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3 years ago

How does the de broglie wavelength of an electron change if its momentum increases?

Physics

1 answer:

Virty [35]3 years ago

3 0

Answer:

The De Broglie wavelength decreases

Explanation:

The relationship between the De Broglie wavelength of a particle and its momentum is given by

$\lambda=\frac{h}{p}$

where

$\lambda$ is the De Broglie wavelength of the particle

h is the Planck constant

p is the momentum of the particle

As we see from the formula, there is an inverse relationship between the De Broglie's wavelength and the momentum. Therefore, we can conclude that:

- if the momentum of the electron increases,

- its De Broglie wavelength will decrease

and vice-versa.

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How can we determine the strength of a sonic boom?

tino4ka555 [31]

<span>There are several factors that can influence sonic booms - weight, size, and shape of the aircraft or vehicle, plus its altitude, attitude and flight path, and weather or atmospheric conditions.</span>

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3 years ago

A swinging pendulum has a total energy of <img src="https://tex.z-dn.net/?f=E_i" id="TexFormula1" title="E_i" alt="E_i" align="a

Zolol [24]

Answer:

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$ (for small oscillations)

Explanation:

The total energy of the pendulum is equal to:

$E_{1} = m\cdot g \cdot (1-\cos \theta)\cdot L$

For small oscillations, the equation can be re-arranged into the following form:

$E_{1} \approx m\cdot g \cdot (1-\theta) \cdot L$

Where:

$\theta = \frac{A}{L^{2}}$ , measured in radians.

If the amplitude of pendulum oscillations is increase by a factor of 4, the angle of oscillation is $4\theta$ and the total energy of the pendulum is:

$E_{2} \approx m\cdot g \cdot (1-4\theta)\cdot L$

The factor of change is:

$\frac{E_{2}}{E_{1}} \approx \frac{1 - 4\theta}{1-\theta}$

$\frac{E_{2}}{E_{1}} \approx 1 -\frac{3\theta}{1-\theta}$

3 0

3 years ago

Match the terms to the correct descriptions. Question 1 options:

faust18 [17]

1) Refraction
2)Reflection
3)Concave
4)Convex

I took the test and got this right so you can believe me :)
Hope this helps

4 0

3 years ago

Read 2 more answers

An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete

Ksju [112]

<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:

R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s

the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>

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2 years ago

Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t

nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

Tsinθ = F ( F is force of repulsion between two charges sphere)

Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )

21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0

3 years ago