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Lyrx [107]
2 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]2 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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Plz like if it helped.
7 0
3 years ago
a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its
LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

4 0
3 years ago
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
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8_murik_8 [283]
<span>Heat from the Sun is transferred to the sand without direct contact. This heat is then transferred to your feet by direct contact.</span>
8 0
2 years ago
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