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2 years ago

6

What is a combination circuit? A combination circuit:

1 answer:

Anon25 [30]2 years ago

7 0

Answer:

Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.

Example:

The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.

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Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi

qaws [65]

Answer:

$\Delta V = 209.151\,L$ , $\Delta C = 217.517\,USD$

Explanation:

The drag force is equal to:

$F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A$

Where $C_{D}$ is the drag coefficient and $A$ is the frontal area, respectively. The work loss due to drag forces is:

$W = F_{D}\cdot \Delta s$

The reduction on amount of fuel is associated with the reduction in work loss:

$\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s$

Where $F_{D,1}$ and $F_{D,2}$ are the original and the reduced frontal areas, respectively.

$\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s$

The change is work loss in a year is:

$\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)$

$\Delta W = 2.043\times 10^{9}\,J$

$\Delta W = 2.043\times 10^{6}\,kJ$

The change in chemical energy from gasoline is:

$\Delta E = \frac{\Delta W}{\eta}$

$\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}$

$\Delta E = 6.81\times 10^{6}\,kJ$

The changes in gasoline consumption is:

$\Delta m = \frac{\Delta E}{L_{c}}$

$\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }$

$\Delta m = 154.772\,kg$

$\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }$

$\Delta V = 209.151\,L$

Lastly, the money saved is:

$\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )$

$\Delta C = 217.517\,USD$

4 0

3 years ago

Express 2/16 in thirty-seconds

mafiozo [28]

Answer:

$\frac{2}{16}$ = $\frac{4}{32}$ in thirty seconds.

Explanation:

one thirty second is one part out of 32 equal section . It is used to describe amounts accurately.

$\frac{2}{16}$ can be easily expressed as $\frac{4}{32}$

3 0

2 years ago

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

4 0

3 years ago

Determine the maximum intensity ww of the uniform distributed load that can be applied to the beam without risk of causing the s

AnnyKZ [126]

Answer:

Please see attachment

Explanation:

Please see attachment

3 0

3 years ago

Which of the following lists the steps of a process in the correct order?

Vladimir79 [104]

Answer: C. Problem identification, solution design, implementation, evaluation.

Explanation:

4 0

2 years ago

Read 2 more answers