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3 years ago

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Write down an equation describing a sinusoidal traveling wave (in 1-D). Tell us (words and/or equations) what in your equation t

ells us the speed and direction of the wave? [Hint: you can google this if you do not know the answer. Be sure you understand it though!]

1 answer:

Natalija [7]3 years ago

6 0

Answer:

Explanation:

The standard equation of the sinusoidal wave in one dimension is given by

$y = A Sin\left ( \frac{2\pi }{\lambda }\left ( vt-x \right )+\phi \right )$

Here, A be the amplitude of the wave

λ be the wavelength of the wave

v be the velocity of the wave

Φ be the phase angle

x be the position of the wave

t be the time

this wave is travelling along positive direction of X axis

The frequency of wave is f which relates with velocity and wavelength as given below

v = f x λ

The relation between the time period and the frequency is

f = 1 / T.

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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

7 0

3 years ago

Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.

Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

$distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m$

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

$displacement = +1.0 m-0 m=+1 m$

8 0

3 years ago

Electric Circuits Homework

nirvana33 [79]

<h2>Sorry, But I don't know!!</h2>

3 0

2 years ago

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

STatiana [176]

Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

$P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2$

$1\ N/m^2=\frac{1}{6894.757}\ lb/in^2$

$3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2$

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0

3 years ago

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the

alexandr402 [8]

Answer:

r=6.05km/hr

z=59.1 degree to the horizontal

Explanation:

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place

can be solved using pythagoras theorem

r^2=o^2+a^2

r^2=5.2^2+3.1^2

r^2=36.65

r=6.1km/hr is te birds velocity relative to the ground

tanz=5.2/3.1

z=tan^-1(5,2/3.1)

z=59.1 degree to the horizontal

7 0

3 years ago

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