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Paladinen [302]
2 years ago
7

The telescope that allowed astronomers to discover most of the planets found with the transit method was called:.

Physics
1 answer:
kvasek [131]2 years ago
4 0

Answer:This quest took a huge leap forward in 2000 when Hubble studied the exoplanet HD 209458 b, the first extrasolar planet known to make “transits” across the face of its star. Hubble became the first telescope to directly detect an exoplanet's atmosphere and survey its makeup.

Explanation:

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What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio
pantera1 [17]
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),
 
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite. 

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite. 
3 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
The difference between the speed of sound n air at 0°C and the speed of sound in air at 20°c is that... A. Cooler air molecules
sergejj [24]
Heat, like sound, is kinetic energy. Molecules at higher temperatures heave more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly. 

So the answer is A.
7 0
3 years ago
Read 2 more answers
Stars fuse most of the hydrogen in the entire star before they die.<br> 2 poir<br> True<br> False
Margaret [11]
Once all hydrogen is depleted the star begins to die
3 0
2 years ago
A 15 kg object and a 18 kg object are connected by a massless compressed spring and
MatroZZZ [7]

Hi there!

This is an example of a recoil collision.

Using the conservation of momentum:

p_i = p_f

The initial momentum is 0 kgm/s (objects start from rest), so:

p_f = 0

We are given that the 15 kg block has a velocity of 12 m/s to the left, so:

m_1v_1' + m_2v_2' = 0 \\\\15(-12) + 18v_2' = 0 \\\\

Solve for v2':

18v_2' = 180 \\\\v_2' = \boxed{10 m/s}

5 0
2 years ago
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