All categories

2 years ago

Which best describes the way a sound wave is sent trough the radio?

Physics

2 answers:

Fittoniya [83]2 years ago

5 0

Answer:

Explanation:

horsena [70]2 years ago

4 0

Sound wave > electric signal > radio wave > sound wave

You might be interested in

A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li

seraphim [82]

Answer: 18.27°

Explanation:

Given

Index of refraction of blue light, n(b) = 1.64

Wavelength of blue light, λ(b) = 440 nm

Index of refraction of red light, n(r) = 1.595

Wavelength of red light, λ(r) = 670 nm

Angle of incident, θ = 30°

Angle of refraction of red light is

θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1

So that,

θ(r) = sin^-1 [(1 * sin 30) / 1.595]

θ(r) = sin^-1 (0.5 / 1.595)

θ(r) = sin^-1 0.3135

θ(r) = 18.27°

4 0

3 years ago

A center-seeking force related to acceleration is _______ force.

Gennadij [26K]

<span>A center-seeking force related to acceleration is centripetal force. The answer is letter A. The rest of the choices do not answer the question above.</span>

4 0

3 years ago

A train is traveling away from you at 120 km/h. it blows its whistle, and you hear a tone of 0.400 kHz. Take the speed of sound

Vaselesa [24]

Answer:B)439.21 Hz

Explanation:

Given

velocity of train $v=120 km/h\approx 120\times \frac{5}{18}=33.33 m/s$

frequency of train $f_a=0.400 kHz\approx 400 Hz$

speed of sound $c=340 m/s$

frequency from Doppler effect can be calculated by

$f_a=(\frac{c}{c+v})\cdot f$

$400=(\frac{340}{340+33.33})\cdot f$

$400=\frac{340}{373.33}\cdot f$

$f=400\times 1.098$

$f=439.21 Hz$

5 0

3 years ago

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

Ghella [55]

Answer:

Total electric potential, $V=1.32\times 10^6\ volts$

Explanation:

It is given that,

First charge, $q_1=3\ \mu C=3\times 10^{-6}\ C$

Second charge, $q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C$

Distance of first charge from origin, $r_1=1.15\ cm=0.0115\ m$

Distance of second charge from origin, $r_2=2\ cm=0.02\ m$

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

$V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$

$V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})$

$V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})$

V = 1321826.08 V

$V=1.32\times 10^6\ volts$

So, the total electric potential at the origin is $1.32\times 10^6\ volts$ . Hence, this is the required solution.

3 0

2 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

3 years ago