Answer:
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
Answer:
Disaggregation
Explanation:
In a company it is a way to create operational plans that are focused, either by time or by section.
Answer: 383.22K
Explanation:
L = 3m, w = 1.5m
Area A = 3 x 1.5 = 4.5m2
Q' = 750W/m2 (heat from sun) ,
& = 0.87
Q = &Q' = 0. 87x750 = 652.5W/m2
E = QA = 652.5 x 4.5 = 2936.25W
T(sur) = 300K, T(panel) = ?
Using E = §€A(T^4(panel) - T^4(sur))
§ = Stefan constant = 5.7x10^-8
€ = emmisivity = 0.85
2936.25 = 5.7x10^-8 x 0.85 x 4.5 x (T^4(panel) - 300^4)
T(panel) = 383.22K
See image for further details.
so people dont die whaddya think?
Answer:
2.7 W/m^2K
Explanation:
Area of pane = 5 m x 6 m = 30 m^2
Solar irradiation Gs = 900 W/m2
Heat rate on panel = Gs x area = 900 x 30 = 27000 W
absorptivity to solar irradiation αs = 0.92
Therefore, absorbed heat is
0.92 x 27000 = 24840 W
For heat gain,
From E = §AT^4
Where § = stefan's constant = 5.7x10^-8 Wm^-2K^-1
T = temperature of panel
24840 = 5.7x10^-8 x 30 x T^4
24840 = 1.71x10^-6 x T^4
1.453x10^10 = T^4
T = 347.167 K
For net heat gain,
From E = §A(T^4 - T^4sur)
24840 = 5.7x10^-8 x 30 x (T^4 - T^4sur)
24840 = 1.71x10^-6 x (1.453x10^10 - T^4sur)
24840 = 24846.3 - 1.71x10^-6(T^4sur)
-6.3 = -1.71x10^-6(T^4sur)
3684210.526 = T^4sur
Tsur = 43.81 K
Also for convective heat,
E = Ah(T - Tsur)
24840 = 30h(347.167 - 43.81)
24840 = 30h x 303.357
81 = 30h
h = 2.7 W/m^2K
