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Effectus [21]
4 years ago
12

The primary difference between integrated/concurrent engineering development and functional/sequential development is: Integrate

d development spends more money up front to order to save money in later stages. Sequential development is more thorough in addressing development steps. In sequential development, multiple functions co-manage design and testing. Integrated development usually leads to higher sustaining and warranty costs.
Engineering
1 answer:
gogolik [260]4 years ago
7 0

Answer:

Integrated development spends more money up front to order to save money in later stages.

Explanation:

The primary difference between integrated/concurrent engineering development and functional/sequential development is: Integrated development spends more money up front to order to save money in later stages.

Cost of integrated development is higher in the initial stages where setup has be made and proper training and education of employees are required and once it is done. It will save money in later stages.

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5. Assume that you and your best friend ench have $1000 to invest. You invest your money
Bezzdna [24]

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

<u>Answer:</u>

<u>(d) She has $10 more than you do</u>

<u>Explanation</u>:

Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

<u>My compound interest is:</u>

= 1000 [ (1-0.1)^1-1

= $1000

$1,000 + $1,000 invested= $2,000 total amount received.

<u>My friend's simple interest is;</u>

To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

5 0
3 years ago
Answer back to question for la ,lot points
Lerok [7]

Answer:

yes

Explanation:

yes

5 0
3 years ago
Vẽ thủ tục cho một cuộc gọi thuê bao
shusha [124]

Lo siento, no sé qué estás diciendo.

8 0
3 years ago
A reversible compression of 1 mol of an ideal gas in a piston/cylinder device results in a pressure increase from 1 bar to P2 an
Mashutka [201]

Answer:

attached below

Explanation:

6 0
3 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0
3 years ago
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