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Nastasia [14]
3 years ago
14

Which characteristics of Earth’s orbit are in agreement with Kepler’s second law? Check all that apply.

Physics
2 answers:
Nataly [62]3 years ago
6 0

Here is a picture of the correct answers.

artcher [175]3 years ago
6 0

Explanation :  

The angular velocity of the planet in the elliptical orbit will vary. Its shows that the planet  travel slower when farther from the sun, then faster when closer to the sun.

1. Yes, Earth experiences the strongest gravitational force when it is closest to the Sun.

2. Yes,  Earth moves the greatest distance in thirty days when it is closest to the Sun.

3. Yes,  Earth travels with the slowest tangential speed when it is farthest from the Sun.

4. Yes,  Earth sweeps out the same area in the same time frame anywhere in its orbit

Kepler's second law : Kepler describe the motion of the planets around the sun. the line joining the planet and the sun sweeps out  the same area in the same time.

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I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove
Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
     1/2 ( 0.466697 + 0.307499) = 0.387 098  AU
 
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3 0
4 years ago
Convert 1.4×10^9km^3 into cubic meters
Vadim26 [7]
1km=10^3 m,1km^3=10^9cubic metres answer is 1.4x10^18cubic meters
3 0
3 years ago
Read 2 more answers
A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
Doss [256]

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

5 0
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iren2701 [21]

Answer:

d

Explanation:

According to me answer is d but gas expand more than others

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All of the electromagnetic energy radiated from the sun (and from
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