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larisa86 [58]
2 years ago
6

In a two-slit experiment using coherent light, the distance between the slits and the screen is 1.10 m, and the distance between

the slits is 0.0400 mm. If the second order bright fringe is measured to be 4.20 cm from the centerline on the screen, what is the wavelength of light?
A) 620 nm
B) 200 nm
C) 401 nm
D) 763 nm
E) 381 nm
Physics
1 answer:
Paul [167]2 years ago
3 0

Answer:

D) 763 nm

Explanation:

Calculation for the wavelength of light

Using this formula

Wavelength of light=Delta Y*Distance / Length

Where,

Delta Y represent the 2nd order bright fringe

Length represent the distance between both the slits and the screen

Distance represent the Distance between the slits

Let note that cm to m = (4.2) x 10^-2 and mm to m= ( 0.0400x 10^-3)

Now Let plug in the formula

Wavelength of light=[(4.2 x 10^-2m)(0.0400 x 10^-3m) / 2(1.1m)]*10^-7 meters

Wavelength of light=[(0.042m) (0.0004m)/2.2m]*10^-7 meters

Wavelength of light =(0.0000168m/2.2m)*10^-7 meters

Wavelength of light =7.63 *10^-7 meters

Wavelength of light =763 nm

Therefore the Wavelength of light will be 763 nm

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A small ranger vehicle has a soft, ragtop roof. When the car is at rest, the roof is flat. When the car is cruising at highway s
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roof bow upwards

Explanation:

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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to
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Answer:

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Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

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The lesser the mass, the greater the acceleration for the given value of force. Why does this happen?
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Newtons law states F=ma
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