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2 years ago

9

A wave travels at a constant speed.How does the frequency change if the wavelength is reduced by a factor of 3 The frequency dec

reases by a factor of 3 the frequency increases by a factor of 3 the frequency increases by a factor of 9 the frequency does not change

1 answer:

nata0808 [166]2 years ago

8 0

Answer:

The frequency increases by a factor of 3.

Explanation:

The relation between speed, wavelength and frequency of a wave is given by :

$v=f\lambda$

or

$f\propto \dfrac{1}{\lambda}$

A wave travels at a constant speed. If the wavelength is reduced by a factor of 3, it would mean that the frequency increases by a factor of 3 because there is an inverse relationship between wavelength and frequency.

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Answer:

100N

Explanation:

if Jane pushes on the rock with 100 Newtons of force, then the rock pushes on Jane with 100 Newtons of force.

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The most common type of lighting unit utilized in street lighting is:

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3 years ago

At a distance of 14.0 m from a point source, the intensity level is measured to be 80 dB. At what distance from the source will

Ivan

Answer:

$d_2=19.8m$

Explanation:

The problem can be solved easily, because the relationship between intensities $I_1$ and $I_2$ at distances $d_1$ and $d_2$ respectively can be written as:

$\frac{I_2}{I_1} =\frac{d_1^2}{d_2^2}$

Where:

$I_1=Intensity\hspace{3}1\\I_2=Intensity\hspace{3}2\\d_1=Distance\hspace{3}1\\d_2=Distance\hspace{3}2$

In this case:

$I_1=80dB\\I_2=40dB\\d_1=14m\\d_2=?$

Solving for $d_2$

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2 years ago

Read 2 more answers

A 200-Ω resistor is connected in series with a 10-µF capacitor and a 60-Hz, 120-V (rms) line voltage. If electrical energy costs

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Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}$

f = frequency (60 Hz)

c= capacitor (10 µF = $10^-6$ )

$\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }$

Substitute the given values

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}$

$\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}$

$\mathrm{x}_{\mathrm{c}}=265.39 \Omega$

Given that, R = 200 Ω

$X^{2}=R^{2}+X c^{2}$

$X^{2}=200^{2}+265.39^{2}$

$X^{2}=40000+70431.85$

$X^{2}=110431.825$

$x=\sqrt{110431.825}$

X = 332.31 Ω

$\text { Current }(I)=\frac{V}{R}$

$\text { Current }(I)=\frac{120}{332.31}$

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,

$\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200$

$\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200$

$\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}$

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12

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Answer:

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The drift velocity is due to the applied electric field across the conductor.

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