The energy flux associated with solar radiation incident on the outer surface of the earth’s atmosphere has been accurately meas
ured and is known to be 1368 W/m2. The diameters of the sun and earth are 1.39 × 109 and 1.27 × 107 m, respectively, and the distance between the sun and the earth is 1.5 × 1011 m. (a) What is the emissive power of the sun? (b) Approximating the sun’s surface as black, what is its temperature? (c) At what wavelength is the spectral emissive power of the sun a maximum? (d) Assuming the earth’s surface to be black and the sun to be the only source of energy for the earth, estimate the earth’s surface temperature.
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Answer:
w=2.25
Explanation:
It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.
The surface of the cross-section of the stapes was determined:
A_ab= 10 mm^2
A-cd= 15 mm^2
The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.
σ_ab = F_ab/A_abσ_allow
σ_cd = F_cd/A_cdσ_allow
In the next step we will determine the static size: Picture b).
We apply the conditions of equilibrium:
∑F_x=0
∑F_y=0
∑M=0
∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0
==> F_cd = 2*w*k*N
∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0
==> F_ab = w*k*N
Now we determine the load w
<u>Sector AB: </u>
σ_ab = F_ab/A_ab σ_allow=300 KPa
= w/10*10^-6 σ_allow=300 KPa
w_ab = 3*10^-3 kN/m
<u>Sector CD: </u>
σ_cd = F_cd/A_cd σ_allow=300 KPa
= 2*w/15*10^-6 σ_allow=300 KPa
w_cd = 2.25*10^-3 kN/m
w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}
==> w=2.25 * 10^-3 kN/m
<u>The solution is: </u>
w=2.25 N/m
note:
find the attached graph
