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lilavasa [31]
3 years ago
10

The wave function of the electron within the well is of the form a cos( 2πx / λ ) where a is a normalization constant. what is t

he approximate value of λ, within about 10%? [note: although you can really only estimate it to within about this accuracy, the current version of smartphysics will require higher accuracy.] λ =
Mathematics
1 answer:
Arada [10]3 years ago
6 0

Answer:

The answer is y is 0.05.

Explanation:

At cos(2x) and the left and right boundaries are at 400 pm.

At 0 pm = cos(2π × x ÷ ∧) = 1

So,

A has to be 0.075.

it looks like at 300 pm, y = 0.05.

0.05 = 0.075 × cos(2π × 300 pm ÷ ∧)

= 0.05.

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Atmospheric pressure decreases by about 11.8% for every 1000 meters you climb. The pressure at sea level is 1013 millibars (a un
erica [24]
At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I

so, the inital amount is 1013, when t = 0,

\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.


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