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olya-2409 [2.1K]
3 years ago
5

In engineering economic analysis, it is permissible to sum dollar amounts without conversion only when __________ .

Engineering
1 answer:
Mashutka [201]3 years ago
6 0

Answer:

<u>they occur at the same time</u>

Explanation:

What this means is that if let's say on a particular cash record we have two transactions on the same date: For example,

  • 11th/20/2020 received $50 from Jude and received $950 from Paul to pay his credit balance.
  • 12th/20/2020

Only the two transactions on the 11th are permissible to sum the dollar amounts without conversion.

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Tires can be recycled instead of thrown out.<br> True<br> False
Arisa [49]

Answer:

True :)

Explanation:

You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.

Hope it helped have a nice day! :)

8 0
3 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
Risk Management in a Business ModelLearning Objectives and OutcomesCreate a report documenting various aspects of how risk mana
zubka84 [21]

Answer:

The question is explained in detailed way in explanation section and in attached files.

Explanation:

The HIPAA Security Rule is designed to be flexible and appropriate for our organization’s particular size, structure, and inherent risks to business and personal information. Risk analysis is meant to be an ongoing process, during which we regularly review our records to track access to business and personal systems and data. With this in mind, I recommend that we expand our information security strategy to include more than just what is required in HIPAA. Just as a reminder below is the HIPAA ecompliance and implementation strategy that we came up with last week as given in attached file 1.

There are several areas in IT security that the above is incomplete or insufficient in. We recommend implementing several more complete or alternative controls in order to protect our systems, patients, employees, contractors, vendors, and assets beyond the HIPAA minimum requirements. The below section describes what the Centers for Medicare and Medicaid Services (CMS) recommend as additional areas to focus on in the effort to increase an organization's security. (See the attached file # example of some of the areas that we should monitor beyond what HIPAA requires are given in attached file # 03.

3 0
3 years ago
A strip ofmetal is originally 1.2m long. Itis stretched in three steps: first to a length of 1.6m, then to 2.2 m, and finally to
andreyandreev [35.5K]

Answer:

strains for the respective cases are

0.287

0.318

0.127

and for the entire process 0.733

Explanation:

The formula for the true strain is given as:

\epsilon =\ln \frac{l}{l_{o}}

Where

\epsilon = True strain

l= length of the member after deformation

l_{o} = original length of the member

<u>Now for the first case we have</u>

l= 1.6m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{1.6}{1.2}

\epsilon =0.287

<u>similarly for the second case we have</u>

l= 2.2m

l_{o} = 1.6m   (as the length is changing from 1.6m in this case)

thus,

\epsilon =\ln \frac{2.2}{1.6}

\epsilon =0.318

<u>Now for the third case</u>

l= 2.5m

l_{o} = 2.2m

thus,

\epsilon =\ln \frac{2.5}{2.2}

\epsilon =0.127

<u>Now the true strain for the entire process</u>

l=2.5m

l_{o} = 1.2m

thus,

\epsilon =\ln \frac{2.5}{1.2}

\epsilon =0.733

6 0
3 years ago
Determine the mass density of an oil if 0.3 tonnes of the oil occupies a volume of 4m.
Yuki888 [10]

Answer:

Do you mean 4m^3 and 3.0 tones?

Explanation:

solution:

Mass = m = 3.0 tones

- 1 ton = 1,000 kg

= 3.0 × 1,000

= 3,000 kg

volume = v = 4m^3

Required:

Mass density of oil = p = ?

We know that;

p =  \frac{mass}{volume} =  \frac{m}{v} =  \frac{3000}{4} = 750kg |m^{3} ans

The answer is:

750kg / m^3

8 0
3 years ago
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