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3 years ago

Please help is due tonight

Engineering

2 answers:

OverLord2011 [107]3 years ago

8 0

The answer to your question is B.

Kipish [7]3 years ago

7 0

Answer:

tHE answer is b

Explanation:

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

8. Describe and correct the error in stating the domain. Xf * (x) = 4x ^ (1/2) + 2 and g(x) = - 4x ^ (1/2) The domain of (f + g)

konstantin123 [22]

Answer:

Explanation:

4 0

2 years ago

A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K

Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

$\Delta S=\dfrac{Q}{T}$

Now by putting the values

$20.3=\dfrac{Q}{400}$

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0

2 years ago

I'm so confused to this question <br>What is the difference between Science and Engineering?

muminat

Explanation:

Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process that solves a problem and fulfills a need (i.e. a technology).

4 0

2 years ago

Read 2 more answers

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using

serious [3.7K]

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

6 0

3 years ago