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Taya2010 [7]
2 years ago
10

Let us be two cylindrical conductors connected in parallel, to which a potential difference of V = 170V is applied. The two cond

uctors are made of the same material, but the first is 6 times the length of the second, and the radius of the second. The resistance of the second is R2 = 469Ω. Determine the equivalent resistance.
Physics
1 answer:
PSYCHO15rus [73]2 years ago
3 0

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

<h3>Resistance</h3>

Resistance is a measure of the opposition to flow of electric current. It is measured in ohms.

It is given by the formula:

R=\rho\frac{l}{A} \\\\where\ l=length,A=area,\rho=resistivity

Given that R₂ = 469 ohm, hence:

R_2=\rho\frac{l_2}{A_2} \\\\469=\rho\frac{l_2}{\pi r_2^2}

But l₁ = 6l₂, r₁ = (1/5)r₂, hence:

R_1=\rho \frac{l_1}{A_1}=\rho *\frac{6l_2}{[\pi (1/5)r_2]^2} =150 * \rho \frac{l_2}{[\pi r_2]^2}=30*469=70350\ ohm

The equivalent resistance (R) is:

R=\frac{R_1R_2}{R_1+R_1}=\frac{469*70350}{469+70350}  =466\ ohm

The equivalent resistance of the two cylindrical conductors connected in parallel is 466 ohm.

Find out more on resistance at: brainly.com/question/17563681

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Answer:

37.42 m/s

Explanation:

We know that apparent frequency, \bar f is given by

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440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s

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3 years ago
Each year, an average person in the United States is exposed to a radiation level of _____.
Anna11 [10]

Answer:

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Explanation:

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Have a nice day!

3 0
3 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
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Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

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n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

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c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

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