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2 years ago

I have a question.What does DIY mean?

Engineering

2 answers:

tino4ka555 [31]2 years ago

8 0

Answer:

DO IT Yourself

Explanation:

alexdok [17]2 years ago

6 0

Answer: It means "Do it yourself".

Explanation: You're welcome!

You might be interested in

What are some common ways of converting fuel to useful energy? Check all that apply.

Rainbow [258]

Answer: Burning it.

Explanation: When your car moves its due to the burning of fuel, works like a turbine the molecules evaporate and turn a turbine like object.

3 0

3 years ago

Read 2 more answers

According to the decreasing order of toughness. list the following materials (note: the steels are assumed to have no cold work

fiasKO [112]

Answer:

1090 Steel >1040 Steel > Pure aluminium >Diamond.

Explanation:

Toughness:

Toughness can be define as the are of load -deflection diagram up to fracture point.

Modulus of toughness can be defines as the area of stress-strain diagram up to fracture point.Modulus of toughness is the property of material.

So the decreasing order of toughness can be given as follows

1090 Steel >1040 Steel > Pure aluminium >Diamond.

8 0

3 years ago

A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast

insens350 [35]

Answer:

a) The final equilibrium temperature is 83.23°F

b) The entropy production within the system is 1.9 Btu/°R

Explanation:

See attached workings

8 0

3 years ago

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

Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

$\Delta S=-0.287\ln \dfrac{100}{15}$

ΔS = -0.544 KJ/Kg.K

3 0

3 years ago