Answer:
Explanation:
Specific Volume 
Absolute Pressure (a) 
Giving
(b) 
Giving
(a)
Generally the equation for quality of Steam X is mathematically given by
(b)
Generally the equation for quality of Steam X is mathematically given by
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:
Explanation:
1) Force Friction = Normal Force * Coefficient of Friction
Force Friction = Mass * Gravity * Coefficient of Friction
2) F = ma
Force = mass * acceleration
Force Friction (from #1) = mass * acceleration
acceleration = Force Friction / Mass
solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
Acceleration, a = (v - u)/t
where v is the final velocity, u is the initial velocity, and t is the time.
This formula on a velocity time graph represents the slope of the graph.
