Answer:
birds-eye view perspective
Explanation:
If someone asked me to design an office building, I would draw it from a birds-eye view perspective. I would draw it this way so I could map out where everything in the office would go and make sure I have enough space for everything. I would also draw it this way in order to clearly see where everything would go in the office. For instance, cubicles/desks could go in the bottom left corner, while the boss's office could go in the top right. It would be easier to organize and it would be easier for me to look back on when I need to actually design the office later.
(i'm not sure if this is what your question is asking for so i just made my best guess)
Answer:
Jordan has more green paints
Explanation:
Given
Required
Which paint does he have more?
For better understanding, it's better to convert both measurements to decimal.
For the green paint:
For the blue paint:
By comparison:
<em>This means that Jordan has more green paints</em>
Answer:
a) ∝ and β
The phase compositions are :
C
= 5wt% Sn - 95 wt% Pb
C
= 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C = 5wt% Sn - 95 wt% Pb
C = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
Answer:
Explanation:
Mountain roads often zigzag across a mountain with a series of sharp turns called. switchbacks.
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
