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3 years ago

6

What method has detected the most extrasolar planets so far?

1 answer:

Sedaia [141]3 years ago

5 0

<h2>Answer: The Transit method</h2>

Detecting extrasolar planets by direct observation (with a telescope) is a complicated task. This is because any planet constitutes an extremely dim light source compared to the star around which it orbits.

So, to detect this extremely dim source is quite difficult due to the glare of the star's light that dulls it.

In this sense, scientists and astronomers have made several methods to find these extrasolar planets, among which the most successful has been the transit method.

This method is based on <u>astronomical transit</u>, a phenomenon in which a body (a planet in this case) passes in front of a larger one (the star), blocking (eclipsing) its vision to some extent.

It should be noted that this is the method currently used in the search for extrasolar planets. Space agencies such as ESA (Europe) and NASA (USA) have put into orbit satellites with extremely sensitive photometric sensors to observe even the smallest variations of intensity of a star due to the passage of a planet.

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What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

masya89 [10]

Answer: $0.0058 m/s^{2}$

Explanation:

Centripetal acceleration $a_{C}$ is calculated by the following equation:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s$ is the Earth's orbital speed

$r=1.5(10)^{11} m$ is the orbital radius

$a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}$

$a_{C}=0.0058 m/s^{2}$

4 0

3 years ago

Sound waves with a constant frequency of 250 hertz are traveling through air at stp. what is the wavelength of the sound waves

Deffense [45]

Answer:

wave length is 1.2m

Explanation:

since formula of wave length is v/f

v(speed of sound in air at stp is 300ms^-1)

f(frequency 250hertz)

then wave length is 300÷250 which give 1.2m

5 0

1 year ago

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0

3 years ago

A student investigated how the mass of water in an electric kettle affected the time taken for the water to reach boiling point.

Sedbober [7]

Answer:

1.because of the heat produced by the socat

2. they should have control how they placed the heater

3. because the water is to much

4.because is different from the question

5. because that is how the question is

4 0

2 years ago

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Nikitich [7]

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8 0

3 years ago