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Irina-Kira [14]
3 years ago
13

A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of t

he cone is well insulated. The plate and the cone, whose surfaces are black, are located in a large, evacuated enclosure whose walls are at 300 K.
(a) What is the temperature of the conical surface, T2?
(b) What is the electrical power that would be required to maintain the circular plate at 600K?
Physics
1 answer:
drek231 [11]3 years ago
5 0
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K

For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
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Answer:

V=IR

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total resistance in parallel=

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since we have two resistor in parallel

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The outer surface of a skier’s clothes of emissivity 0.7000.700 is at a temperature of 5.505.50 °C. Find the rate of radiation i
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Answer:

121.0 W

Explanation:

We use the equation for rate of heat transfer during radiation.

Q/t = σεA(T₂⁴ - T₁⁴)

Since temperature of surroundings = T₁ = -20.0°C = 273 +(-20) = 253 K, and temperature of skier's clothes = T₂ = 5.50°C = 273 + 5.50 = 278.5 K.

Surface area of skier , A = 1.60 m², emissivity of skier's clothes,  ε = 0.70 and σ =  5.67 × 10⁻⁸ W/m²K⁴ .

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Q/t = σεA(T₂⁴ - T₁⁴) = (5.67 × 10⁻⁸ W/m²K⁴ ) × 0.70 × 1.60 m² × (278.5⁴ - 253⁴) = 6.3054 × (1918750544.0625) × 10⁻⁸ W = 1.2098 × 10² W = 120.98 W ≅ 121.0 W

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Answer:

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As a result, the combined motion of the projectile has a curved trajectory (parabolic, more specifically). So the following options are correct:

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