36 is what percent of 150?
2 answers:
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Answer:
solution:
dT/dx =T2-T1/L
&
q_x = -k*(dT/dx)
<u>Case (1) </u>
dT/dx= (-20-50)/0.35==> -280 K/m
q_x =-50*(-280)*10^3==>14 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (2)
dT/dx= (-10+30)/0.35==> 80 K/m
q_x =-50*(80)*10^3==>-4 kW
Case (3)
q_x =-50*(160)*10^3==>-8 kW
T2=T1+dT/dx*L=70+160*0.25==> 110° C
Case (4)
q_x =-50*(-80)*10^3==>4 kW
T1=T2-dT/dx*L=40+80*0.25==> 60° C
Case (5)
q_x =-50*(200)*10^3==>-10 kW
T1=T2-dT/dx*L=30-200*0.25==> -20° C
note:
all graph are attached
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ = ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀ ) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀ ) Q / (1.25R)²
E = (1 /4πε₀ ) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀ ) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B