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:
D
that's should be the answer
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as

Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is


At the same time since Newton's second law we know that:

Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:




But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:


Therefore there is a weight loss of 0.3N every kilometer.
Answer:
The magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F
Explanation:
Given;
magnitude of the magnetic force is F at 14.0⁰ with respect to the field.
To determine the magnitude of the magnetic force at 32.5⁰ to the field, we apply the following formula and solve with proportion;
f*Sin(14.0⁰) = F
f*Sin(32.5⁰) = ?

Therefore, the magnitude of the magnetic force at 32.5⁰ to the field, is 2.22 F
Answer:
The work input during this process is -742 kJ
Explanation:
Given;
Initial temperature of nitrogen T₁ = 250 K
final temperature of nitrogen T₂ = 450 K
mass of nitrogen, m = 5 kg

The work input during the process is calculated as;

where;
R is gas constant = 0.2968 kJ/kgK
substitute given values in above equation.

Therefore, the work input during this process is -742 kJ