Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Answer:
189.15cy
Explanation:
To understand this problem we need to understand as well the form.
It is clear that there is four wall, two short and two long.
The two long are 
The two long are 
The two shors are 
The height and the thickness are 14ft and 0.83ft respectively.
So we only calculate the Quantity of concrete,
![Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3](https://tex.z-dn.net/?f=Q_c%20%3D%20%5B%282%2A122.08%29%2B%282%2A86-375%29%5D%2A14%2A0.833%5C%5CQ_c%3D4864.02ft%5E3)
That in cubic yards is equal to 
Hence, we need order 5% plus that represent with the quantity

Https://www.slader.com/discussion/question/an-insulated-rigid-tank-is-divided-into-two-equal-parts-by-a-partition-initially-one-part-contains-4/
there will be the answer
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
<em>Values are gotten from the table named: blackbody radiati</em>on functions
<u>a) Calculate the band emission fractions for the visible region</u>
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm