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lubasha [3.4K]
3 years ago
11

The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light y

ears. A light year is the distance light travels in one year. Light travels at a speed of 3.00 × 108 m/s. Calculate the distance from Earth to Sirius in miles. (1 mi = 5280 ft) g
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:5.11(10)^{13}miles

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.  

This unit is equivalent to 5.879(10)^{12}miles, which mathematically is expressed as:

1Ly=5.879(10)^{12}miles

Doing the conversion:

8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles  This is the distance from Earth to Sirius in miles.

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A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the ope
MariettaO [177]

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

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