The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
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
The distance quantity/ measurement must be squared.
It’s D. An enlargement (hope this helps!)
