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muminat
3 years ago
5

Where are the ar manufacturers not fitting the engine in the high end sport cars

Engineering
1 answer:
fomenos3 years ago
8 0

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

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Can someone help me with this maze shown below.
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3 years ago
Select the option below that contains the correct order or sequence of events. A) 1) Add host to subscription, 2) Use host as re
butalik [34]

The  option that has the correct order or sequence of events are;

1) Scan host.

2) Add host to subscription.

3) Use host as report source.

<h3>What are host assets?</h3>

Host assets are known to be a kind of IP addresses in a person's account. This type of assets is one that can be used as a kind of  scan, map and also to report targets.

A person can view the hosts in their account by going the  Assets panel to  the Host Assets. One can add new hosts by the use of the New menu. The person can also only carry out actions like scanning, mapping and reporting on one's hosts.

Learn more about IP addresses from

brainly.com/question/24930846

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The electrical panel schedules are located on EWR Plan number ___.
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3 years ago
Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure
Klio2033 [76]

Answer:

A) ΔS_refrigerant = 0.70754 Kj/K

B) ΔS_space = -0.68441 Kj/K

C) ΔS_total = 0.02313 Kj/K

Explanation:

A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C

Converting to degree kelvin yields;

T = -18.77 + 273 = 255.23 K

Formula for entropy change of refrigerant is given as;

ΔS_refrigerant = Q_in/T_refrigerant

We are given Q = 180 KJ

Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K

B) Formula for entropy change of cooled space is given as;

ΔS_space = Q_out/T_s pace

T_space = -10°C = 273 - 10 = 263K

Thus, ΔS_space = -180/263 = -0.68441 Kj/K

C) the total entropy change would be;

ΔS_total = ΔS_refrigerant + ΔS_space

Thus,

ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K

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