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koban [17]
2 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]2 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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8 0
2 years ago
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

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3 0
2 years ago
A 2-bit positive-edge triggered register has data inputs d1, d0, clock input clk, and outputs q1, q0. Data inputs d1d0 are 01 an
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Answer:

  q1q1 ⇒ 01

Explanation:

The outputs of a positive edge triggered register will match the inputs after a rising clock edge.

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The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u
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Answer:

F = 0.0022N

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Given:

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Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

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what do we call a landslide in which fine-grained soil moves cohesively but with extensive internal shearing?
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