Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
(a) dynamic viscosity = 
(b) kinematic viscosity = 
Explanation:
We have given temperature T = 288.15 K
Density 
According to Sutherland's Formula dynamic viscosity is given by
, here
μ = dynamic viscosity in (Pa·s) at input temperature T,
= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
= reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
= 291.15
when T = 288.15 K
For kinematic viscosity :
Answer:
Explanation:
Using the kinematics equation 
to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
Then:
The distance traveled by car B in the given time (t) is expressed as:
For car A, the needed time (t) to come to rest is:
Also, the distance traveled by car A in the given time (t) is expressed as:
Relating both velocities:
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
The answer is False!
The answer is false
Answer:
Explanation:
This question will be solved with the help of diagram (see attachment)
Given:
Δ∅ = 0.50 degrees (correction)
BC = 800 mm
AC = 600 mm
Solution:
We use coordinate system with point C as origin (0,0)
Hence,
Point A = (600,0)
Point B = (0,800)
The change in length or displacement can be calculated BB' :
BB' = BC * tan (Δ∅) = (800 mm) * tan (0.5) = 6.9815 mm
Hence, Point B' = (-6.9815,800) and we calculate distance A and B
We calculate distance A and B'
Normal Strain in AB is:
The solution is :
