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2 years ago

Energy and work are measured in

Physics

2 answers:

bekas [8.4K]2 years ago

5 0

Answer:

i'll help you bro

Explanation:

Advocard [28]2 years ago

5 0

Answer:

Energy and work are measured in Joule.

Hope it will help :)

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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

S_A_V [24]

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):

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

7 0

3 years ago

Which are characteristics of electromagnetic waves?check all that apply.

emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, $c=3\cdot 10^8 m/s$ )

travel only through matter --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space --> FALSE, electromagnetic waves can also travel through matter

can bend around objects --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0

3 years ago

Read 2 more answers

We sometimes conform to others because they belong to a certain ____

Andrew [12]

A.) reference group

"A reference group includes individuals or groups that influence our opinions, beliefs, attitudes and behaviors. They often serve as our role models and inspiration"(study.com).

4 0

3 years ago

Read 2 more answers

PLEASE HELP! **

Eduardwww [97]

It moves to 56 km per hours

8 0

3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

love history [14]

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.