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MArishka
2 months ago
14

The 3rd bright fringe of a double slit interference pattern is 30.0 cm above the central bright fringe. If the angle from the ho

rizontal to this 3rd bright fringe is 12.0 degrees, what is the distance (in meters) between the double slits and the viewing screen
Physics
1 answer:
Keith_Richards [3.2K]2 months ago
5 0
12 degrees = (π / 180) x 12 radians =.2093 radians; the position of the third bright fringe is determined as 3λ D/d, where λ represents the light wavelength, D is the distance to the screen, and d is the slit separation. Thus, the equation results in 3λ D/d = 30 x 10⁻², and from the angular separation, we deduce 3λ/d =.2093. Substituting the two equations, we find.2093 D = 30 x 10⁻², leading to D = 30 x 10⁻² /.2093 = 1.43 m.
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Competitive forces model

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
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\text{heat loss} = 24864.05 \ W/m^2

Clarification:

If

  • T_1, T_2 represent the temperatures of gases and liquids in Kelvins,
  • t_1 and t_2 denote the thicknesses of the gas layer and steel slab in meters,
  • h_1, h_2 are the convection coefficients for gas and liquid in W/m^2 \cdot K,
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then: part(a):

\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

by employing known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Utilizing the rate equation:

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature is T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Correspondingly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature profile is depicted in the image provided

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