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pogonyaev
3 years ago
10

The astrometric technique of planet detection works best for

Physics
1 answer:
denpristay [2]3 years ago
6 0

The astrometric technique of planet detection works best for massive planets around nearby stars.

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_____ in the uterus has to occur for a woman to confirm her suspicions that she is pregnant.
vlabodo [156]
Conception is the correct answer
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3 years ago
As the ambulance got closer, Marge noticed that the pitch of the siren got higher. This happened because
Neko [114]

Explanation :

As the ambulance got closer, Marge noticed that the pitch of the siren got higher. This is due to the Doppler's effect.

The siren of the ambulance has a higher pitch when it approaches Marge. When the waves move away from him, it will have a lower pitch.

As the waves come closer to us, the waves become compressed and hence the frequency of the wave increases.

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3 years ago
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A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed of the wave? Group of answer choices
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3 years ago
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio
bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

3 0
3 years ago
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