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3 years ago

9

2.44mW of incident 520 nm light is directed through a1 cm sample cuvette and 0.68 mW of Plight exits the sample what is the abso

rbance of the sample?

1 answer:

Flura [38]3 years ago

7 0

Answer:

?/?

Explanation:??

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Solid Isomorphous alloys strength

Sphinxa [80]

Answer:

Explanation:

ℎ

3 0

3 years ago

A mixing basin in a sewage filtration plant is stirred by a mechanical agitator with a power input/WF L T=. Other parameters de

MakcuM [25]

Answer: π= G[√(u.V/W)]

STEP 1

Given parameters:

Power Input W= FL/T,

Absolute Viscosity u= FT/L²

Basin volume V= V/L³

Velocity gradient G= V/L³

STEP 2

We start by expressing the velocity gradient G as a function of W, u, V

G= G(W,u,V)

To get the pii terms, we use the dimension number formula n=k - r

where n and k are natural numbers representing number of fundamental dimensions and variable present respectively.

n= 4-3=1

STEP 3:

We expressed the pii terms as

π= G.W^a.u^b.V^c

The three fundamental F L T

We can write as

Fⁿ.Lⁿ.Tⁿ= 1/T. (FL/T)^a.(FT/L²)^b.(L³)

Using the exponential rule and by comparing coefficient on both sides;

Fⁿ.Lⁿ.Tⁿ= F^a+b. L^a-2b+3c. T^-a+b-1

Fⁿ= F^a+b = a+b= 0..............I

Lⁿ= L^a-2b+3c=0 = a-2b+3c=0...........ii

Tⁿ=L^-a+b-1=0. -a+b-1=0............iii

From the above equations we have,

a+b =0: b=-a...........iv

putting eq. iv into iii , we have

-a-a-1=0: -2a-1=0: a= -1/2

substituting the above value of a into eq iv, we have

b= 1/2

substituting the value of b above into eq 2, we have,

-1/2-2(1/2)+3c=0

c=1/2.

Lastly, from the pii terms given above we can obtain dimensionless relationship,

π=G(W^-1/2.u^1/2.V^1/2)

We can write this as

π= G[ √1/W.√u. √1/2] = G[(√u.V/√W)] or G[√(u.V/W)].... final answer.

5 0

4 years ago

Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

3 0

3 years ago

If it is desired to lay off a distance of 10,000' with a total error of no more than Â± 0.30 ft. If a 100' tape is used and the

Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0

3 years ago

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

load current in primary = 31.38 < 25.84 A

<u>C) find the input power factor </u>

power factor = 0.9323 leading

<em></em>

<em>attached below is the detailed solution </em>

8 0

3 years ago