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3 years ago

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 must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward

Yuri [45]

Answer:

$q = 2.067 \times 10^{-5}\ C$

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

$q =\dfrac{mg}{E}$

$q =\dfrac{0.00141 \times 9.81}{670}$

$q = 2.067 \times 10^{-5}\ C$

Charge of the particle is equal to $q = 2.067 \times 10^{-5}\ C$

6 0

3 years ago

There is up to 30 times more gold in a tons of old mobile phones than in a tons of gold ore. true or false

FrozenT [24]

Answer:

true

Explanation:

dbgkdodocofkci ifkdcl k kfododocp v

3 0

2 years ago

A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan

spin [16.1K]

The ideal mechanical advantage of the pulley system is 3

7 0

3 years ago

Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120

igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter.

As you have the circumference of B, you divide it by Pi to get the Diameter.

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter.

As' radius and Diameter will be 3x greater than B.

38.2 x 3 = 114.6

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0

3 years ago

The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

vitfil [10]

The expression of the electric flux is

$\Phi = \frac{Q}{\epsilon_0}$

Here,

Q = Total charge enclosed in the closed surface

$\epsilon_0$ = Permittivity due to free space

Rearranging to find the charge,

$Q = \epsilon_0 \Phi$

Replacing with our values we have finally

$Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)$

$Q = 1.6284*10^{-8} C$ $(\frac{10^9nC}{1C})$

$Q = 0.1684nC$

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0

3 years ago