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polet [3.4K]
3 years ago
15

What is 76 to the 3rd power

Mathematics
2 answers:
ivann1987 [24]3 years ago
7 0
438976 !!
hope this helped
marishachu [46]3 years ago
6 0

Answer:

438976

Step-by-step explanation:

76^3

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A = 3.2 because -4.5 + 1.3 = 3.2
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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
Hitman42 [59]

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
I need to find the area of this but I need help.
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The area of the parallelogram is the product of its base length and height, (12 ft)*(12 ft) = 144 ft^2.

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Answer:

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