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Luden [163]
1 year ago
13

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

oduce maxima (in different orders) that coincide.
a. What is the order (m) of each of the two overlapping lines?
b. At what angle does this overlap occur?

Physics
1 answer:
Harlamova29_29 [7]1 year ago
8 0

Answer:

a) m = 3 for λ1 and m= 2 for λ2 will overlap since they have the same values

b) At what angle does this overlap occur = 34.54 degree

Explanation:

The detailed steps and appropriate formula is as shown in the attachment.

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The answer would be C. It will decrease with descent. Hope this helps!
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An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
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Find the given attachment for solution

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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

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2 years ago
The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f
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Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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