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Vlad [161]
3 years ago
14

A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su

n. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? The planet's orbital period will be four times Earth's orbital period. The planet's orbital period will be one-half Earth's orbital period. The planet's orbital period will be one-fourth Earth's orbital period. The planet's orbital period will be equal to Earth's orbital period. The planet's orbital period will be twice Earth's orbital period.
Physics
1 answer:
aniked [119]3 years ago
7 0

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

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In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
2 years ago
A juggler is practicing with one ball. It takes 2.4 seconds for the ball to leave her hand and return to her hand. How fast did
nikklg [1K]

The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

  • Time t = 2.4 s

To find

  • Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

          y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

   

With the ball in hand, its position is zero

         0 = 0 + v₀ t - ½ g t²

         v₀ t - ½ g t² = 0

         v₀ = ½ g t

Let's calculate

         v₀ = ½ 9.8 2.4

         v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

5 0
2 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 634-nm
eimsori [14]

Answer:

minimum thickness of the coating = 122.868 nm

Explanation:

Given data

lens index of refraction = 1.29

wavelength = 634 nm

glass index of refraction = 1.53

to find out

minimum thickness of the coating

solution

we have given non reflective coating

so

we know that minimum thickness of the coating formula

minimum thickness of the coating = Wavelength / 4n

here n is coating index of refraction

so put here both value to get thickness

minimum thickness of the coating = Wavelength / 4n

minimum thickness of the coating = 634 / 4 ( 1.29 )

so minimum thickness of the coating = 122.868 nm

5 0
3 years ago
Two electric motors drive two elevators of equal mass in a three-story building 10 meters tall. Each elevator has a mass of 1,00
algol [13]

Answer:

The power output of the first motor is,  P = 2.0 x 10⁴ watts

Explanation:

Given data,

The height of the building, h = 10 m

The mass of the elevator, m = 1000 kg

The time duration of the motor to do this work, t = 5.0 s

The force acting on the elevator,

                                  F = m x g

                                    = 1000 x 9.8

                                     = 9800 N

The work done by the elevator,

                                W = F  x h

                                     = 9800 x 10

                                     = 98000 J

The power output of the first motor,

                                P = W / t

                                   = 98000 / 5

                                   = 19600 watts

                                   = 1.96 x 10⁴ watts

Hence, the power output of the first motor is, P = 2.0 x 10⁴ watts

3 0
3 years ago
The current in an electric hair dryer is 10
Vikki [24]
First let's convert the time in seconds:
\Delta t= 5 min= 5 min \cdot (60 s/min)= 300 s

The current is defined as the quantity of charge flowing through a certain section of a circuit per unit of time:
I= \frac{Q}{\Delta t}
Using I=10 A, and \Delta t=300 s, we can find the amount of charge flown through the hair dryer in this time:
Q=I \Delta t=(10 A)(300 s)=3000 C

The charge of a single electron is q=1.6 \cdot 10^{-19} C, so the number of electrons flown through the hair dryer is the total charge divided by the charge of a single electron:
N= \frac{Q}{q}= \frac{3000 C}{1.6 \cdot 10^{-19} C} =1.88 \cdot 10^{22}
8 0
3 years ago
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