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podryga [215]
1 year ago
5

You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen

gth.
Physics
1 answer:
Oksanka [162]1 year ago
3 0

The de Broglie wavelength \lambda = 4.0\times 10^{-30}m

We know that

de Broglie wavelength = \lambda = \frac{h}{mv}\lambda = \frac{6.63\times 10^{-34}}{0.11\times 1.5 \times 10 ^{-3}}

\lambda = 4.0\times 10^{-30}m

<h3>What is de Broglie wavelength?</h3>

According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

Therefore, whether a particle is tiny or macroscopic, it will have a wavelength when examined.

The wave nature of matter can be seen or observed in the case of macroscopic objects.

To learn more about de Broglie wavelength with the given link

brainly.com/question/17295250

#SPJ4

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A force of 5 N is applied to the end of a lever that has a length
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The answer is 10Nm

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X = v xo t. x = (10.0 m/s)(3.53 s) x = ????
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x = V<em>x</em> * t

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Read 2 more answers
A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

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3 years ago
2.2.4 Quiz: Conservation of Energy
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Answer:

it is the judicious use of energy to prevent wastage

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