Answer:
The value is ![V = 2 V](https://tex.z-dn.net/?f=V%20%20%3D%20%202%20V)
Explanation:
From the question we are told that
The length of the wire is ![l = 10 \ m](https://tex.z-dn.net/?f=l%20%3D%2010%20%5C%20m)
The current density is ![J = 4*10^6 \ A/m^2](https://tex.z-dn.net/?f=J%20%3D%20%204%2A10%5E6%20%5C%20%20A%2Fm%5E2)
The conductivity is
Generally conductivity is mathematically represented as
![\sigma = \frac{l}{RA}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Cfrac%7Bl%7D%7BRA%7D)
Here R is the resistance which is mathematically represented as
![R = \frac{V}{I}](https://tex.z-dn.net/?f=R%20%3D%20%20%5Cfrac%7BV%7D%7BI%7D)
Here I is the current which is mathematically represented as
![I = J * A](https://tex.z-dn.net/?f=I%20%20%3D%20%20J%20%2A%20A)
So
![R = \frac{V}{ J * A}](https://tex.z-dn.net/?f=R%20%3D%20%20%5Cfrac%7BV%7D%7B%20%20J%20%2A%20A%7D)
And
![\sigma = \frac{l}{\frac{V}{ J * A} * A}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Cfrac%7Bl%7D%7B%5Cfrac%7BV%7D%7B%20%20J%20%2A%20A%7D%20%2A%20A%7D)
=> ![\sigma = \frac{l}{\frac{V}{J}}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Cfrac%7Bl%7D%7B%5Cfrac%7BV%7D%7BJ%7D%7D)
=> ![V = \frac{l * J}{\sigma }](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bl%20%2A%20J%7D%7B%5Csigma%20%7D)
=> ![V = \frac{10 * 4*10^6}{2*10^{7} }](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B10%20%20%2A%204%2A10%5E6%7D%7B2%2A10%5E%7B7%7D%20%20%7D)
=> ![V = 2 V](https://tex.z-dn.net/?f=V%20%20%3D%20%202%20V)
Answer:
The answer is A/B, they're the same answer anyways.
Explanation:
Chromatic aberration is the result when the lens fail to focus all the colors on the same point. The light then focuses in different points,and could lead to causing two images at once. The main culprit of this is usually dispersion.
Answer:
Its inductance L = 166 mH
Explanation:
Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR
R = V/I = 5.62/0.698 = 8.052 Ω
Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from
V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'
Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω
WE now find the reactance X of the coil from
Z² = X² + R²
X = √(Z² - R²)
= √(97.5² - 8.05²)
= √(9506.25 - 64.8025)
= √9441.4475
= 97.17 Ω
Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.
L = X/2πf
= 97.17/2π(93.1 Hz)
= 97.17 Ω/584.965 rad/s
= 0.166 H
= 166 mH
Its inductance L = 166 mH
Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh