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Alex777 [14]
3 years ago
5

The Skeleton does a surprising number of things. What are the three main Ones? lol, it's due today at 12:59 pm I need help.

Physics
1 answer:
ICE Princess25 [194]3 years ago
5 0

Answer:

Support

Posture

Protection

Explanation:

Support – the skeleton keeps the body upright and provides a framework for muscle and tissue attachment.

Posture – the skeleton gives the correct shape to our body.

Protection – the bones of the skeleton protect the internal organs and reduce the risk of injury on impact.

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Please check my answers: The voltage across a 10-ohm resistor carrying 3 amps must be? I got 30 volts-v=?, I=3 amps, R= 16 ohm.
Igoryamba
1) 30 volts is correct
2) Rt = (4x6)/(4+6)=2.4 Ω
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Two male moose charge at each other with the same speed and meet on a icy patch of tundra. as they collide, their antlers lock t
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3 years ago
Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi
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Dark matter does not affect our view, humans can see through them.

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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
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Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

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

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

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

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

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

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

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

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

}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}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

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