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alexandr402 [8]
4 years ago
4

Use the data from Table D of your Student Guide to answer the questions. Be sure to round your answers to the nearest tenth.

Physics
1 answer:
katen-ka-za [31]4 years ago
6 0

Answer:

table??

Explanation:

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What is the frequency corresponding to a period of 8.01 s?
masha68 [24]
The frequency of an oscillation is equal to the reciprocal of the period:
f= \frac{1}{T}
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f is the frequency
T is the period

In our problem, the period is T=8.01 s, therefore the corresponding frequency is
f= \frac{1}{8.01 s}=0.12 s^{-1} = 0.12 Hz
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The steps in the scientific process must be followed in order. True False
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Compare and contrast the three <br> classes of levers.
Xelga [282]
<em>Class I .</em> . .
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3 years ago
Yellow light of wavelength 590 nm passes through a diffraction grating and makes an interference pattern on a screen 80 cm away.
riadik2000 [5.3K]

Answer:B

Explanation:

Given

Wavelength of light \lambda =590\ nm

Screen distance L=80\ cm

First fringe is at a distance y_1=1.9\ cm

No of lines per mm is given by N

N=\frac{1}{d}

where d=slit width

From N-slits Experiment

\sin \theta _m=\frac{m\lambda }{d}

d=\frac{m\lambda }{\sin \theta _m}-----1

Position of bright fringe is given by

y=\tan \theta _m\cdot L

\tan \theta _m=\frac{y}{L}

\theta _m=\tan^{-1}(\frac{y}{L})

Put the value of \theta _m  in eq. 1

d=\frac{m\lambda }{\sin (\tan^{-1}(\frac{y}{L}))}

Therefore N=d^{-1}

N=\frac{\sin (\tan^{-1}(\frac{y}{L}))}{m\lambda }

for m=1

N=\frac{\sin (\tan^{-1}(\frac{1.9\times 10^{-2}}{0.8}))}{1\times 590\times 10^{-9}}

N=40243\ line/m

N=40\ line/mm

   

4 0
4 years ago
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