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Leya [2.2K]
3 years ago
12

Narrow, bright fringes are observed on a screen behind a diffraction grating. The entire experiment is then immersed in water. D

o the fringes on the screen get closer together, get farther apart, remain the same, or disappear? Explain.
Physics
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:

 n (a sin θ) =  m λ₀

n> 1, therefore the fringes move away from each other

Explanation:

The diffraction experiment the constructive interference fringes is described by

          a sin θ = m λ₀

in this equation it is assumed that the experiment emptied the air n = 1

When the same experiment is performed in water, the wavelength changes

           λₙ = λ₀ / n

execution for constructive interference

            a sin θ = m λₙ

we substitute

           a sin θ = m λ / n

           n (a sin θ) =  m λ₀

the refractive index of water is n = 1.33, so for the same wavelength the separation of the spectrum is multiplied by n> 1, therefore the fringes move away from each other

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xenn [34]

Answer:

the third one is incorrect

Explanation:

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3 years ago
An object is traveling at 15 m/s and then after 5 seconds ends up traveling at 10 m/s. Calculate the speeds acceleration rate. W
SSSSS [86.1K]

Answer:

1) a = -1 m/s²

2) v = 12 m/s

Explanation:

Given,

The initial velocity of the object, u = 15 m/s

The final velocity of the object, v = 10 m/s

The time taken by the object to travel is, t = 5 s

Using the first equation of motion

                               <em>v = u + at</em>

                               a = (v - u) / t

Substituting the values

                                a = (10 - 15) / 5

                                    = -1 m/s²

The negative sign indicates the body is decelerating

The acceleration of the object is, a = -1 m/s²

The speed of the object after 2 seconds

From the above equations of motion

                                  v = 15 + (-1) 2

                                     = 12 m/s

Hence, the speed of the object after 2 seconds is, v = 12 m/s

7 0
4 years ago
Un contenedor de 1800 N está en reposo sobre un plano inclinado a 28°, el coeficiente de fricción entre el contenedor y el plano
babunello [35]

Answer:

F = 1480.77N

Explanation:

In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:

F-Wsin\alpha-F_r=0\\\\F-Wsin\alpha-N\mu cos\alpha=0\\\\F-Wsin\alpha-W\mu cos\alpha=0     (1)

F: required force = ?

W: weight of the container = 1800N

N: normal force = weigth

α: angle of the incline = 28°

g: gravitational acceleration = 9.8m/s^2

μ: coefficient of friction = 0.4

You solve the equation (1) for F and replace the values of the other parameters:

F=W(sin\alpha+\mu cos\alpha)\\\\F=(1800N)(sin28\°+(0.4)cos28\°)=1480.77N

The required force to push the container for the incline with a constant velocity is 1480.77N

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3 years ago
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Answer:

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Explanation:

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