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3 years ago

A process generates 250 watts of useful energy and 600 watts of waste energy. WhT is the efficiency of the process?

Physics

1 answer:

Cloud [144]3 years ago

3 0

The efficiency of the process is 42%

From the question given above, the following data were obtained:

Useful energy = 250 watt

Wasteful energy = 600 watt

<h3>Efficiency =? </h3>

The efficiency of the process can be obtained as illustrated below:

$efficiency \: = \frac{useful \: energy}{wasteful \: energy} \times 100 \\ \\ efficiency \: = \frac{250}{600} \times 100 \\$

<h3>Efficiency = 42%</h3>

Therefore, the efficiency of the process is 42%

Learn more: brainly.com/question/2300041

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3 years ago

Batteries are rated in terms of ampere-hours (A·h). For example, a battery that can produce a current of 2.00 A for 3.00 h is ra

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(a) 423 J

The power of the battery is the ratio between the total energy stored (E) and the time elapsed (t):

$P=\frac{E}{t}$

However, the power is also the product of the voltage (V) and the current (I):

$P=VI$

Linking the two equations together,

$\frac{E}{t}=VI\\E=VIt$

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(b) $7.79\cdot 10^{-6}$ dollars

The battery has a total energy of E = 423 J. (2)

1 Watt (W) is equal to 1 Joule (J) per second (s):

$1 W = \frac{1 J}{1 s}$

so 1 kW corresponds to 1000 J/s:

$1 kW = \frac{1000 J}{1 s}$

Multiplying both side by 1 hour (1 h):

$1 kW \cdot h = \frac{1000 J}{1 s} 1 h$

and $1 h = 3600 s$ , so

$1 kWh = \frac{1000 J}{1 s}\cdot 3600 s =3.6\cdot 10^6 J$

So we find the conversion between kWh and Joules. So now we can convert the energy from Joules (2) into kWh:

$1 kWh = 3.6\cdot 10^6 J = x : 423 J\\x=\frac{1 kWh \cdot 423 J}{3.6\cdot 10^6 J}=1.18\cdot 10^{-4}kWh$

And since the cost is $0.0660 per kilowatt-hour, the total cost will be

$C=$0.0660\cdot 1.18\cdot 10^{-4} kWh=7.79\cdot 10^{-6}$ dollars

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