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} }](https://tex.z-dn.net/?f=T%20%3D%20%20%5Csqrt%5B4%5D%7B202.857%20%2A%2010%20%5E%7B8%7D%20%7D%20)
T = 3.774 · 10² =
377.4 KAnswer: Equilibrium temperature is 377.4 K.
Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
The answer should be 35.89 meters in 3.4 minutes. I dont know for sure though
Answer: a) - 437.8° F, b) - 261°c.
Explanation: a) the kelvin and Fahrenheit temperature scale are related by the formulae below.
5 (°F - 32) = 9 (k - 273)
Where °F = temperature in Fahrenheit and k = temperature in kelvin.
For question A, k = 12.0, by substituting to have the value for °F, we have
5(°F - 32) = 9 ( 12 - 273)
5(°F - 32) = 9(-261)
5(°F - 32) = - 2349
°F - 32 = - 2349/5
°F - 32 = - 469.8
°F = - 469.8 + 32
°F = - 437.8
Question B
The centigrade and kelvin scale are related by the formulae below
°c = k - 273
Where °c = temperature in centigrade and k = temperature in kelvin =12
°c = 12 - 273
°c = - 261
