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AURORKA [14]
3 years ago
10

The Kepler Space Telescope is searching for extrasolar planets by the transit method. It is necessary for Kepler to photometrica

lly monitor a large number of stars because ____.
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0

Answer:

The Kepler Space Telescope is searching for extrasolar planets by the transit method. It is necessary for Kepler to photometrically monitor a large number of stars because increase the probability to see a transit.

Explanation:

Photometry is the study of the intensivity of light radiated from a particular object.

In the other hand, the transit method consists in the measured of the dimming on the brightness of a star when a planet is passing in front of it, as long as the star, the planet and the detector (in this case the Kepler Telescope) are in the same line of sign.  

However, that transit has a short duration. So it is necessary that the Kepler Telescope monitorates the brightness of several stars each thirty minutes in order to increase the probability of detection of a transit.

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(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​
Musya8 [376]

Answer:

the answee is

2.968456 ×10^7

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What are the four layers of the earth
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Answer:

earth's crust

meatle

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A 0.171-kilogram hockey puck is struck by a hockey stick. If the force from the hockey stick is 30.2 Newtons, but the puck also
Komok [63]

Answer:

a = 156.14 m/s^2

Explanation:

Using the laws of newton:

∑F = ma  

where ∑F is the sumatory of forces, m the mass and a the aceleration.

so:

F -f_k = ma

where F is the force from the hockey stick and f_k is the force of friction.

Replacing values, we get:

(30.2N) - (3.5N)= (0.171kg)a

Finally, solving for a:

a = 156.14 m/s^2

4 0
3 years ago
PLS HELP ME!
jeyben [28]

Answer:

\underline{ \boxed{ yes}}\\

Explanation:

given : initial \: velocity \: (u )= 40 {ms}^{ - 1}  \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1}  \\ given :   - (acceleration) \: (a_r) = 2 {ms}^{ - 2}  \\ given : distance  \: (s) \: =   \: ? : \\  but \:  {v}^{2}  =  {u}^{2}  + 2( a)s\\  {0}^{2}  =  {40}^{2}  + 2( - 2)s \\  -  {40}^{2}  =  - 4s \\ s =  \frac{ -  {40}^{2} }{ - 4}  \\ s =  \frac{1600}{4}  \\s = 400 \: m

3 0
3 years ago
A charge of +0.001 C is 1 m to your right and another charge of +1000 C is 1 m to your left. You are holding a charge of −1 C. W
almond37 [142]

Answer:

C and D

Explanation:

Charge on right = +0.001 C

Charge on left = 1000 C

Charge held in between = -1C

Ratio between right and left charge:

=\frac{1000}{.001}\\\\=1\times 10^6

Which shows charge on left exerts 1,000,000 times more force then charge on right

Charge on left is 1000 times greater in magnitude than charge held between and charge held between is 1000 times greater in magnitude than charge on right. So the magnitude of the force on the charge you are holding would be the same if it were +1 C instead of −1 C.

8 0
3 years ago
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