Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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PART a)
As we know that gravitational potential energy is given by the formula
here we can see that gravitational potential energy inversely varies with the distance
so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position
So gravitational potential energy is minimum at the nearest point and maximum at the farthest point
PART b)
Since we know that sum of kinetic energy and potential energy is constant here
so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum
So here speed is maximum at the nearest point
Part C)
since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance
Convex mirrors only produce virtual images
Answer:
huh? do you need help on math?
Explanation:
what do you mean?
