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8_murik_8 [283]
3 years ago
11

6. Aaron will run from home at y mph and walk back at x mph. how much distance does he need to travel to spend a total of t hour

s walking and jogging. (A) xt/y (B) (x+t)/xy (C) xyt/(x+y) (D) (x+y+t)/xy (E) (y+t)/x-t/y
Physics
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

Total distance, d=\dfrac{xyt}{(x+y)}

Explanation:

It is given that,

Speed of Aaron from home is y mph and walk back at x mph. Let t is the total time he spend in walking and jogging. Let d is the distance covered.

We he moves from home to destination, time is equal to, \dfrac{d}{x}

Similarly, when he move back to home, time taken is equal to \dfrac{d}{y}

Total time taken is equal to :

\dfrac{d}{x}+\dfrac{d}{y}=t

d(\dfrac{1}{x}+\dfrac{1}{y})=t

d=\dfrac{t}{(\dfrac{1}{x}+\dfrac{1}{y})}

d=\dfrac{xyt}{(x+y)}

So, the distance he speed in walking and jogging is \dfrac{xyt}{(x+y)}. Hence, this is the required solution.

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Answer:

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Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

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Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

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Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

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Kinetic energy of a proton = thermal kinetic energy

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}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

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