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sineoko [7]
3 years ago
10

Select Light for the type of wave, adjust the wavelength so that the light is red, and increase the amplitude of the light to th

e max. Then, select the start button at the source location to begin producing the waves. Light is a form of electromagnetic wave, containing oscillating electric and magnetic fields. The wave amplitude detector mentioned above shows how the electric field oscillates in time at the location of the probe. The amplitude of the wave at the location of the probe is equal to the maximum electric field measured. How does the amplitude of the wave depend on the distance from the source?
Physics
1 answer:
Sergeu [11.5K]3 years ago
5 0

Answer:

here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa

Explanation:

As wee know that the amplitude of the wave will decide the energy of the wave

Here we know that energy density of electromagnetic wave is given as

u = \frac{1}{2}\epsilon_0E_0^2

now we have

\frac{I}{c} = \frac{1}{2}\epsilon_0 E_0^2

so here we can say that intensity of the wave at the given distance from the source is given by formula

I = \frac{P}{4\pi r^2}

so here as we increase the distance the intensity will decrease and hence the amplitude of the electric field will decrease and vice-versa.

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Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

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Width of slit(b)=0.210

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(b)Width of the first order bright fringe

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Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from
kodGreya [7K]

Answer:

ax = 2.60m/s^{2}, t = 26.92s

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x (1)

Where (vx)f^{2} is the final velocity, (vx)i^{2} is the initial velocity, ax is the acceleration and  \Lambda x is the distance traveled.

Equation (1) can be rewritten in terms of ax:

(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x

2ax \Lambda x = (vx)f^{2} - (vx)i^{2}

ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}  (2)

Since the plane starts from rest, its initial velocity will be zero ((vx) = 0):

Replacing the values given in equation 2, it is gotten:

ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}

ax = \frac{4900m/s}{2(940m)}

ax = \frac{4900m/s}{1880m}

ax = 2.60m/s^{2}

So, The acceleration of the plane is 2.60m/s^{2}    

Now that the acceleration is known, the next equation can be used to find out the time:

(vx)f = (vx)i + axt (3)

Rewritten equation (3) in terms of t:

t = \frac{(vx)f - (vx)i}{ax}

t = \frac{70m/s - 0m/s}{2.60m/s^{2}}

t = 26.92s

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

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