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Archy [21]
3 years ago
7

Find the probability density of a particle moving in an interval of 101o the box. The length of the one-dimensional box is 20x10

-10 m (a) 0.2 (b) 0.3 (c) 0.4 (d) None of the above
Physics
1 answer:
zalisa [80]3 years ago
3 0

Answer:

The probability density function is 0.25\times 10^{8} m^{- 1}

Solution:

Position of the particle in the box, S = 10^{- 10} m

Length of the box, L = 20\times 10^{- 10} m

Now, the probablity is given by:

P = \frac{10^{- 10}}{20\times 10^{- 10}} = \frac{1}{20}

Now,

The probability density, \psi = \frac{P}{L}

\psi = \frac{\frac{1}{20}}{20\times 10^{- 10}} = 0.25\times 10^{8} m^{- 1}

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\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

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\displaystyle \vec{v_2}=

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If the vectors are given as a magnitude and an angle (M\ ,\ \theta ), each component can be found as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The first vector has a magnitude of 3.14 m and an angle of 30°, so

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=

The second vector has a magnitude of 2.71 m and an angle of -60°, so

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=

The sum of the vectors is

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