Answer:
(a) ![K = \frac{I}{2\pi a}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7BI%7D%7B2%5Cpi%20a%7D)
(b) ![J = \frac{I}{2\pi as}](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7BI%7D%7B2%5Cpi%20as%7D)
Explanation:
(a) The surface current density of a conductor is the current flowing per unit length of the conductor.
![K = \frac{dI}{dL}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7BdI%7D%7BdL%7D)
Considering a wire, the current is uniformly distributed over the circumferenece of the wire.
![dL = 2\pi r](https://tex.z-dn.net/?f=dL%20%3D%202%5Cpi%20r)
The radius of the wire = a
![dL = 2\pi a](https://tex.z-dn.net/?f=dL%20%3D%202%5Cpi%20a)
The surface current density ![K = \frac{I}{2\pi a}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7BI%7D%7B2%5Cpi%20a%7D)
(b) The current density is inversely proportional
......(1)
k is the constant of proportionality
![I = \int\limits {J} \, dS](https://tex.z-dn.net/?f=I%20%3D%20%5Cint%5Climits%20%7BJ%7D%20%5C%2C%20dS)
........(2)
substituting (1) into (2)
![I = \frac{k}{s} \int\limits\, dS](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7Bk%7D%7Bs%7D%20%5Cint%5Climits%5C%2C%20dS)
![I = k \int\limits^a_0 \frac{1}{s} {s} \, dS](https://tex.z-dn.net/?f=I%20%3D%20k%20%5Cint%5Climits%5Ea_0%20%5Cfrac%7B1%7D%7Bs%7D%20%20%7Bs%7D%20%5C%2C%20dS)
![I = 2\pi k\int\limits\, dS](https://tex.z-dn.net/?f=I%20%3D%202%5Cpi%20k%5Cint%5Climits%5C%2C%20dS)
![I = 2\pi ka](https://tex.z-dn.net/?f=I%20%3D%202%5Cpi%20ka)
![k = \frac{I}{2\pi a}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7BI%7D%7B2%5Cpi%20a%7D)
substitute ![J = \frac{k}{s}](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7Bk%7D%7Bs%7D)
![J = \frac{I}{2\pi as}](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7BI%7D%7B2%5Cpi%20as%7D)
Answer:
For Xenon fluoride, the average bond energy is 132kj/mol
For tetraflouride,the average bond energy is 150.5kj/mol.
For hexaflouride, the average bond energy is 146.5 kj/mol
Explanation:
For xenon fluoride
105/2 = 52.5
For F-F
159/2 = 79.5
Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole
For tetraflouride
284/4 = 71
For F-F
159/2 = 79.5
Average bond energy = 79.5 + 71 = 150.5kj/mol
For hexaflouride
402/6 = 67
F-F = 159/2 = 79.5
Average bond energy = 67 + 79.5 = 146.5kj/ mol
The pressure drop in pascal is 3.824*10^4 Pascals.
To find the answer, we need to know about the Poiseuille's formula.
<h3>How to find the pressure drop in pascal?</h3>
- We have the Poiseuille's formula,
![Q=\frac{\pi r^4P}{8\beta l}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5Cpi%20r%5E4P%7D%7B8%5Cbeta%20l%7D)
- where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
- By substituting values and rearranging we will get the pressure drop as,
![P=3.284Pascals](https://tex.z-dn.net/?f=P%3D3.284Pascals)
Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.
Learn more about the Poiseuille's formula here:
brainly.com/question/13180459
#SPJ4
Answer:
<h2>The amount of torque put on the car is 33,000Nm</h2>
Explanation:
Formula for calculating torque is expressed as T = rFsin
where;
r is the radius of the of the arm of the jack = 3m
F is the force exerted = 11000
is the angle of rotation = 90°
On substituting;
![T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm](https://tex.z-dn.net/?f=T%20%3D%203%2A11000sin90%5E%7Bo%7D%20%5C%5CT%20%3D%203%2A11000%20%28sin90%5E%7Bo%7D%20%3D1%29%5C%5CT%20%3D%2033000Nm)
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía