Electric Charge andStatic ElectricityIt was known as early as 600BCEthat objects could become charged by rubbing fur on various
1 answer:
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To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation
Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by
Then replacing our values we have that
We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,
The uncollided flux density at the outer surface of the tank nearest the source is
Answer:
130.165636364°C
Explanation:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant = 0.082 L atm/mol K
From ideal gas law we have
The initial temperature is
Answer:
a) For y = 102 mA, R = 98.039 ohms
For y = 97 mA, R = 103.09 ohms
b) Check explanatios for b
Explanation:
Applied voltage, V = 10 V
For the first measurement, current
According to ohm's law, V = IR
R = V/I
Here,
For the second measurement, current
b)
A linear equation is of the form y = Gx
The nominal value of the resistance = 100 ohms
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)]