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Lunna [17]
3 years ago
12

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

fifth-order maximum of the unknown wavelength exactly overlaps the third-order maximum of the red light.
What is the unknown wavelength? (in nm)
Physics
1 answer:
marusya05 [52]3 years ago
4 0

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

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F = 3x10^8m/s / 6x10^-7m = Hz
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Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot
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Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = \frac{kQq}{r^{2} }

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

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for the y-component

N_{y} = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

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The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

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Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
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Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

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so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

5.8=g(t_1+t_2)(t_1-t_2))

and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

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3 years ago
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Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>
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3 years ago
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