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Alexandra [31]
3 years ago
7

At high noon, the sun delivers 1150 w to each square meter of a blacktop road. if the hot asphalt loses energy only by radiation

, what is its equilibrium temperature?
Physics
1 answer:
Pie3 years ago
3 0
Stephen`s Law:
P = (Sigma) · A · e · T^4
P in = P out
e = 1 for blacktop;
1150 W = (Sigma) · T^4
(Sigma) = 5.669 · 10 ^(-8) W/m²K^4
T^4 = 1150 : ( 5.669 · 10^(-8) )
T^4 = 202.875 · 10^8
T =  \sqrt[4]{202.857 * 10 ^{8} }
T = 3.774 · 10² = 377.4 K
Answer: Equilibrium temperature is 377.4 K. 
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Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

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Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

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