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Arlecino [84]
3 years ago
9

Your boss tells you that she needs a decision by the end of the day about the machine you want to purchase for your new operatio

n. You have four options, and your boss reminds you to be as cost-efficient as possible but that she also wants the best efficiency. Which is your best option?
Option 1: This $5,000 machine is supplied energy at a rate of 1,900 W and does useful work at a rate of 1,786 W.
Option 2: This $4,500 machine is supplied energy at a rate of 1,450 W and does useful work at a rate of 1,300 W.
Option 3: This $4,750 machine is supplied energy at a rate of 1950 W and does useful work at a rate of 1833 W.
Option 4: This $4,750 machine is supplied energy at a rate of 2,000 W and does useful work at a rate of 1,822 W.

    
  A. Option 4
  B. Option 3
  C. Option 1
  D. Option 2
Physics
2 answers:
Dvinal [7]3 years ago
6 0
It might be B. Option 3
KatRina [158]3 years ago
4 0

Explanation:

It is given that, your boss tells you that she needs a decision by the end of the day about the machine you want to purchase for your new operation.

Option 1.

This $5,000 machine is supplied energy at a rate of 1,900 W and does useful work at a rate of 1,786 W.

Efficiency = (output power/ input power) × 100

Here, output power is 1786 W and input power is 1900 W

So, \eta=\dfrac{1786}{1900}\times 100=94\%

Option 2.

This $4,500 machine is supplied energy at a rate of 1,450 W and does useful work at a rate of 1,300 W.

Efficiency = (output power/ input power) × 100

Here, output power is 1300 W and input power is 1450 W

So, \eta=\dfrac{1300}{1450}\times 100=89\%

Option 3.

This $4,750 machine is supplied energy at a rate of 1950 W and does useful work at a rate of 1833 W.

Efficiency = (output power/ input power) × 100

Here, output power is 1833 W and input power is 1950 W

So, \eta=\dfrac{1833}{1950}\times 100=94\%

Option 4.

This $4,750 machine is supplied energy at a rate of 2,000 W and does useful work at a rate of 1,822 W.

Efficiency = (output power/ input power) × 100

Here, output power is 1822 W and input power is 2000 W

So, \eta=\dfrac{1822}{2000}\times 100=91.1\%

Since, both option (3) and (1) is 94 % but in option (c) the cost is less and is giving maximum efficiency. So, the correct option is (b) " option 3."

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