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White raven [17]
2 years ago
7

Derive the following conversion factors: (a) Convert a pressure of 1 psi to kPa (b) Convert a vol ume of 1 liter to gallons (c)

Convert a viscosity of 1 lbf .s/ft^2 to N s/m^2
Engineering
1 answer:
kirill [66]2 years ago
4 0

Answer:

a)6.8 KPa

b)0.264 gallon

c)47.84 Pa.s

Explanation:

We know that

1 lbf=  4.48 N

1 ft =0.30 m

a)

Given that

P= 1 psi

psi is called pound force per square inch.

We know that 1 psi = 6.8 KPa.

b)

Given that

Volume = 1 liter

We know that 1000 liter = 1 cubic meter.

1 liter =0.264 gallon.

c)

1\ \frac{lb.s}{ft^2}=47.84\ \frac{Pa.s}{ft^2}

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6 0
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Clarifying the issues of a problem is the _____ step in the problem solving process.
ratelena [41]
The answer is 2nd Step because the first step is to define the problem and third is to define your goals
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Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r
Evgesh-ka [11]

Answer:

Explanation:

In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.

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4 0
3 years ago
The function below takes a two parameters: a list called a_list and a value called a_value. Complete the function to first check
Alik [6]

Explanation:

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if a_value in list://if statement to

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//nothing

else://else statement for if the

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5 0
3 years ago
In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images
Zarrin [17]

Answer:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

A= \frac{r^5 \rho}{t^2}

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant \rho_{air}.

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= \frac{ML^2}{T^2} A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as \frac{L}{T}

[t]=T represent the time

[\rho]=\frac{M}{L^3} represent the density.

Solution to the problem 

And if we analyze the function for r we got this:

[r]=L=[A]^x [\rho]^y [t]^z

And if we replpace the formulas for each on we got:

[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z

And using algebra properties we can express this like that:

[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

x+y =0 , 2x-3y=1, -2x+z=0

We can solve for x like this x=-y and replacing into quation 2 we got:

2(-y)-3y = 1

-5y = 1

y= -\frac{1}{5}

And then we can solve for x and we got:

x = -y = -(-\frac{1}{5})=\frac{1}{5}

And if we solve for z we got:

z=2x =2 \frac{1}{5}=\frac{2}{5}

And now we can express the radius in terms of the dimensional analysis like this:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}

A= \frac{r^5 \rho}{t^2}

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed \rho =1.2 kg/m^2, and replacing we got:

A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}

And we can convert this into ergs we got:

A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs

And then we know that 1 g of TNT have 4x10^4 erg

And we got:

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

3 0
3 years ago
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