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3 years ago

The frequency of a simple pendulum that makes 120 complete oscillations in 3 minutes is:

Physics

1 answer:

nekit [7.7K]3 years ago

5 0

Answer: 0.67 Hz

Explanation:

Given that:

Number of oscillations = 120

Time taken for oscillations = 3 minutes

Since standard unit of time is seconds, convert 3 minutes to seconds

(If 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Then, recall that frequency is the number of complete cycles or oscillations in one second, and its unit is Hertz.

Thus, frequency = (number of oscillations / time)

F = 120 / 180 seconds

F = 0.67 Hz

Thus, the frequency of a simple pendulum is 0.67Hz

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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

A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

3 years ago

In the nucleus, the a)...... have b)...... charge.

Inga [223]

C:carry
I hope I help

8 0

4 years ago

1. George is traveling to Boston from Springfield. Springfield is 144

denpristay [2]

Answer:

48 kilometers per hour

Explanation:

144/3 = 48 km/h

3 0

2 years ago

10 points!! If you help!!!

marta [7]

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

4 0

3 years ago