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3 years ago

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7. When a wave crosses a surface at an angle from one medium into another, why does it change directions as it moves across the

boundary into the new medium? Use 3 – 4 complete sentences in your own words to explain.

1 answer:

Debora [2.8K]3 years ago

7 0

The changing direction of light is called refraction. This happens because the medium that the wave is going into is either more dense or less dense than the medium the wave was in before. Depending on the density of the second medium, the wave will refract either closer or farther away from the normal (a verticle line in the middle in which you calculate the incident and refracted rays from). If the second medium is denser than the first medium, then the wave will move closer to the normal. If the second medium is less dense, then the wave will move away from the normal.

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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

A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop

postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination $(\theta )=30^{\circ}$

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

$F_{net}=0$

Force applied by cable $=mg\sin \theta$

$F=60\times 9.8\times \sin (30)=294 N$

work done $=F\cdot x$

$W=294\cdot 75=22.125 J$

(b)Power $=F\cdot v$

$P=294\cdot 2.4=705.6 W\approx 0.946 hp$

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3 years ago

How does radiation differ from conduction?

Natali5045456 [20]

Answer:

b

Explanation:

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3 years ago

Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

ch4aika [34]

C longitudnal waves

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3 years ago

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A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee

SOVA2 [1]

Answer:

$\omega=0.12\frac{rad}{s}$

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

$\omega=\frac{2\pi}{T}(1)$

Here T is the period, that is, the time taken to complete onee revolution:

$T=\frac{2\pi r}{v}(2)$

Replacing (2) in (1):

$\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}$

3 0

4 years ago