How were the fossil symbols and mountain bets helpful in deciding where to move the continenets
1 answer:
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Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so
Change in entropy
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]