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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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It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
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<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
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3 years ago
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1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

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For the first bulb:

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Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

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|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

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L = √(4+16+9)

L = √29

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Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

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(c) 2i − k

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L = √(4+0+1)

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(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

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L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

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(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

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(f) i + j + k

The length is

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- Strong nuclear force: it is the strongest of all forces. It is responsible for holding the nucleons together inside the nucleus, and it is attractive. It has a very limited range (10^{-15}m), so it is relevant only at very small scales

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Tommy sings during the procedure because the region of the brain responsible for memory and singing is stimulated by this procedure.

Learn more about brain-stimulating therapies here:

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