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2 years ago

If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would

Physics

1 answer:

steposvetlana [31]2 years ago

3 0

Answer:

B. get smaller

Explanation:

The parallax angle of a star measured with respect to the Earth is inversely proportional to the distance of the star from the Earth:

$\theta = \frac{1}{d}$

where

$\theta$ is the parallax angle, measured in arcsec

d is the distance between the star and the Earth, measured in parsec

Therefore, if the star Alpha Centauri is moved farther from Earth, then d increases, and therefore the parallax angle will get smaller.

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List 4 ways to separate mixtures

Sergeeva-Olga [200]

1.Paper Chromatography. This method is often used in the food industry. ...

2.Filtration. This is a more common method of separating an insoluble solid from a liquid. ...

3.Evaporation. ...

4Simple distillation. ...
Fractional distillation.

7 0

2 years ago

A rope pulls a 82.5 kg skier at a constant speed up a 18.7Â° slope with Î¼k = 0.150. How much force does the rope exert?

Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f = μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0

3 years ago

a person using a machine applies a force of 100 newton's over a distance of 10 Meters to raise a 500 n object 1.5 meters what is

Hoochie [10]

Answer:

Work done by the machine (W) = 500 × 1.5 = 750 J

Work supplied to the machine (W) = 100 × 10 = 1000 J

Here, work supplied to the machine is input work = 1000 J

7 0

3 years ago

suppose an electrically charged ruler transfers some of its charge by contact to a tiny plastic sphere. will the ruler and the s

poizon [28]

Answer:

Here ball and rod will repel each other as they are of similar charges

Explanation:

As we know that the two charges attract or repel each other by electrostatic force

This force is given as

$F = \frac{kq_1q_2}{r^2}$

so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other

So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other

5 0

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