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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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Answer:

Frick is pushing harder

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if Frack weighs more and he was pushing harder they would be moving, but if Frick pushes harder then they wont move

7 0
2 years ago
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A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
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Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

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induced emf =  0.0286538 V

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3 0
3 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
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3 years ago
What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?
lorasvet [3.4K]

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Needed to be pointed out that mechanical advantage is when the distance traveled is traded for force applied

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