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
<span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
Newton's first law of motion states that. an object on the rest or motion is stay the same unless external force applied on it.
Answer:
I. 0 m/s
II. 20 m/s
III. Part BC
Explanation:
I. Determination of the initial velocity.
From the diagram given above,
The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s
II. Determination of the maximum velocity attained.
From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.
III. Determination of the part of the graph that represents zero acceleration.
It important that we know the meaning of zero acceleration.
Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.
From the graph given above, the car has a constant velocity between B and C.
Therefore, part BC illustrates zero acceleration.
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
