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2 years ago

How can gravitational tugs from orbiting planets affect the motion of a star?.

Physics

1 answer:

gayaneshka [121]2 years ago

3 0

Answer:

Gravitational tugs from orbiting planets don't affect the motion of a star. The star, being much larger than the planet, has a much smaller orbit. But it does move slightly. Explain how alien astronomers could deduce the existence of planets in our solar system by observing the Sun's motion.

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dem82 [27]

Answer:both are body waves often associated with seismic waves

Explanation:

Both are body waves often associated with seismic waves as in earthquakes

5 0

3 years ago

Read 2 more answers

New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$

$\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)$

4 0

3 years ago

The highest amount a landlord can charge for rent is an example of

NeTakaya

The correct answer that would best complete the given statement above would be RENT CONTROL. The rent control is the highest amount that a landlord or a property owner can charge for a rent. Rent control is established by the government that gives the maximum amount as to how much a landlord can charge the tenant. Hope this answer helps.

6 0

3 years ago

Read 2 more answers

A solid sphere of diameter D = 26 cm has a charge of Q = 4 nano-coulombs uniformly distributed on it. Calculate the magnitude of

kozerog [31]

Answer:

E = 4.83 N/ C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

$E= \frac{k*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

Where:

K: coulomb constant (N*m²/C²)

a: sphere radius (m)

Q: Total sphere charge (C)

r : Distance from the center of the sphere to the region where the electric field is calculated (m)

Equivalences

1nC=10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=4nC=4 *10⁻⁹C

D = 26 cm = 26*10⁻²m = 0.26m

a = D/2 = 0.13m

r= R+a = 2.6 m+ 0.13m = 2.73m

Problem development

Magnitude of the electric field at r = 2.73m from the center of the sphere

r>a , We apply the Formula (1) :

$E= \frac{k*Q}{r^{2} }$

$E= \frac{9*10^{9}*4*10^{-9} }{2.73^{2} }$

E= 4.83 N/ C

3 0

3 years ago

A friend of yours is loudly singing a single note at 412 Hz while racing toward you at 23.0 m/s on a day when the speed of sound

kobusy [5.1K]

Doppler Effect is required to solve the problem. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. The general formula of observed frequency, due to Doppler's effect is as follows,

$f_0 = (\frac{v+v_0}{v-v_s})f_s$