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3 years ago

What Number Am I?

Engineering

1 answer:

noname [10]3 years ago

5 0

Answer:

The number is 768

Explanation:

The given parameters of the three digits number are;

The number is a multiple of four, therefore, a factor of the number = 4

The value of the first digit = A prime number larger than 5

The sum of the last two digits > 10

The value of the ones (place value) digit > The tens (place value) digit

Let 'xyz' represent the digits of the three digits number, we have;

x = A prime number digit larger than 5

∴ x = 7

y + z > 10

z > y and z > x

Given that 4 is a factor of the number, we have;

4 is a factor of 'z'

4 is also a factor of yz

Therefore, the digit, z > 7 and z is divisible by 4, therefore, z = 8

y + z = y + 8 > 10

∴ y > 10 - 8

y > 2

The multiples of 4 larger than 728 with the last digit equal to 8 are found using Microsoft Excel and presented as follows;

748, 768, 788

The numbers that have a larger ones place value are; 748 and 768

The sum of the las two digits are therefore, 4 + 8 = 12 and 6 + 8 = 14

The number that has the largest ones place value for the sum of the last two digits is 768

The number is 768

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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

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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

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STEP 2

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

G= G(W,u,V)

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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

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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

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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.

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4 years ago

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lisabon 2012 [21]

Answer:

Explanation:

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Then

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p(y) = SGw * (h - y)

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$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

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1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

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3 years ago

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Anna [14]

Maximum shear stress in the pole is 0.

<u>Explanation:</u>

Given-

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3 0

3 years ago