1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Paha777 [63]
3 years ago
12

Which of the following is an example of an iterative process?

Engineering
1 answer:
Jlenok [28]3 years ago
3 0
Answer:

-testing the solution and redesigning the prototype
You might be interested in
6. Question
valkas [14]

Answer:

Check  the 2nd, 3rd and 4th statements.

Explanation:

4 0
3 years ago
Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca
Andru [333]

Answer:

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

Explanation:

<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

Capacitance of the second capacitor C_{2}= 7.4цF  = 7.4 x 10-6 F

Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

Capacitance of a series combination is given by:

C_{s}=\frac{Q}{V}

Then the charge stored in the series combination is:  

Q=C_{s} V

Energy stored in the series combination is:  

U_{c}=\frac{1}{2}  V^{2} C_{s}

<u>Solve the Problem:  </u>

Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

<u>b)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

U_{c}=\frac{1}{2} V^2C_{p}

<u>Solve the Problem: </u><em>  </em>

Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J

5 0
3 years ago
Read 2 more answers
If you are convicted of D.U.I. a second time in five years, your license may be revoked for up to __________ year/s.
LuckyWell [14K]
If you are convicted of a dui for a second time in five years, your license may be revoked for up to (2) years
5 0
3 years ago
1. A pipeline constructed of carbon steel failed after 3 years of operation. On examination it was found that the wall thickness
jek_recluse [69]

Answer:

check the explanation

Explanation:

1.

Thickness Loss = t =\frac{t_{o}-t_{i}}{2} = \frac{114.3-102.3}{2} = 2mm

t_{f} = \frac{1}{2}*6 = 3mm

Hence Rate of Corrosion = 6*\frac{1-0.5}{3} = 1mm/year = 0.03 inches per year

2.

As the expected future life is 7 years,

40 carbon steel pipe has to be replaced every 3 years as given in the question,

Cost per unit length is the sum of material cost and installation cost.

Cost of 40 carbon steel = (5 dollars + 16.5 dollars) * 3 = 64.5 dollars

For 80 carbon steel pipe, first calculate the thickness loss,

\frac{114.3-97.2}{2} = 8.55mm

The critical thickness is given to be 3mm, Hence change in thickness is 8.55-3 = 5.5mm

This 80 carbon steel pipe has to be replaced one more time

Hence, Cost per unit length is the sum of material cost and installation cost.

Cost of 80 carbon steel = (8.3 dollars + 16.5 dollars) * 2 = 49.6 dollars

The best is of stainless steel which does not undergo corrosion at all and thus it needs to be replaced only once throughout the plant operation. Its cost = 24.8 dollars + 16.5 dollars = 41.3 dollars

Hence, stainless steel is the recommended pipe to be used.

3 0
3 years ago
Oleg is using a multimeter to test the circuit branch you just installed. After turning off the current to the circuit at the se
irakobra [83]

Answer:

Option D, Ground Wire

Explanation:

Oleg is a mustimeter or multi tester or a basic voltage tester which is used to test the ground. The main purpose of the tester is to ensure that the outlet is connected to the ground of the circuit, thereby ensuring its proper functioning.

Hence, option D is correct

7 0
3 years ago
Other questions:
  • 5. Identify the pros and cons of<br> manufactured siding.
    12·1 answer
  • Ayuda con este problema de empuje y principio de arquimedes.
    6·1 answer
  • Mobility refers to the ability to?
    12·1 answer
  • Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra
    12·1 answer
  • 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/
    15·1 answer
  • (a) Sabbir usually (sit)______ in the front bench.
    13·1 answer
  • Some General Motors flex fuel vehicles do not use a fuel sensor to measure the percentage of ethanol in the fuel. These vehicles
    5·1 answer
  • What is the output of the following program fragment. Choose appropriate data-types of variables to match output.
    10·1 answer
  • What computer program can you use to publish and share a research project with others?
    7·1 answer
  • Which one of the following best defines hardness: (a) energy absorbed by a material when an object strikes its surface, (b) resi
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!