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kozerog [31]
3 years ago
12

What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87 m?​

Physics
1 answer:
Deffense [45]3 years ago
7 0

v = λf

v = 87 x 2 = 174 m/s

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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq
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Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}

Given;

\frac{dP}{dt} (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( \frac{dv}{dt});

(600 x 40) + (150 x \frac{dv}{dt}) = 0

\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min

Therefore, the volume is decreasing at 160 cm³/min

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Explain how you can work out the
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Answer:

The speed of light changes as it moves between media. This causes refraction. Angles of refraction can be calculated using known speeds or wavelengths. Beyond the critical angle, light is reflected.

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3 years ago
What happens to air as it warms ,expands and become less dense
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A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v
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3 years ago
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

3 0
3 years ago
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