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Lyrx [107]
3 years ago
6

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an

angle of 62.5°?
Physics
1 answer:
8090 [49]3 years ago
4 0

Answer:

The distance is d =  1.747 *10^{-6} \ m  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   \lambda  =  775 nm  = 775 * 10^{-9}  \ m

         The angle is  \theta =  62.5^o

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          dsin \theta  =  m *  \lambda

where  d is  the distance between the lines on a  diffraction grating

     So  

            d =  \frac{m *  \lambda  }{sin (\theta  )}

substituting values  

           d =  \frac{2  *  775 *1^{-9} }{sin ( 62.5  )}

          d =  1.747 *10^{-6} \ m

   

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A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of
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Answer:

Explanation:

Given

mass of spring m=100\ gm

extension in spring x=5\ cm

downward velocity v=70\ cm/s

Position in undamped free vibration is given by

u(t)=A\cos \omega _0t+B\sin \omega _0t

where \omega _0^2=\frac{k}{m}

also \frac{k}{m}=\frac{g}{L}

\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}

\omega _0=14

u(t)=A\cos(14t)+B\sin(14t)

it is given

u(0)=0

u'(0)=70\ cm/s

substituting values we get

A=0

u(t)=B\sin (14t)

u'(t)=14B\cos (14t)

70=14B

B=\frac{10}{2}

B=5

u(t)=5\sin (14t)

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3 years ago
A Rankine cycle with one closed feedwater heater with its drain cascaded backward has a water mass flow rate through the steam g
mamaluj [8]

Answer:Draw a T-s diagram for the ideal Rankine Cycle

Explanation:

6 0
3 years ago
We can model a lightning bolt as a very long, straight wire. If a lightning bolt carries a current of 30 kA, and you are unfortu
Liula [17]

Answer:

Magnetic field experienced = 4.5 × 10⁻⁴ T

Explanation:

The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by

B = (μ₀I)/(2πr)

B = ?

I = 20 KA = 20000 A

r = 8.9 m

μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A

B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T

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How many times is larger than a centigram is a dekagram
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A decagram is 1000 times bigger than a centigram
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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
11111nata11111 [884]

Answer:

1.62\times 10^{-8}\ \text{s}

Explanation:

\epsilon_0 = Vacuum permittivity = 8.854\times 10^{-12}\ \text{F/m}

A = Area = 10\times 2\times 10^{-4}\ \text{m}^2

d = Distance between plates = 1 mm

V_c = Changed voltage = 60 V

V = Initial voltage = 100 V

R = Resistance = 1000\ \Omega

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}

We have the relation

V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}

The time taken for the potential difference to reach the required level is 1.62\times 10^{-8}\ \text{s}.

5 0
3 years ago
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