Answer:
Gross building area
Explanation:
The Gross building area refers to the entire area of a building covering all the floors. The measurement is expressed in square feet. The Gross building area also includes basements, penthouses, and mezzanines. It is calculated by estimating the exterior dimension of the building. Storage rooms, laundries, staircases are also a part of the gross building area.
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
Answer:
v₀ = 2,562 m / s = 9.2 km/h
Explanation:
To solve this problem let's use Newton's second law
F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v
F dx = m v dv
We replace and integrate
-β ∫ x³ dx = m ∫ v dv
β x⁴/ 4 = m v² / 2
We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max
-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)
x_max⁴ = 2 m /β v₀²
Let's look for the speed that the train can have for maximum compression
x_max = 20 cm = 0.20 m
v₀ =√(β/2m) x_max²
Let's calculate
v₀ = √(640 106/2 7.8 104) 0.20²
v₀ = 64.05 0.04
v₀ = 2,562 m / s
v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)
v₀ = 9.2 km / h
The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.
<u>Explanation</u>:
<u>Given</u>:
tensile stress is applied parallel to the [100] direction
Shear stress is 0.5 MPA.
<u>To calculate</u>:
The magnitude of applied stress in the direction of [101] and [011].
<u>Formula</u>:
zcr=σ cosФ cosλ
<u>Solution</u>:
For in the direction of 101
cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)
cos λ = 1/√2
The magnitude of stress in the direction of 101 is 12.25 MPA
In the direction of 011
We have an angle between 100 and 011
cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)
cosλ = 0
Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.
Answer:
For aluminum 110.53 C
For copper 110.32 C
Explanation:
Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:
Where
q: heat transferred
k: conduction coeficient
A: surface area
th: hot temperature
tc: cold temperature
d: thickness of the plate
Rearranging the terms:
d * q = k * A * (th - tc)
The surface area is:
If the pan is aluminum:
If the pan is copper:
