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solmaris [256]
2 years ago
13

The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the

mathematical relationship between amplitude and energy. Use your equation to find the energy if the amplitude is 6 units. Amplitude Energy 1 unit 2 units 2 units 8 units 3 units 18 units 4 units 32 units A. 66 units B. 60 units C. 72 units D. 74 units​
Physics
2 answers:
Maslowich2 years ago
3 0

The energy of a wave is directly proportional to the square of the amplitude of the wave.

<h3>What is the relationship between energy and amplitude?</h3>

There is direct relationship between energy of the wave and the amplitude of the wave. The energy transported by a wave is directly proportional to the square of the amplitude of the wave. This means if energy is increase the amplitude of wave becomes double and vice versa.

Energy = (amplitude)2

So we can conclude that the energy of a wave is directly proportional to the square of the amplitude of the wave.

Learn more about energy here: brainly.com/question/13881533

#SPJ1

Svetradugi [14.3K]2 years ago
3 0

Energy = (amplitude)^{2}

<h3>What is energy and amplitude?</h3>

Energy is define as the capacity for doing work.

Amplitude is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position.

So , the relationship between energy of the wave and the amplitude of the wave is described as the energy transported by a wave is directly proportional to the square of the amplitude of the wave.

Here , If energy is increase the amplitude of wave becomes double.

Energy = (amplitude)^{2}

Hence , the energy of a wave is directly proportional to the square of the amplitude of the wave.

By using the above equation if the amplitude is 6 Units then :

Energy = (6)^{2} = 36 units

learn more about amplitude :

brainly.com/question/8662436

#SPJ1

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denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
3 years ago
If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples
kvasek [131]

Answer is :

D. Quadruples

7 0
3 years ago
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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
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