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garri49 [273]
3 years ago
11

Zack's new home is progressing well, and the foundation work is finished. The general contractor stops by Zack's present home to

let him know that the lumber order will be dropped off in two days. Almost as a side comment, the contractor asks if Zack has decided what he wants for wall sheathing. Zack says he is not sure what would be best and asks for a recommendation. What do you think the contractor will offer as options? Which option would be best?
Engineering
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

hahajajskekjekdjwjhjdjdjsiahshjsjs

thank me later

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How is the minimum circuit ampacity for an air-conditioning condensing unit calculated?
horsena [70]
The answer is 1. Hopefully it helps!
5 0
3 years ago
Smith wants to get a job in network systems. Which major should he choose to give him the best chance of getting a job?
frez [133]
Computer engineering
7 0
3 years ago
Read 2 more answers
sa kasalukuyang panahon posible pakayang mag karoon ng panibagong kilusang propaganda at lilusang katipunan oo?at bakit​
gladu [14]

Answer:

?

Explanation:

6 0
3 years ago
Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/
kap26 [50]

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of  vaporisation .

Sensible heat for water Q

Q=mC_p\Delta T

For water

C_p=4.178\ KJ/Kg.K

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

8 0
4 years ago
A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para
mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

V_p=\frac{3810.5}{\sqrt{3} }=2200 V

The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA

The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

The phase voltage at the sending end is:

V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

The magnitude of the line voltage at the source end of the line (V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V

b) The Total real and reactive power loss in the line is:

S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0
4 years ago
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