1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Liula [17]
3 years ago
14

Which of the following has to happen before the eardrum begins to vibrate with the same frequency as the source of the sound wav

es? A. Vibrations have to be transmitted to the fibers of the auditory nerve. B. The brain has to interpret the direction from which the sounds are being delivered. C. Impulses have to be sent to the brain through the auditory nerve. D. The outer ear has to collect the sound waves and lead them to the middle ear.
Physics
2 answers:
777dan777 [17]3 years ago
8 0
I just took the test it is D




I am 100% sure
Sladkaya [172]3 years ago
6 0

Answer:

D. The outer ear has to collect the sound waves and lead them to the middle ear.

Explanation:

Here when sound propagates through air then first the sound is collected by the outer part of the ear.

then it is propagated to the middle ear and then this will hit the ear drum situated at the inner ear.

Then this vibrations received on the inner part of year is converted into electrical signals and carried by the nerves to the brain.

then brain interpret these signals into sound

So here we will have

D. The outer ear has to collect the sound waves and lead them to the middle ear.

You might be interested in
A set is any collection of objects. true or false.
Semenov [28]
It can be, but set also means many other things.
3 0
3 years ago
Definition of distance in physics
solniwko [45]

Answer:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Explanation:

8 0
3 years ago
Explain how an alpine glacier can change the topography of a mountainous area
igor_vitrenko [27]
<span>An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process.  In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.</span>
7 0
3 years ago
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
CHAPTER 6: KINETICS OF A PARTICLE
vekshin1

Explanation:

Work done by winch = kinetic energy of car

∫ T ds = ½ mv²

∫ 225s ds = ½ mv²

225/2 s² = ½ mv²

225 s² = mv²

v = 15s / √m

Given s = 10 m and m = 2500 kg:

v = 15 (10) / √2500

v = 3 m/s

8 0
3 years ago
Other questions:
  • A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st
    15·1 answer
  • A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t
    9·1 answer
  • A train travels from over city to another its initial velocity is lower than its final velocity this is a example of
    8·1 answer
  • A 833 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 561 kg. The two cars lock up
    12·1 answer
  • Why can't you trust the law of gravity? (RIDDLE)
    10·2 answers
  • What is the mass of the lunar landing module on the moon? kg
    8·1 answer
  • what is the difference in gravitation potential energy when a 3.5 of box is raised from a height of 1.2m to a height of 4.0m?​
    6·1 answer
  • Who speaks the line "Lord, what fools these mortals be"?
    13·1 answer
  • Most of the motion generated by joints in the human body are examples of levers. An example of a third class lever is the forear
    15·1 answer
  • Quien me ayuda en una tarea de matemáticas de una figura geométrica que no se cómo se llama​
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!