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dalvyx [7]
4 years ago
7

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

r because it is an extra assignment.
4. How far away from the sun was Mercury at aphelion? On what day did aphelion occur?
5. How far away from the sun was Mercury at perihelion? On what day did it occur?
6. How many days elapsed between aphelion and perihelion? What percent of the time to complete one orbit was this?
7. What was the average radius of Mercury in its orbit that you calculated? How does it compare to the accepted value of 0.387 AU? Calculate the percent error using the following equation. Show your work.

If someone could explain it to me I would really appreciate it!
Physics
1 answer:
Hoochie [10]4 years ago
3 0

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
 
This also happens to be 1/2 of the major axis of the elliptical orbit.


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A 75.0 kg students sits 1.15 m away from a 68.4 kg student. What is the force of gravitational attraction between them?
AVprozaik [17]

Answer:

F=2.589×10⁻⁷ Newtons

Explanation:

The gravitational force of attraction between two bodies is given by

F= Gm₁m₂/r²

Where m₁ and m₂ are the masses of the two bodies, G is the universal gravitational constant while r is the distance between the two bodies.

G=6.67408 × 10⁻¹¹m³kg⁻¹s⁻²

r=1.15m

m₁=75.0kg

m₂=68.4kg

Therefore F= (6.67408×10⁻¹¹m³kg⁻¹s⁻²×75.0kg×68.4kg)/(1.15m)²

F=2.589×10⁻⁷ Newtons

7 0
4 years ago
Convert 45km to miles show your work
amid [387]

Answer:

distance = 27.95 [mi]

Explanation:

in order to solve this problem, we must use the appropriate conversion factor, i.e. a conversion factor that relates the kilometers to Miles.

1 [km] = 0.6214 [mill]\\45[km]*0.6212[\frac{mill}{1km} ]=27.95 [mill]

7 0
3 years ago
0.00432 km = _____ mm
melomori [17]
Hello there

<span>0.00432 km =4320mm

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8 0
3 years ago
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Compare the properties of sodium chloride and sand
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I am 11 year old and I don't known the answer to this question
6 0
3 years ago
A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal su
nataly862011 [7]

10.7 rad/s is the final angular velocity of the stick.

Given:

Mass of the stick = 4.42 kg

Length of the stick = 1.23m

Force of impulse (I) = 12.8 N s

The linear velocity of the stick, v=\frac{I}{m}

                                                  v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}

                                                  v  = 2.89 m/s

Therefore, the final linear velocity of the stick is 2.89 m/s

∴w=\frac{12 Ir}{ml^{2} }

w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}

w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}

w=10.7   rad/s

Therefore, 10.7 rad/s is the final angular velocity of the stick.

Learn more about linear velocity here:

brainly.com/question/15154527

#SPJ4

                                                 

5 0
2 years ago
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