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koban [17]
3 years ago
5

Suppose that units of force, length, and time are chosen such that the density of water and the acceleration of gravity are both

of unit magnitude. If the pound is taken as the unit of force, what are the sizes of the units of length and time?
Engineering
1 answer:
k0ka [10]3 years ago
3 0

Answer: So a 1 pound = 0.453592kg

the density of the water is D= 997 kg/m^{3} so we need to use replace the kg with pounds with 2.20462 pounds = 1kg.

then D = 2.20462*997 pounds/m^{3}

and we want to make this equal to one changing the metters for other thing.

we need to replace x^{3} with 2.020462*997 meters qube.

then our new metters variable x = 12.6meters.

now g = 9.8 meters for second square.

in our new lenght variable x.

g = 12.6*9.8 x for second square.

so if y is our new time variable

y^{2} = 12.6*9.8 seconds square. so y = 11.1seconds

so the units of length are x = 12.6 meters and for time y = 11.1 seconds.

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See explaination

Explanation:

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please kindly see attachment for the step by step solution of the given problem.

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A screw is a simple machine that can be described as an inclined plane wrapped around a cylinder.
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3 years ago
An LED camping headlamp can run for 18 hours, powered by three AAA batteries. The batteries each have a capacity of 1000 mAh, an
KIM [24]

Answer:

a) the power consumption of the LEDs is 0.25 watt

b) the LEDs drew 0.0555 Amp current

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<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------

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a) the power consumption of the LEDs

I_battery = 1000 mAh / 18hrs    { for 18 hrs}

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we substitute

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P = 0.25 watt

Therefore the power consumption of the LEDs is 0.25 watt

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5 0
3 years ago
The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint
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Answer:

volume of biological sludge = 28.566 m³ per day

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mass of solid = 7240 kg/day

initial moisture content = 78%

solution

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% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = \frac{100}{100 - P} M kg  per day

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mass of sludge produced = \frac{100}{100 - 78} 7240 kg

mass of sludge produced = 32909.09 kg

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and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}

S sludge = 1.152

so that

density of sludge = S sludge × density of water

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density of sludge = 1152 kg/m³

so that

volume of biological sludge = \frac{mass sludge produce}{\rho sludge}

volume of biological sludge = \frac{32909.09}{1152}

volume of biological sludge = 28.566 m³ per day

6 0
3 years ago
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