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valina [46]
3 years ago
5

Do you think there are other planets outside of our solar system? Support your response with facts

Physics
2 answers:
Agata [3.3K]3 years ago
6 0


The habitable zone is the range of distances from a star where a planet’s temperature allows liquid water oceans, critical for life on Earth. The earliest definition of the zone was based on simple thermal equilibrium, but current calculations of the habitable zone include many other factors, including the greenhouse effect of a planet’s atmosphere. This makes the boundaries of a habitable zone "fuzzy."



Astronomers announced in August 2016 that they may have found such a planet orbiting Proxima Centauri. The newfound world, known as Proxima b, is about 1.3 times more massive than Earth, which suggests that the exoplanet is a rocky world, researchers said. The planet is also in the star's habitable zone, just 4.7 million miles (7.5 million kilometers) from its host star. It completes one orbit every 11.2 Earth-days. As a result, it's likely that the exoplanet is tidally locked, meaning it always shows the same face to its host star, just as the moon shows only one face (the near side) to Earth.



The young sun would have had a very strong magnetic field, whose lines of force reached out into the disk of swirling gas from which the planets would form. These field lines connected with the charged particles in the gas, and acted like anchors, slowing down the spin of the forming sun and spinning up the gas that would eventually turn into the planets. Most stars like the sun rotate slowly, so astronomers inferred that the same “magnetic braking” occurred for them, meaning that planet formation must have occurred for them. The implication: Planets must be common around sun-like
A Canadian team discovered a Jupiter-size planet around Gamma Cephei in 1988, but because its orbit was much smaller than Jupiter’s, the scientists did not claim a definitive planet detection. “We weren’t expecting planets like that. It was different enough from a planet in our own solar system that they were cautious," Matthews said.
Most of the first exoplanet discoveries were huge Jupiter-size (or larger) gas giants orbiting close to their parent stars. That's because astronomers were relying on the radial velocity technique, which measures how much a star “wobbles” when a planet or planets orbit it. These large planets close in produce a correspondingly big effect on their parent star, causing an easier-to-detect wobble.
Before the era of exoplanet discoveries, instruments could only measure stellar motions down to a kilometer per second, too imprecise to detect a wobble due to a planet. Now, some instruments can measure velocities as low as a centimeter per second, according to Matthews. “Partly due to better instrumentation, but also because astronomers are now more experienced in teasing subtle signals out of the data.”

Today, there are more than 1,000 confirmed exoplanets discovered by a single telescope: the Kepler space telescope, which reached orbit in 2009 and hunted for habitable planets for four years. Kepler uses a technique called the “transit” method, measuring how much a star's light dims when a planet passes in front of it.

Kepler has revealed a cornucopia of different types of planets. Besides gas giants and terrestrial planets, it has helped define a whole new class known as “super-Earths”: planets that are between the size of Earth and Neptune. Some of these are in the habitable zones of their stars, but astrobiologists are going back to the drawing board to consider how life might develop on such worlds.

In 2014, Kepler astronomers (including Matthews’ former student Jason Rowe) unveiled a “verification by multiplicity” method that should increase the rate at which astronomers promote candidate planets to confirmed planets. The technique is based on orbital stability — many transits of a star occurring with short periods can only be due to planets in small orbits, since multiply eclipsing stars that might mimic would gravitationally eject each other from the system in just a few million years.

While the Kepler (and French CoRoT) planet-hunting satellites have ended their original missions, scientists are still mining the data for discoveries, and there are more to come. MOST is still operating, and the NASA TESS (Transiting Exoplanet Survey Satellite), Swiss CHEOPS (Characterizing ExOPlanets Satellite) and ESA’s PLATO missions will soon pick up the transit search from space. From the ground, the HARPS spectrograph on the European Southern Observatory's La Silla 3.6-meter telescope in Chile is leading the Doppler wobble search, but there are many other telescopes in the hunt.

With almost 2,000 to choose from, it’s hard to narrow down a few. Small solid planets in the habitable zone are automatically standouts, but Matthews singled out five other exoplanets that have expanded our perspective on how planets form and
Readme [11.4K]3 years ago
3 0

Answer:  Yes

Explanation: There was life on Mars at one point, so yes. Also, Every large non-microscopic organism on Earth breathes either oxygen or carbon dioxide (plants), so there could possibly be life in the water below the surface of Europa.

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A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per
FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0
2 years ago
Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0
3 years ago
Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

\eta = 1-\frac{T_C}{T_H}

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

T_C = 270^{\circ}C+273=543 K

T_H = 450^{\circ}C+273=723 K

Therefore the maximum efficiency is

\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

T_H = 270^{\circ}C+273=543 K is the high-temperature reservoir

T_C = 150^{\circ}C+273=423 K is the low-temperature reservoir

If we apply again the formula of the efficiency

\eta = 1-\frac{T_C}{T_H}

The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

8 0
3 years ago
A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
 So the distance between the red and violet rays is 0.07038 cm.
7 0
2 years ago
If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t
Oxana [17]

Answer:

The answer to your question is:      V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

                          P1V1 = P2V2

                          V2 = (P1V1) / P2

                          V2 = (200 x 1.5) / 300

                          V2 = 1 l

6 0
3 years ago
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