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3 years ago

15

james wants to qualify for icp are and licensure. Which degree would be required in order to qualify for a two year master of ar

chitecture degree?

1 answer:

ololo11 [35]3 years ago

4 0

Answer:

A degree in architecture with 60 credit hours.

Explanation:

The requirements need for a student to qualify for a two year master of architecture degree are;

60 credit hours in architecture
Complete 60 credit hours in related area of profession such as; planning, landscape architecture ,public health and others.
45 credit hours in architecture course at the level of 500/600

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A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in

Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, $\sigma=\frac {F}{A}$

where F is applied force and A is original area

Also, engineering strain, $\epsilon=\frac {\triangle l}{l}$ where l is original area and $\triangle l$ is elongation

We also know that Hooke's law states that $E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

$\triangle l= 0.1 mm= 0.1\times 10^{-3} m$

By substitution we obtain

$E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa$

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3 years ago

A rotor in a compressor stage has a mean blade radius of 0.285 m and an angular rotor velocity of 8500 RPMs. The static temperat

pantera1 [17]

Answer:

0,285 is the answer

Explanation:

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3 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

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3 years ago

Who plays blox burg???

xeze [42]

Answer:

I don't have robux

Explanation:

but i love adopt me

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3 years ago

Read 2 more answers

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago