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sineoko [7]
4 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]4 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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Answer:

Specific heats

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3 years ago
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straig
kupik [55]

Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s =  19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

4 0
4 years ago
Helen is driving her car if the kinetic energy of the car is 8.10 E4 J and the mass of the car and its contents is 1010 kg Helen
larisa86 [58]
( 8.10) x 10^4 J = ½(1010kg)V² 
<span>......................= 505V² </span>
<span>( 8.10) x 10^4 J =[(.0505) x 10^4] V² ........<==10^4 cancels </span>
<span>8.10 = .0505 V² </span>
<span>.........______ </span>
<span>V = ²√ 160.4 </span>

<span>V= 12.6 m/s </span>

<span>(that's 12.6m/sec(1km/1000m)(3600secs/1hr) = 45.36kph) </span>
<span> my diagram is wrong, it is just an ordinary car, </span>
6 0
3 years ago
Which term refers to the rate at which the velocity of a moving object changes? A.acceleration B.displacement C.resonance D.turb
Inessa [10]
A. Acceleration is the answer to you question. When an object changes velocity it can accelerate.<span />
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4 years ago
A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V
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Answer:

655128 ohm

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V₀ = Voltage of the battery = 9 Volts  

V = Potential difference across the battery after time "t" = 4.20 Volts  

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Potential difference across the battery after time "t" is given as  

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Time constant is given as  

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5.11 = (7.8 x 10⁻⁶) R  

R = 655128 ohm

3 0
3 years ago
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