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liberstina [14]
3 years ago
9

Each part and area of the ear have specific functions in the hearing process. In which of the following parts of the ear are sou

nd waves changed into a signal the brain can understand?
Physics
1 answer:
V125BC [204]3 years ago
6 0

There are no "following" parts of the ear listed.

Sound waves are changed into a signal the brain can understand, consisting of nerve impulses, in the "hair cells" in the lining of the cochlea ... that spiral thing deep in the inner ear.

(In fact, that's as deep as you can get in the inner ear, because it's literally the end of the ear.  After those hair cells inside the spiral cochlea, there's no more ear.  The sound has now changed to electrical signals that go from there to the brain, in the form of electrical currents flowing in nerves.)

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A man wishes to row the shortest possible distance from north to south across a river which is flowing at 2 km/hr from the east.
sweet-ann [11.9K]

From the diagram we have that

sin\theta = \frac{2}{4}

\theta = sin^{-1} (\frac{1}{2})

\theta = 30\°

Therefore the direction is 30° from east of south

8 0
3 years ago
A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr
xxTIMURxx [149]
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
3 0
3 years ago
A centripetal force of 210 N acts on a 1,600-kg satellite moving with a speed of 5,500 m/s in a circular orbit around a planet.
Dafna1 [17]

Answer:

<h2>230476.19km</h2>

Explanation:

Step one:

given

Force F= 210N

mass m= 1600kg

velocity v=5500m/s

Step two

Required is the radius r

the expression for the force is

F_c = \frac{mv^2}{r}

substitute

210=1600*5500^2/r

cross multiply we have

210r=48400000000

divide both side by 210

r=230476190.476m

r=230476.19km

4 0
3 years ago
Show that for a projectile d2 (v2) / dt2 = 2g2
stich3 [128]
<span>Ok, you need to derive it. Derive v^2</span>
8 0
2 years ago
Please help!?!? needs to show work too.
ankoles [38]

In order to solve problems like this, you need to know that whenever
there are waves involved ...

          <u>(wavelength) x (frequency)  = speed of the wave </u>.

Both of these problems are about sound, so you'll need to know the speed
of sound.  I'm going to use 340 meters per second. You should look it up to
see if that's a reasonable number.

1). (wavelength) x (frequency)  = speed of the wave

The wavelength is given, and I picked a number for the speed.

         (0.667 m) x (frequency) = (340 m/s)

Divide each side by 0.667 m :    Frequency = (340 m/s) / (0.667 m) = <em>509.7 Hz</em>.


2).  I picked a number for speed, and two frequencies are given.
We have to find the wavelength for each frequency.

<u>20 Hz:</u>
(wavelength) x (frequency)  = speed of the wave

       (wavelength) x (20 Hz) = 340 m/s

Divide each side by  20 Hz:        Wavelength = (340 m/s) / (20 Hz) = <em>17 m</em>

<u>16,000 Hz:</u>
(wavelength) x (frequency)  = speed of the wave

   (wavelength) x (16,000 Hz) = 340 m/s

Divide each side by 16,000 Hz:  Wavelength = (340 m/s)/(16,000 Hz) = <em>2.125 cm </em>


I also want to tell you how much I like the wavy appearance of these
questions about waves in the picture you attached !


7 0
3 years ago
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