Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
Answer:
the correct answer is C, 281 lb
Explanation:
This is an exercise in fluid mechanics specifically on Pascal's principle, which states that the pressure in a fluid is equal to all points that are at the same depth
P =
= 
If we use the subscript 1 for the piston diameter mentor (d = 1.00vm) which is r = 0.05 cm, the force is F1 = 45 lb
F₂ = \frac{F_{1} }{A_{1} } F₁
in area of a circules
A = π r²
we substitute
F₂ =
let's calculate
F2 = 2.50² / 1.00² 45
F2 =281 lb
therefore the correct answer is C
Voltage is the difference in charge between two points.
Current is the rate the charge flows
Resistance is the tendency a material has to resist the flow of charge (current)
Combining voltage resistance and current Ohm developed the formula
V (Voltage)= I (Current) x R (Resistance)
Answer:
The value is 
Explanation:
From the question we are told that
The frequency of each sound is
The speed of the sounds is 
The distance of the first source from the point considered is 
The distance of the second source from the point considered is 
Generally the phase angle made by the first sound wave at the considered point is mathematically represented as
![\phi_a = 2 \pi [\frac{a}{\lambda} + ft]](https://tex.z-dn.net/?f=%5Cphi_a%20%3D%20%202%20%5Cpi%20%5B%5Cfrac%7Ba%7D%7B%5Clambda%7D%20%20%2B%20ft%5D)
Generally the phase angle made by the first sound wave at the considered point is mathematically represented as
Here b is the distance o f the first wave from the considered point
Gnerally the phase diffencence is mathematically represented as
=> ![\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }](https://tex.z-dn.net/?f=%5CDelta%20%20%5Cphi%20%20%20%3D%20%20%20%5Cfrac%7B2%5Cpi%20%5B%20a%20-%20b%5D%7D%7B%20%5Clambda%20%7D)
Gnerally the wavelength is mathematically represented as

=> 
=> 
=> ![\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }](https://tex.z-dn.net/?f=%5CDelta%20%20%5Cphi%20%20%20%3D%20%20%20%5Cfrac%7B2%2A%203.142%20%5B%204.40%20-%204.0%20%5D%7D%7B%20%200.611%20%20%7D)
=> 
Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."