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olga55 [171]
3 years ago
6

Why are there few effective HCI standards?

Engineering
1 answer:
Oliga [24]3 years ago
3 0

Answer:

<em>There are few effective Human Computer Interaction (HCI) standards because for one, standards are more suitable for hardware than software because they are relatively unstable. ... Some software product standards have been in place long before any formal standard documents were published.</em>

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. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298
andriy [413]

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

5 0
3 years ago
A(n)_____ is a device that provides the power and motion to manipulate the moving parts of a valve or damper used to control flu
Lesechka [4]

Answer:

Out of the four options provided

option A. actuator

is correct

Explanation:

An actuator is the only device out of the four mentioned devices that provides power and ensures the motion in it in order to manipulate the movement of the moving parts of the damper or a valve used whereas others like ratio regulator are used to regulate air or gas ratio and none mof the 3 remaining options serves the purpose

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3 years ago
Is a jeep cherokee faster than a bmw 325i
soldier1979 [14.2K]

Answer:

Yes

Explanation:

8 0
3 years ago
Read 2 more answers
You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______
stich3 [128]
I believe it’s D) 20 feet
3 0
3 years ago
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
3 years ago
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