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Arte-miy333 [17]
3 years ago
9

Students create a standing wave with three loops on a slinky 3.75 m longThey time 20 oscillations in 6.73 . What frequency would

create a standing wave with 5 loops?
Physics
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

f = 4.95\ hz

Explanation:

Given,

Student create wave of three loop on a slinky 3.75 m.

number of oscillations= 20

Time, t = 6.73 s

Frequency = \dfrac{20}{6.73}

                  = 2.97 Hz

We need to find frequency when 5 loops is created.

f = \dfrac{5}{3} \times 2.97

f = 4.95\ hz

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Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

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The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

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The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

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Hope this helps! :)

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