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qwelly [4]
3 years ago
12

What are the primary methods by which planets have been found around other stars in our galaxy

Physics
1 answer:
Andrei [34K]3 years ago
6 0

<u>Answer</u>:

The primary method by which the planets have been found around other stars in the galaxy is Kepler Space Telescope.

<u>Explanation</u>:

The primary method by which planets have been found around other stars in our beautiful Galaxy is the Kepler space telescope. Kepler launched Kepler Telescope in 2009 to help to find terrestrial planets super-earth etc.

This is the telescope which measures the amount of light coming from a star and detects a slight variation in brightness as a planet passes in front. Using this technique NASA's Kepler Mission has founded thousands of candidate planet in this world.                      

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What is the volume of a 1.2kg and displaced 1.0g/cm3
Zinaida [17]
Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!
4 0
3 years ago
How would you characterize William's relationship with his father? In what ways did his father help shape William's outlook?
Aleksandr [31]

William's father shaped William's outlook on life when he was younger due to his bravery.

<h3>What was the relationship between William and his father?</h3>

The impact of significant others in the life of a person can not be over emphasized. These significant others often serve as role models and help to develop the individuality and personality of people.

The life of William was shaped quite well by his father because the both of them had a cordial relationship with each other. William's father shaped William's outlook on life when he was younger due to his bravery.

Learn more about William and his father:brainly.com/question/4066623

#SPJ1

3 0
1 year ago
An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal
alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

        V = IR

     and volage

   V = \epsilon - Ir

now,

   IR = \epsilon - Ir

   I(R+r) = \epsilon

   I= \dfrac{\epsilon}{R+r}

inserting the values

   I= \dfrac{12}{0.05+0.01}

      I = 200 A

b) Voltage

   V = I R

   V = 200 x 0.05

   V = 10 V

c) Power

    P = I V

    P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

 I= \dfrac{\epsilon}{R+r}

   I= \dfrac{12}{0.14+0.01}

     I = 80 A

     V = 80 x 0.05 = 4 V

     P = 4 x 80 = 320 W

6 0
3 years ago
Read 2 more answers
The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit
Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, P=2.2\times 10^5\ Pa

Area of each tire, A=0.023\ m^2

Area of 4 tires,  A=0.092\ m^2

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=2.2\times 10^5\times 0.092

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0
3 years ago
A lamp is rated for 240v,2.5A.what is the cost of using the lamp for 3hrs if 1kWh of electricity cost #12​
marshall27 [118]

Answer:

≈ 22¢

Explanation:

240 / 1000 = 0.240 kV

0.240 kV(2.5 A)(3 hr) = 1.8 kW•hr

1.8 kW•hr($0.12/kW•hr) = $0.216

6 0
3 years ago
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