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hichkok12 [17]
3 years ago
11

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

e between these wavelengths?
Physics
1 answer:
Strike441 [17]3 years ago
6 0

The solution would be like this for this specific problem:

 

Given:  

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

 

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m <span>

Whenas n = 1, maxima (grating equation) sinθ = λ/d 
Grant distance of each maxima from centre = y .. 
<span>As sinθ ≈ y/D  y/D = λ/d λ = yd / D </span>

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D 
∆λ = (d/D) [y2 -y1] 

<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>

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Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc
sammy [17]
If it takes t seconds to reach the car, then the distance d is 3.9t.

The bear's distance from the tourist's starting point is
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3.9t=6t-23
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so the distance is
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6 0
3 years ago
What is the gravitational force on a 35.0 kg object standing on the Earth's surface?
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An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0
3 years ago
A laser source has a bandwidth of 30GHz (a) Calculate the coherence length of the source b) What is the time separation of secti
abruzzese [7]

Explanation:

It is given that,

Bandwidth of a laser source, f=30\ GHz=30\times 10^9\ Hz

(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :

T=\dfrac{1}{f}

T=\dfrac{1}{30\times 10^9}

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l=c\times T

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l=3\times 10^8\times 3.33\times 10^{-11}

l = 0.0099 m

Hence, this is the required solution.

6 0
3 years ago
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