is related to it luminosity thought the following equation:

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b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

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