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sveticcg [70]
3 years ago
11

How many D-cell batteries would it take to power a human for 1 day?

Engineering
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

it would take approximately 232 to 258 D cell batteries to power a human for 1 day.

Explanation:

<em>Estimate the recommended daily food intake (in Food Calories).</em>

An average adult man requires between 2000 to 3000 calories per day.

<em>Estimate the daily energy (in joule) a human needs.</em>

As we know 1 food calorie is equal to 4.184 Joules of energy

2000*4.184 to 3000*4.184

8368 to 12552 Joules

But for engineering calculations 1 food calorie is equal to 1000 engineering calories, so

8368*1000 to 12552*1000

8368000 to 12552000 Joules

<em>Estimate the voltage (in volt) of one D cell battery</em>.

The voltage of a D cell battery is around 1.5 Volts

<em>Estimate the charge (in amp-hours) of one D cell battery.</em>

The amp-hours of a D cell battery varies with the manufacturing company, the typical amp-hours are in the range of 6 to 10 amp-hours.

<em>Estimate the energy of one D cell battery (in watt-hour).</em>

Energy in watt-hour is given by

voltage*amp-hour

1.5*6 to 1.5*10

9 to 15 watt-hour

<em>Estimate the number of D-cell batteries it takes to power a human for 1 day.</em>

First let us calculate the energy in a D cell battery,

1 watt-hour is equal to 3600 Joules

9*3600 to 15*3600

32400 to 54000 Joules

The number of D cell batteries required is found by dividing the energy need of a human by the energy stored in a D cell battery.

12552000/54000 to 8368000/32400

232 to 258 batteries

Therefore, it would take approximately 232 to 258 D cell batteries to power a human for 1 day.

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Answer:

Using the above algorithm matches one pair of Ghostbuster and Ghost. On  each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are  the same, so use the algorithm recursively on each side of the line to find pairings. The  worst case is when, after each iteration, one side of the line contains no Ghostbusters  or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P(n^{2} lg n)-  time algorithm.

4 0
3 years ago
A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of
yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0
3 years ago
When a retaining structure moves towards the soil backfill, the stress condition is called:__________.
Alecsey [184]

Answer:

(C) passive state.

Explanation:

The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:

a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.

b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest

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6 0
2 years ago
How many types of engineers are there
Fiesta28 [93]
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6 0
3 years ago
Read 2 more answers
Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three
vesna_86 [32]

Answer:

the evaluation in SI unit will be 2.69\times 10^{-5}sec^{2}

Explanation:

We have evaluate \frac{(204mm\times 0.00457kg)}{34.6N}

We know that 1 mm =10^{-3}m

So 240 mm =204\times 10^{-3}m

Newton can be written as kgm/sec^2

So \frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}

So the evaluation in SI unit will be 2.69\times 10^{-5}sec^{2}

4 0
3 years ago
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