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Hitman42 [59]
3 years ago
14

A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s

tar. What is the planet’s orbital period in Earth years? Round your answer to the nearest whole number. __ Earth years
Physics
2 answers:
nadezda [96]3 years ago
7 1

Answer:

16 years just got it wrong for choosing 4.6 believe me if u want

Explanation:

kondaur [170]3 years ago
4 1

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

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nataly862011 [7]
Metals are the best conductors.
8 0
3 years ago
Read 2 more answers
Question 2 (1 point)
mixer [17]

Answer:

The initial vertical velocity is zero, u = 0 m/s

Explanation:

Given;

height of the table, h = 0.55 m

horizontal distance traveled by the tennis, x = 0.12 m

Apply the following kinematic equation;

h = ut + ¹/₂gt²

where;

u is the initial vertical velocity = 0, since the tennis ball rolled off the edge of a table.

h = ¹/₂gt²

The time to fall from the vertical height is given by;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2(0.55)}{9.8} }\\\\t = 0.335 \ s

The initial horizontal velocity of the tennis is given by;

x = vₓt

vₓ = x / t

vₓ = (0.12) / (0.335)

vₓ = 0.358 m/s

Therefore, the initial vertical velocity is zero, u = 0 m/s and initial horizontal velocity, vₓ is 0.358 m/s

3 0
3 years ago
What feature of an object does not affect air resistance?
lakkis [162]
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7 0
3 years ago
A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
If a FM radio station broadcasts at 80. 3 MHz (megahertz), what is its wavelength in m (speed of light 3. 0 x 108 m/s)
hammer [34]

Answer:

Wavelength = 3.74 m

Explanation:

In order to find wavelength in "metres", we must first convert megahertz to hertz.

1 MHz = 1 × 10⁶ Hz

80.3 Mhz = <em>x</em>

<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz

The formula between wave speed, frequency and wavelength is:

v = fλ  [where v is wave speed, f is frequency and λ is wavelength]

Reorganise the equation and make λ the subject.

λ = v ÷ f

λ = (3 × 10⁸) ÷ (8.03 × 10⁷)

λ = 3.74 m [rounded to 3 significant figures]

8 0
2 years ago
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