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2 years ago

What are the importance of crystallography in geology

1 answer:

8 0

Answer:Crystallography is useful in phase identification. When performing any process on a material, it may be desired to find out what compounds and what phases are present in the material. Each phase has a characteristic arrangement of atoms.

Explanation:

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How does warming up the tires on a car increase grip with the pavement?

ICE Princess25 [194]

Answer:

because burning rubber increases the grip power

8 0

3 years ago

I need help with this question

Ad libitum [116K]

Answer:

LOL where is the question, that u need help with?

Explanation:

5 0

3 years ago

Briefly discuss if it would be better to operate with pumps in parallel or series and how your answer would change as the steepn

Aleksandr [31]

Answer:

1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.

Explanation:

1) Pump in series:

When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.

In the first diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3.

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.

- Point 3 is the point where the system is operating when both pumps are running.

Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) Pump in parallel:

When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.

In the second diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.

- Point 3 is the point where the system is operating when both pumps are running.

In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.

5 0

3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$

$w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um$

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

3 years ago

The cubic capacity of a four-stroke over-square spark-ignition engine is 245cc. The over-square ratio is (1.1) The clearance vol

pashok25 [27]

Answer:

Bore = 7 cm

stroke = 6.36 cm

compression ratio = 10.007

Explanation:

Given data:

Cubic capacity of the engine, V = 245 cc

Clearance volume, v = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

V = $\frac{\pi}{4}D^2L$

V = $\frac{\pi}{4}\frac{D^3}{1.1}$

on substituting the values, we have

245 = $\frac{\pi}{4}\frac{D^3}{1.1}$

D = 7.00 cm

Now,

we have

D/L = 1.1

thus,

L = D/1.1

L = 7/1.1

L= 6.36 cm

Now,

the compression ratio is given as:

$\textup{compression ratio}=\frac{V+v}{v}$

on substituting the values, we get

$\textup{compression ratio}=\frac{245+27.2}{27.2}$

Compression ratio = 10.007

4 0

4 years ago