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2 years ago

Kepler's 3rd law: harmonic law

Physics

1 answer:

Amanda [17]2 years ago

7 0

<span>orbital velocities to their mean distances from the Sun.</span>

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a bird flies 25.0 m in the direction 55° east of south to its nest. the bird then flies 75.0 m in the direction 55° west of nort

son4ous [18]

The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.

<h3>Displacement of the bird</h3>

The displacement of the bird is the change in the position of the bird.

<h3>Vertical component of the bird's displacement </h3>

Vy₁ = -25 m x sin(55)

Vy₁ = -20.48 m

Vy₂ = 75 m x sin(55)

Vy₂ = 61.44 m

Total vertical displacement = 61.44 m - 20.48 m = 40.96 m

<h3>Horizontal component of the bird's displacement </h3>

Vx₁ = -25 m x cos(55)

Vx₁ = -14.34 m

Vx₂ = 75 m x cos(55)

Vx₂ = 43.02 m

Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m

Learn more about displacement here: brainly.com/question/2109763

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8 0

1 year 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

A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric

MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

$J = \frac{I}{\pi R^2}$

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

$J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}$

This enclosed current is now to be used in Ampere’s Law.

$\mu_o I_{enc} = \int {B} \, dl$

Here, $\int \, dl$ represents the circular path of radius r. So we can replace the integral with the circumference of the path, $2\pi r$ .

As a result, the magnetic field is

$B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}$

5 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$