All categories

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.

You might be interested in

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

(pleases show work)

Feliz [49]

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

$v=u+at$

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s is the initial velocity

$a=+3 m/s^2$ is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

$v=16 + (3)(10)=46 m/s$

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

$s=ut+\frac{1}{2}at^2$

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s is the initial velocity

$a=+3 m/s^2$ is the acceleration

t = 10 s is the time

Substituting into the equation,

$s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m$

4 0

3 years ago

An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni

Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, $v=(2\times 10^6i+3\times 10^6j)\ m/s$

Magnetic field, $B=(0.030i-0.15j)\ T$

Charge of electron, $q_e=-1.6\times 10^{-19}\ C$

(a) Let $F_e$ is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

$F_e=q_e(v\times B)$

$F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]$

$F_e=-1.6\times 10^{-19}\times (-390000)(k)$

$F_e=6.24\times 10^{-14}k\ N$

(b) The charge of electron, $q_p=1.6\times 10^{-19}\ C$

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0

3 years ago

A resistor, capacitor, and switch are all connected in series to an ideal battery of constant terminal voltage. Initially, the s

erma4kov [3.2K]

Answer:

The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.

Explanation:

This is because when a capacitor is charged no current or voltage flows through it so it will have a voltage equal to the terminal voltage of the battery

6 0

3 years ago

Does Pascals law apply to solids?

monitta

No it does not . That is the answer

5 0

2 years ago