Answer:
The temperature of the Aluminium plate 44.84⁰C
Explanation:
Number of transistors = 4
Since the heat dissipated by each transistor is 12W
Total heat dissipated, Q = 4 * 12 = 48 W
Q = 48 W
Cross sectional Area of the Aluminium plate, A = 2(l * b)
l = Length of the aluminium plate = 22 cm = 0.22 m
b = width of the aluminium plate = 22 cm = 0.22 m
A =2( 0.22 * 0.22 )
A = 0.0968 m²
From the heat balance equation, Q = hAΔT
h = 25 W/m²·K
A = 0.0968 m²
ΔT = T - T(air)
T(air) = 25°C
ΔT = T - 25°C
Q = 25 * 0.0968 * ( T - 25)
Q = 2.42 (T - 25)
Substitute Q = 48 into the equation above
48 = 2.42 (T - 25)
T - 25 = 19.84
T = 25 + 19.84
T = 44.84 ⁰C
Answer:
Explanation:
Width of central diffraction peak is given by the following expression
Width of central diffraction peak= 2 λ D/ d₁
where d₁ is width of slit and D is screen distance and λ is wave length.
Width of other fringes become half , that is each of secondary diffraction fringe is equal to
λ D/ d₁
Width of central interference peak is given by the following expression
Width of each of bright fringe = λ D/ d₂
where d₂ is width of slit and D is screen distance and λ is wave length.
Now given that the central diffraction peak contains 13 interference fringes
so ( 2 λ D/ d₁) / λ D/ d₂ = 13
then ( λ D/ d₁) / λ D/ d₂ = 13 / 2
= 6.5
no of fringes contained within each secondary diffraction peak = 6.5
Answer:
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