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: B) The ability of steel to be shaped
Answer:
The sun
Explanation:
In this system the energy of the sun heats the water in the pipe, producing a high pressured steam, which is used for moving a turbine and producing electricity, is a transformation of energy from solar to thermal, then to mechanical to electrical.
Answer:
Opposition of passing a electric circuit
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
