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3 years ago

12

Consider a Young's two-slit experiment in which the wavelength of light is 37% smaller than the distance between the slits. How

many complete dark fringes would be seen on a distant screen? (Assume for all these questions that the screen is as large as it needs to be.)

1 answer:

Diano4ka-milaya [45]3 years ago

4 0

Answer: The answer is 3

Explanation:

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A man pulls a sled at a constant velocity across a horizontal snow surface. if a force of 80 n is being applied to the sled rope

balandron [24]

A constant velocity implies the two forces must be equal and opposite.
Friction acts horizontal to the ground, therefore we must find the force applied to the sled rope that acts horizontal to the ground.
Do this by resolving:
Force = 80cos53
The force opposing this is equal, and so also = 80cos53 = 48 N (2 sig. fig.)

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3 years ago

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if i got an C+ 77% in my science class but i got an 3 F's 0% recently on a science fair project, What will my grade go down to?

Crank

Your grade will probably go down to a D 68% or little higher than that

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2 years ago

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Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

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1 year ago

As a mercury atom absorbs a photon of

Allisa [31]

The energy absorbed by photon is 1.24 eV.
This is the perfect answer.

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2 years ago

A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring

Dennis_Churaev [7]

Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

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2 years ago