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1 year ago

Solve for A, find the exact value of each expression below.

Mathematics

1 answer:

nataly862011 [7]1 year ago

4 0

$\begin{gathered} \theta=\frac{7\pi}{6} \\ 2\sin (\frac{7\pi}{6})=2(\frac{-1}{2})=-1 \\ 2\sin (\frac{7\pi}{6})=-1 \\ \\ \sin ^2(\frac{7\pi}{6})=(\frac{-1}{2})^2=\frac{1}{4} \\ \sin ^2(\frac{7\pi}{6})=\frac{1}{4} \\ \\ \sin (-\frac{7\pi}{6})=-\sin (\frac{7\pi}{6})=-(\frac{-1}{2})=\frac{1}{2} \\ \sin (-\frac{7\pi}{6})=\frac{1}{2} \end{gathered}$

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

Are you sure you wanted that "b" in the second equation?

If so, here's how we eliminate y:

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3 years ago

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Nicholas is adding by place value from right to left. When he adds the hundredths, he gets 0.46. Describe what his next step sho

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3 years ago

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th

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Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

$P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx$

$P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx$ (1)

The solution of the intregral of equation (1) is:

$P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}$

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}$

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}$

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}$

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5$

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

$P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}$

I hope it helps you!

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4 years ago