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Nostrana [21]
2 years ago
15

A train is approaching the station. Explain what happens to the frequency of the sound as the train draws closer. What do you he

ar? (4 points)
Physics
1 answer:
lisabon 2012 [21]2 years ago
8 0

If I am at the station that the train is approaching, and if the train is making some kind of steady sound, then as the train draws closer, I hear the amplitude of the sound steadily increasing ... it gets louder and louder as the train gets closer and closer to me.

(The frequency of the sound I hear is higher than the frequency of the sound the train is actually making, but I don't know that, because I have no idea what frequency sound the train is actually making.)

=================================

If anybody tries to tell you that the frequency of the sound from the train keeps getting higher and higher, please tell that person to come see me, and I will straighten him out.  It doesn't.

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A submarine is 2.99 ✕ 102 m horizontally from shore and 1.00 ✕ 102 m beneath the surface of the water. A laser beam is sent from
Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of  incidence  beam striking the water is \theta = 49.63^o

c  the angle of  refraction  beam striking the water is r = 59.7^o

d The angle the refracted beam make with respect to the horizontal is = 30.3^o

e The height of the target above sea level is  

                   h= 125.05m

Explanation:

From the diagram we see that the angle of the beam striking the water is

                   tan \theta = \frac{100}{85}

                        \theta = tan^{-1}(\frac{100}{85} )

                           = 49.63^o

According to Snell's law

                   \mu_{water} *sin(i) = \mu_{air}  *sin(r)

Where \mu_{water } is the refractive index of water =  1.333

           i is the angle of incidence

          \mu_{air} is the refractive index of air  = 1

            r is the angle of refraction

 Substituting values accordingly

          1.33 * sin (40.37) = 1 * sin(r)

    Making r the subject of the formula

                       r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})

                          = 59.7^o

looking at the diagram we can see that to  obtain the angle the refraction beam makes with the horizontal   by subtracting the angle refraction from 90°

                 i.e  90 -59.7 = 30.3 °

From the diagram we see that the height  target above sea level can be obtained by this relation

                   tan \theta = \frac{h}{214}\\

Where h is the is the height

                   tan(30.3) = \frac{h}{214}

                         h = 214 * tan (30.3)

                            =125.05m

           

4 0
3 years ago
A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point
pogonyaev

Answer:

L = 1.545 m

Explanation:

Let the total length of the rod is L

now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum

so we will have

mg \times d = F(L - d)

so we have

325\times 9.8 \times 0.266 = F(L - 0.266)

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. Methylated spirits have ethanol as a base but may include methyl alcohol (methanol) as part of the denaturing process.
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ira [324]

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Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

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Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

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T_A=T_B

because the two boxes has the same acceleration because they are attached together:

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So the magintude of the acceleration for both objects is 3.11m/s^2

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