3 years ago

Please help:)

Physics

1 answer:

timofeeve [1]3 years ago

5 0

Answer:

tqm-apwk-wix. join join

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Molodets [167]

Answer:

I = 11.26 mA

Explanation:

given,

V = 0.7 V length = 80 m

diameter = 0.2 mm = 0.02 cm

radius = 0.01 × 10⁻² m

$resistance = \dfarc{\rho l}{A}$

ρ for gold wire = 2.44 × 10⁻⁸ ohm-m at 20 °C

A = cross sectional area = π r² = π (0.01 × 10⁻² )²

= 31.4× 10⁻⁹ m²

$resistance = \dfarc{2.44\times 10^{-8} \times 80}{31.4\times 10^{-9}}$

R = 62.165 Ω

$I = \dfrac{V}{R}$

$I = \dfrac{0.7}{62.165}$

I = 11.26 mA

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3 years ago

Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

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Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$

Applying the limits and solving for A we get

$I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k$

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Answer:

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Answer:

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Compression and rarefraction, the other guy's answer it's wrong

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