Answer:
Explanation:
The options are:
- In an isometric drawing, multiple angles and axes can be shown in one sketch.
- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.
- Only one sketch will be needed since all other previous designs will no longer be necessary.
- Computer programs will not be necessary to create the exact dimensions of the design.
Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view. Isometric drawings are perspective views from different angles.
So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."
Answer:
See attachment for completed question
Explanation:
Given that; Brainly.com
What is your question?
mkasblog
College Engineering 5+3 pts
The dry unit weight of a soil sample is 14.8 kN/m3.
Given that G_s = 2.72 and w = 17%, determine:
(a) Void ratio
(b) Moist unit weight
(c) Degree of saturation
(d) Unit weight when the sample is fully saturated
See complete solving at attachment
load every electric circuit,regardless of where it is or how large or small, has four basic parts: an energy source (ac or dc),a conductor (wire), an electrical load (device), and at least one controller(switch)
How to create a personal hot spot on an iPhone?
Go to Settings | Cellular | Personal Hotspot.
Tap the slider next to Allow Others to Join. ...
Your Wi-Fi Password will be shown right underneath the Allow Others to Join option. ...
Now, on another device, such as a laptop, go to the Wi-Fi section and search for nearby networks.
Answer:
A) 282.34 - j 12.08 Ω
B) 0.0266 + j 0.621 / unit
C)
A = 0.812 < 1.09° per unit
B = 164.6 < 85.42°Ω
C = 2.061 * 10^-3 < 90.32° s
D = 0.812 < 1.09° per unit
Explanation:
Given data :
Z ( impedance ) = 0.03 i + j 0.35 Ω/km
positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km
A) calculate Zc
Zc =
=
=
= 282.6 < -2.45°
hence Zc = 282.34 - j 12.08 Ω
B) Calculate gl
gl =
d = 500
z = 0.03 i + j 0.35
y = j4.4*10^-6 S/km
gl = 
= 
= 0.622 < 87.55 °
gl = 0.0266 + j 0.621 / unit
C) exact ABCD parameters for this line
A = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
B = Zc sin h (gl) Ω = 164.6 < 85.42°Ω ( as calculated )
C = 1/Zc sin h (gl) s = 2.061 * 10^-3 < 90.32° s ( as calculated )
D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )
where : cos h (gl) = 
sin h (gl) = 