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xenn [34]
2 years ago
15

The amplitude of a mechanical wave shows What?

Physics
1 answer:
Anuta_ua [19.1K]2 years ago
5 0
I’m not sure what you are asking
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A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside
vitfil [10]

Answer:

The value  is  B =  0.0452 \  T

Explanation:

From the question we are told that

   The number of turns is  N  =  1000

    The length is  L =  50 cm =  0.50 m  

    The radius is  r =  2.0 cm  =  0.02 m

     The current is I  =  18.0 A

   

Generally the magnetic field is mathematically represented as

         B = \mu_o  * \frac{N }{L}  *  I

Here \mu_o is the permeability of free space with value  

     \mu_o  =  4\pi * 10^{-7} N/A^2

So

     B =  4\pi * 10^{-7}  *   \frac{1000}{0.50} *  18.0

=>   B =  0.0452 \  T

3 0
3 years ago
When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

L=\dfrac{N\phi}{I}.........(2)

From equation (1) and (2) we get :

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

N=\dfrac{\epsilon I}{\phi (dI/dt)}

N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0
2 years ago
If a photon has frequency = 2.00 x 1014s-1 and the speed of light = 3.00 x 108ms-1, then what is its wavelength?
butalik [34]

Answer:

The photon has a wavelength of 1.5x10^{-6}m

Explanation:

The speed of a wave can be defined as:

v = \nu \cdot \lambda (1)

Where v is the speed, \nu is the frequency and \lambda is the wavelength.

Equation 1 can be expressed in the following way for the case of an electromagnetic wave:

c = \nu \cdot \lambda (2)              

 

Where c is the speed of light.    

Therefore, \lamba\lambda can be isolated from equation 2 to get the wavelength of the photon.

\lambda = \frac{c}{\nu} (3)

\lambda = \frac{3.00x10^{8}m/s}{2.00x10^{14}s^{-1}}

\lambda = 1.5x10^{-6}m

Hence, the photon has a wavelength of 1.5x10^{-6}m        

<em>Summary:  </em>

Photons are the particles that constitutes light.

3 0
3 years ago
Matter that emits no light at any wavelength is called
lorasvet [3.4K]

Matter that emits no light at any wavelength is called DARK MATTER.

4 0
3 years ago
An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a
prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611

Converting to SI units

10\ ft/s=10\times \dfrac{1}{3.281}

32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}

2\ ft=2\times \dfrac{1}{3.281}

Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

7 0
3 years ago
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