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aksik [14]
3 years ago
11

For a flow in the xy plane, the x component of velocity is given by u = Ax2y2, where A = 0.4 m-3 · s-1, and x and y are measured

in meters. (a) Find a possible y component for steady, incompressible flow. (b) Is it also valid for unsteady, incompressible flow? (c) How many possible y components are there? (d) Determine the equation of the streamline for the simplest y component of velocity (use C as constant).

Engineering
1 answer:
makvit [3.9K]3 years ago
5 0

Answer: v=(-2/3)*(A*x*y³)

              The equation is not valid for the unsteady flow, as there is no dependence upon time neither on x or y component of the velocity,

The corresponding stream function:-

                                             v(x,y)=  (1/3)(A*x²*y³)

Coordinates through the streamline:- (1,4) and (2,4)

                                             y(x) = Cx^(-2/3)

Explanation:

Use continuity equation:-

                                             \frac{du}{dx} +\frac{dv}{dy}  = 0\\\\\frac{dv}{dy}=-\frac{du}{dx}=-2Axy^2\\\\

Integrate V to obtain to obtain:-

v= -2/3*A*x*y^3

                             

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

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3 years ago
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Bezzdna [24]

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

<u>Answer:</u>

<u>(d) She has $10 more than you do</u>

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Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

<u>My compound interest is:</u>

= 1000 [ (1-0.1)^1-1

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$1,000 + $1,000 invested= $2,000 total amount received.

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To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

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An example of a career advancement is when an employee progresses from entry-level position to management and transits from an occupation to another.

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