Answer: 0.3872m
Explanation:
q= 100nC --> 100x10^-9 C
k= 9x10^9 Nm^2/C^2
E= 6kN/C --> 600 N/C
r=?
--> 
Despejas "r"
Resuelves
<h3>
(la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>
Answer:
a. Zin = 41.25 - j 16.35 Ω
b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c. Pin = 216 w
d. PL = Pin = 216 w
e. Pg = 478.4 w , Pzg = 262.4 w
Explanation:
a.
Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]
βl = 2π / λ * 0.15 λ = 54 °
Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]
Zin = 41.25 - j 16.35 Ω
b.
I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶
V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)
V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c.
Pin = ¹/₂ * Re * [V₁ * I₁]
Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)
Pin = 216 w
d.
The power PL and Pin are the same as the line is lossless input to the line ends up in the load so
PL = Pin
PL = 216 w
e.
Pg Generator
Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)
Pg = 478.4 w
Pzg dissipated
Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50
Pzg = 262.4 w
Answer:
1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]
Explanation:
1. We know that the density is defined by the following expression.
![Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]](https://tex.z-dn.net/?f=Density%20%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cmass%3D90%5Bkg%5D%5C%5Cvolume%3D0.075%5Bm%5E%7B3%7D%20%5D%5C%5Cdensity%3D%5Cfrac%7B90%7D%7B0.075%7D%20%5C%5Cdensity%3D1200%5B%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%20%7D%5D)
2. First we need to convert the units to meters.
wide = 35[cm] = 35/100 = 0.35[m]
long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]
Thick = 15[mm] = 15/1000 = 0.015[m]
Now we can find the density using the expression for the density.
![density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]](https://tex.z-dn.net/?f=density%3D%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5Cwhere%3A%5C%5Cvolume%20%3D%20wide%2Along%2Athick%5C%5Cvolume%3D0.35%2A1.1%2A0.015%20%3D%200.005775%5Bm%5E3%5D%5C%5C%5C%5Cmass%3D%20density%2Avolume%20%3D%202700%2A0.005775%20%3D%2015.59%5Bkg%5D)
Answer:
Explanation:
From the question we are told that
Generally the equation for momentum is mathematically given by
Therefore
T-Joe momentum 
Answer:
21.3 V, 1.2 A
Explanation:
1.
These resistors are in series, so the net resistance is:
R = R₁ + R₂ + R₃
R = 20 + 30 + 45
R = 95
So the current is:
V = IR
45 = I (95)
I = 9/19
So the voltage drop across R₃ is:
V = IR
V = (9/19) (45)
V ≈ 21.3 V
2.
First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:
1/R₂₃ = 1/R₂ + 1/R₃
1/R₂₃ = 1/10 + 1/10
R₂₃ = 5
Now we find the overall resistance by adding the resistors in series:
R = R₁ + R₂₃ + R₄
R = 10 + 5 + 10
R = 25
So the current through R₁ is:
V = IR
30 = I (25)
I = 1.2 A
