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Nonamiya [84]
3 years ago
10

The spreading of waves behind an aperture is more for long wavelengths and less for short wavelengths.Less for long wavelengths

and more for short wavelengths. Which is true and why?
Physics
1 answer:
Allisa [31]3 years ago
5 0

Answer:

Increase in wavelength of incident wave also increases the spread angle or spread of the interference pattern.

Explanation:

Solution:-

- The diffraction occurs when light bends in the same medium. The bending is the result of light waves "squeezing" through small openings or "curving" around sharp edges.

- Moreover, waves diffract best when the size of the diffraction opening (or grting or groove) corresponds to the size of the wavelength. Hence, light diffracts more through small openings than through larger openings.

- The formula for diffraction shows a direct relationship between the angle of diffraction (theta) and wavelength:

                                         d sin (θ) = m λ

Where,

     λ : Wavelength , θ : The spread angle , d : Slit opening or grating

- We can see that the wavelength λ and spread angle θ are related proportionally. So if we increase the wavelength of incident wave we also increase the spread angle or spread of the interference pattern.

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A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
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Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
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5 0
3 years ago
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s
DedPeter [7]

Answer:

212.8 m/s^{2}

Explanation:

Time taken by stone to cover horizontal distance

t=\sqrt{\frac {2h}{g}} where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

t=\sqrt{\frac {2*2.1}{9.81}}= 0.654654 seconds

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Velocity, v= distance/time

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v=15.3 m/s

a=\frac {v^{2}}{r} where r is radius of circle, substituting r with 1.1m

a=\frac {15.3^{2}}{1.1}

a=212.8 m/s^{2}

Therefore, centripetal acceleration is 212.8 m/s^{2}

8 0
3 years ago
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