ogla has mistakenly created a connection between the black and white wire. What has she created? A) open circuit B) SHORT CIRCUI
1 answer:
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Answer:
Explanation:
Low frequency gain is= 40db= 20logK=>100 poles at 2MHz,20MHz
Zero at -200MHz, zero at infinity.
A) A(s) = 100FH(s)
B) Poles (1): 2 pi × 2 × 10^6= 4pi × 10^6MHz
(2): 2pi × 20 × 10^6= 4pi × 10^6 MHz
Zeroed: 2pi × 10^6 × 200= 400pi × 10^6, at infinity.
T/(S) = (1 + S/400π × 10^6)/S(1 + S/4π × 10^6)(1 + S/4π × 10^6)
Answer:
D = 0.060732 in
Explanation:
given data
sp. wt. = 500 lb/ft³
diameter = 0.036 in
solution
we get here maximum diameter of rod that is express as
D = ......................1
here surface tension of water at 60⁰f = 5.03 ×
lb/ft and y = 500 lb/ft³
so put here value and we will get
D =
D = 0.005061 ft
D = 0.060732 in
Answer:
V = 30 V
vector (E) = -10*a_p + 17.32*a_Q - 24*a_z
mag(E) = 31.24 V/m
dV / dN = 31.2 V /m
a_N = 0.32*a_p -0.55*a_Q + 0.77*a_z
p_v = 276 pC / m^3
Explanation:
Given:
- The Volt Potential in cylindrical coordinate system is given as:
- The point P is at p = 3 , Q = 60 , z = 2
Find:
values at P for:
a.) V
b.) E
c.) E
d.) dV/dN
e.) aN
f.) rhov in free space
Solution:
a)
Evaluate the Volt potential function at the given point P, by simply plugging the values of position P. The following:
b)
To compute the Electric field from Volt potential we have the following relation:
E = - ∀.V
Where, ∀ is a del function which denoted:
∀ =
Hence,
Plug in the values for point P:
c)
The magnitude of the Electric Field is given by:
E = √((-10)^2 + (17.32)^2 + (24)^2)
E = √975.9824
E = 31.241 N/C
d)
The dV/dN is the Field Strength E in normal to the surface with vector N given by:
dV / dN = | - ∀.V |
dV / dN = | E | = 31.2 N/C
e)
a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:
f)
The charge density in free space:
p_v = E.∈_o
Where, ∈_o is the permittivity of free space = 8.85*10^-12
p_v = 31.24.8.85*10^-12
p_v = 276 pC / m^3