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serg [7]
3 years ago
9

Which of these passages is an example of cohesive writing?

Engineering
2 answers:
aksik [14]3 years ago
5 0

Answer:

It should be number 1

Oksana_A [137]3 years ago
3 0
The answer mostly likely is: “1” due to the fact it is well ordered, and follows a similar thought process, best of luck!
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5 cause of a electrical problem​
cupoosta [38]

Answer:

1. Poor circuit protection

2.Grounding issue

3. lighting problem

4. Electrical shocks

5. High electricity bills

Explanation:

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3 years ago
Meaning of perfomance bond
nekit [7.7K]

Answer:

a bond issued by a bank or other financial institution, guaranteeing the fulfilment of a particular contract

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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
Was rachel and ross on a break
yaroslaw [1]

Answer:

yes they were on a break

Explanation:

3 0
3 years ago
Please help fast
Vitek1552 [10]
The balloon could not rise. The air inside would not be an lighters than the air
outside.
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3 years ago
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