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2 years ago

A tungsten wire is 4.5 m long and 0.55 mm in diameter. what is its resistance?

1 answer:

3 0

First you need to know the formula for resistance, which is:

$R=$ ρ $\frac{L}{A}$ --- (A)

Where,
ρ = Resistivity
L = length
A = Area

ρ = Resistivity of tungsten = 5.28 x $10^{-8}$ Ωm
L = Length = 4.5 m
A = Area = 0.55 * $10^{-3}$ * π

Plug-in the values in equation(A), you would get:

$Resistance = \frac{5.28 *10^{-8} * 4.5}{0.55 * 10^{-3} * \pi }$

$Resistance = 1.375 * 10^{-4}$ Ω

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2 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

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Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

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3 years ago

If the tension of the cable is 25.0 N what is the mass of the ball

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Explanation:

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2 years ago

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What are the units, if any, of the particle in a box wavefunction. What does this mean?

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Answer:

$[\psi]= [Length^{-3/2}]$
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Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

$\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \, dy \, dx$

must be dimensionless, as represents a probability.

As the differentials has units of length

$[dx]=[dy]=[dz]=[Length]$

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

$[\psi]^2 = [Length^{-3}]$

taking the square root this gives us :

$[\psi] = [Length^{-3/2}]$

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2 years ago