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
chubhunter [2.5K]
3 years ago
12

If the energies sound in the device is microphone what is the energy out

Physics
1 answer:
Katen [24]3 years ago
5 0

Answer:

Microphones are a type of transducer - a device which converts energy from one form to another. Microphones convert acoustical energy (sound waves) into electrical energy (the audio signal). Different types of microphone have different ways of converting energy but they all share one thing in common: The diaphragm.

Explanation:

You might be interested in
A wire 25.0cm long lies along the z-axis and carries a current of 9.00A in the positive +z-direction. The magnetic field is unif
marysya [2.9K]

The expression of the magnetic force and solving the determinant allows to shorten the result for the value of the magnetic force are:

  • In Cartesian form  F = 2.46 i ^ - 0.605 j ^
  • In the form of magnitude and direction F = 2.53 N and θ = 346.2º

Given parameters.

  • Length of the wire on the z axis is: L = 25.0 cm = 0.25 m.
  • The current i = 9.00 A in the positive direction of the z axis.
  • The magnetic field B = (-0.242 i ^ - 0.985 j ^ -0.336 k ^ ) T

To find.

  • Magnetic force.

The magnetic force on a wire carrying a current is the vector product of the direction of the current and the magnetic field.

          F = i L x B

Where the bold letters indicate vectors, F is the force, i the current, L a vector pointing in the direction of the current and B the magnetic field.

The best way to find the force is to solve the determinant, in general, a vector (L) is written in the form of the module times a <em>unit vector</em>.

         F= i |L| \left[\begin{array}{ccc}i&j&k\\L_x&L_y&L_z\\B_x&B_y&B_z\end{array}\right]  

Let's calculate.

       F= 9.00  \ 0.25 \ \left[\begin{array}{ccc}i&j&k\\0&0&1\\-0.242&-0.985&-0.336\end{array}\right]  

       F = 2.26 \ ( - 1 B_y i  \   + 1 B_x j  \ )  

       F = 2.5 (0.985 i ^ - 0.242 j ^)

       F = ( 2.46 i ^ - 0.605 j^ ) N

To find the magnitude we use the Pythagorean theorem.

        F = \sqrt{F_x^2 + F_y^2}  

        F = \sqrt{2.46^2 + 0.605^2}  

        F = 2.53 N

Let's use trigonometry for the direction.

        Tan θ ’= \frac{F_y}{F_x}  

        θ'= tan⁻¹ \frac{F_y}{F_x}  

        θ'= tan⁻¹1 (\frac{-.605}{2.46} )

        θ’= -13.8º

To measure this angle from the positive side of the x-axis counterclockwise.

        θ = 360- θ'

        θ = 360 - 13.8

        θ = 346.2º

In conclusion using the expression of the magnetic force and solving the determinant we can shorten the result for the value of the force are:

  • In Cartesian form    F = 2.46 i ^ - 0.605 j ^
  • In the form of magnitude and direction  F = 2.53 N and θ = 346.2º

Learn more here:  brainly.com/question/2630590

5 0
3 years ago
PLEASE HELP!! XD<br> Why are the noble gases not very reactive?
Colt1911 [192]
The atom's most outer shell is full.
3 0
3 years ago
Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

v_{f}=1721.1m/s

Explanation:

Given data

q = 3.1 µC

m = 47 mg

r1 = 0.83 mm

r2 = 2.5 mm.

As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
A book weighing 18.0 N is lifted from the floor and placed on a shelf 1.5 m high. What is the minimum amount of work required to
algol13
B 12j is the correct answer
6 0
3 years ago
A spring is hung from the ceiling. When a coffee mug is attached to its end, it stretches 2.5 cm before reaching its new equilib
densk [106]

Answer:

Explanation:

In equilibrium , weight of mug is equal to restoring force .

mg = kx where m is mass of mug , k is spring constant and x is extension .

k / m = g / x = 9.8 ms⁻² / .025 m

= 392

frequency of oscillation n = \frac{1}{2\pi}\sqrt{\frac{k}{m} }

n=\frac{1}{2\pi}\sqrt{392 }

= 4.46 per second.

5 0
3 years ago
Other questions:
  • A fold is a _ in a rock , and a fault is a _in a rock?
    13·2 answers
  • Michael's car ran over a nail, which made a hole in one of the tires. At what point will air stop pouring out of the hole in the
    11·1 answer
  • What instrument is used to measure current?
    10·1 answer
  • Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER
    6·1 answer
  • List any four uses of electricity in our daily life​
    9·2 answers
  • Will mark BRAINLYEST
    15·2 answers
  • Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially
    14·1 answer
  • ILL GIVE U BRAINLIEST!! help me
    8·2 answers
  • An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air
    5·1 answer
  • exhibit 6-5 the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation o
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!