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OLga [1]
3 years ago
13

Which of the following methods has led to the earliest discoveries of massive planets orbiting near their parent stars a. detect

ing a planet ejected from a binary star system b. detecting the starlight reflected off the planet c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum d. detecting the infrared light emitted by the planet
Physics
1 answer:
Mrac [35]3 years ago
3 0

Answer:

c. detecting the gravitational effect of an orbiting planet (The Wobble"") by looking for the Doppler shifts in the star's spectrum

Explanation:

In a solar system the mass of the star and planets affect each other's orbital  movements. The center of gravity of a star and a planet is inside the star. This causes the star to be closer and farther from the Earth at different times. Due to this wobble the star appears to be red shifted when it is farther and blue shifted when it is closer.

When the mass of the planet is high, like a hot Jupiter it causes more wobble i.e., change in radial velocity. This makes it easier to detect the planet. The earliest hot Jupiter found by this method is the planet 51 Pegasi b.

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a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe
Mariana [72]

Answer:

1.34352 kg

Explanation:

m_w = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

\Delta T = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

m_v = Mass of water in the vessel

Here the potential energy will balance the internal energy

m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg

Mass of the water in the vessel is 1.34352 kg

6 0
3 years ago
A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con
erastova [34]

Let the key is free falling, therefore from equation of motion

h = ut +\frac{1}{2}gt^2..

Take initial velocity, u=0, so

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2.

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

d= v \times t

From above substituting t,

d = v \times \sqrt{\frac{2h}{g} }.

Now substituting all the given values and g = 9.8 m/s^2, we get

d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m.

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0
3 years ago
A person should be able to find all the answers to their science questions in the text
Vsevolod [243]
TRUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
4 0
3 years ago
Read 2 more answers
What types of electromagnetic radiation does the sun emit?
Ksju [112]

Answer:

https://gml.noaa.gov/education/info_activities/pdfs/LA_radiation.pdf

Explanation:

5 0
2 years ago
Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbselectromagnetic wa
FrozenT [24]

Answer:

D. 4000 km

Explanation:

f = Frequency of wave that is being transmitted = 76 Hz

\lambda = Wavelength of wave that is being transmitted

v = The velocity of electromagnetic waves through air = 3\times 10^8\ m/s

Velocity of a wave is given by

v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{3\times 10^8}{76}\\\Rightarrow \lambda=3947368.42105\ m=3947.36842105 km\approx 4000\ km

Hence, the approximate wavelength of the waves is 4000 km

8 0
3 years ago
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