Crest is to high, as through is to

Consider a two-dimensional incompressible velocity potential phi = ???????? cos theta + ????????theta, where B and L are constan

scoundrel [369]

solution:

Increasing the magnification and decreasing the field view

we are given:

a(t)=4*t^2-)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

x(0)=-2 )

x(2) = -20 )

where x(t) = distance travelled as a function of time.

need to find x(4).

from (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1 )

where k1 is a constant, to be determined.

integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2 ) where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 )

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20

16/3 -4 + 2k1 -2 = -20

2k1 = -20-16/3 +4 +2 = -58/3

=>

k1 = -29/3 )

thus substituting (6) and (7) in (5), we get

x(t) = (t^4)/3 - t^2 - 29t/3 + 2 )

which, by putting t=4 in (8)

x(4) = (4^4)/3 - (4^2 - 29*4/3 +2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)

7 0

3 years ago

This is problem 4 from chapter 6 of the course text. Find the Thevenin equivalent seen at the terminal A-B. Select which answer

miv72 [106K]

Answer:

Option C is correct.

Vs = 9 V, Rth = 30 ohms

Explanation:

In the circuit diagram, there are two sources, so, using superposition, we'll find voltage across A and B, Vab, with respect to each of the sources.

Taking the voltage source as primary source.

We will open circuit the current source just like it is presented in the first drawing of the 2nd image I have added to this solution.

By open-circuiting the current source, the three resistors R₁, R₂ and R₃ are in series.

This means that the Vab is the voltage across the R₃ resistor.

Using voltage divider rule,

Vab = (R₃/R₁ + R₂ + R₃) Vs₁

Vab = [40/(60 + 60 + 40)] (12)

Vab = 3 V

Taking the current source as the primary source.

We will now short circuit the voltage source as shown in the 2nd drawing of the 2nd image I have attached to this solution.

As shown in the 2nd drawing of the 2nd file I attached to this solution, the R₂ is in parallel with R₁ and R₃, that is R₂//(R₁ + R₃)

Using the current divider rule

Current in the (R₁ + R₃) branch = [R₂/(R₂ + (R₁ + R₃))] Is₂ = (60/(60 + 60 + 40)) (0.4) = 0.15 A

Vab = voltage across the R₃ resistor = IR₃ = 0.15 × 40 = 6 V

Total voltage across A and B, Vab = Vab due to Vs₁ + Vab due to Is₂ = 3 + 6 = 9 V

And for the Rth, we open-circuit the current source and short-circuit the voltage source simultaneously and look at the resistance of the setup from the AB terminal, as shown in the first drawing of the 3rd file attached to this solution.

It is evident that R₃ is in a parallel combination with (R₁ + R₂)

Rth = (R₁ + R₂)//R₃ = (60 + 60)//40 = 120//40 = (120× 40)/(120 + 40) = 30 ohms

The image of the thevenin equivalent of the circuit as seen from terminals AB is presented in the 2nd drawing on the 3rd image attached.

5 0

3 years ago

Any programmer who writes a Diophantine equation solver must occasionally encounter an infinite loop.

Sergio [31]

Answer: True

Explanation:

An Infinite loop occurs in a computer programming when a sequence of instructions run endlessly without stopping until there is an external intervention, this intervention could be pull or plug. It is also known as endless loop. It occurs due to using of variables that are not properly updated or when there is an error in looping condition.

4 0

4 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

or

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace z with 1/z to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense f(z) into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse Z transform of f(z) is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

Why do engineers play a variety of roles in the engineering process?

Crazy boy [7]

Answer: The engineering design process emphasizes open-ended problem solving and encourages students to learn from failure. This process nurtures students abilities to create innovative solutions to challenges in any subject! In addition to their involvement in design and development, many engineers work in testing, production, or maintenance. These engineers supervise production in factories, determine the causes of a component's failure, and test manufactured products to maintain quality.

Explanation:

6 0

3 years ago