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

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:

a)

Identify the unknown: 

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

List the Knowns:

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

Set Up the Problem:

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:

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

Solve the Problem: 

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

b)

Identify the unknown:

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

Set Up the Problem: 

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:

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

Solve the Problem:

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

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