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damaskus [11]
3 years ago
8

Volume of sale (i.e., the number of parts sold) is a factorwhen determining which

Engineering
1 answer:
nadya68 [22]3 years ago
3 0

Answer: N has to be lesser than or equal to 1666.

Explanation:

Cost of parts N in FPGA = $15N

Cost of parts N in gate array = $3N + $20000

Cost of parts N in standard cell = $1N + $100000

So,

15N < 3N + 20000 lets say this is equation 1

(cost of FPGA lesser than that of gate array)

 Also. 15N < 1N + 100000  lets say this is equation 2  

(cost of FPGA lesser than that of standardcell)

Now

From equation 1

12N < 20000

N < 1666.67

From equation 2

14N < 100000

N < 7142.85

AT the same time, Both conditions must hold true

So N <= 1666 (Since N has to be an integer)

N has to be lesser than or equal to 1666.

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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
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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

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What is Elon Musk mad about?
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Describe each occupation
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Answer:

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5 0
2 years ago
How is engine power expressed?
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Explanation: (I hope this helped!! ^^)

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Suppose values is a sorted array of integers. Give pseudocode that describes how a new value can be inserted so that the resulti
Novosadov [1.4K]

Answer:

insert (array[] , value , currentsize , maxsize )

{

   if maxsize <=currentsize

  {

      return -1

  }

  index = currentsize-1

  while (i>=0 && array[index] > value)

  {

      array[index+1]=array[index]

      i=i-1

  }

 

  array[i+1]=value

  return 0

}

Explanation:

1: Check if array is already full, if it's full then no component may be inserted.

2: if array isn't full:

  • Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.  
  • Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
  • assign new worth to the position that is next to the known position of initial smaller component.
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