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klio [65]
3 years ago
10

A child holds one end of a 33.0-meter long rope in her hand and moves it up-and-down to produce a sinusoidal wave by moving her

hand from 8.00 cm above her shoulder to 8.00 cm below her shoulder at a frequency of 2.00 Hz and a wavelength of 75.0 cm. If the child doubles the amplitude of her hand's motion on that same rope with the same tension in the rope, then what will the wavelength of the wave now be
Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

Wavelength=75 cm.

The wavelength well remain unchanged which is 75 cm.

Explanation:

The formula which will help us to answer the question is:

V=f*λ

Where:

V is the velocity

f is the frequency of wave

λ is the wave length

Now:

λ=V/f    Eq (1)

The equation show's that wavelength is independent of the amplitude but it depends on the frequency and the velocity with which wave is moving.

The wavelength well remain unchanged which is 75 cm.

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What is a cool example of newtons 3rd law?
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Usually wire resistance can be neglected. But we see the effect of it sometimes when a light dims as another high-current device
tangare [24]

Please find the figure attached below:

Answer:

a.Resistance of bulb=R_{2}=192 ohm

b.power consumed by bulb after motor disconnection=74.609 W

Explanation:

<u>a.Resistance of bulb=</u>R_{2}<u>=?</u>

As we know that P=\frac{P^{2} }{R}

putting R as resistance of bulb i.e. R=R_{2}

P=\frac{V^{2}}{R_{2} }\\R_{2}=\frac{V^{2}}{P} \\ R_{2}=\frac{(120)^{2} }{75}\\ R_{2}=192ohm

<u>b.power consumed by bulb after motor disconnection? </u>

from the figure we see that R_{1}\ and\ R_{2} are in series so

R_{eq}=R_{1} +R_{2} \\R_{eq}=192+0.5\\R_{eq}=192.5ohm

current through their resistance is

I_{eq}=\frac{V}{R_{eq} }\\ I_{eq}=\frac{120}{192.5 } \\I_{eq}=0.6233A

Power consumed by bulb is

P_{b} =I^{2}R\\\\ P_{b} =(0.6233)^{2}(192)\\\\P_{b} =74.609W

3 0
4 years ago
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marshall27 [118]

Answer:

the correct answer to your problem I believe would be B

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Consider a point charge qqq in three-dimensional space. Symmetry requires the electric field to point directly away from the cha
bonufazy [111]

Answer:

 E = \frac{1}{4\pi  \epsilon_o } \  r^2

Explanation:

For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere

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the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product

           Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}

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the charge inside the sphere is q = + q

           

we substitute

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