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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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aivan3 [116]

The reason for people to swim easier in salt water than fresh water is because of buoyancy

Explanation:

In fresh water there is lack of minerals and has fresh water alone. The density of fresh water is 1000 kg/m³.  Hence, in fresh water  cannot exert the suitable buoyancy for the swimmer to float easier than that of salt water.

But in Salt water due to enrichment of salts  and minerals it is found that salt water has more density than fresh water. Here the salt water offers more buoyancy to the swimmer to lift him up in the water surface and to swim faster and easier than fresh water.

It is similar to that egg floats in the salt water and sinks inside the fresh water because of its own body weight.

4 0
3 years ago
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5. Convert 0.984 miles per hour to meters per minute
Shtirlitz [24]

Answer:

This is the answer bro hope it will help you

7 0
3 years ago
Read 2 more answers
Should genetic genealogy be used to solve crimes ? Yes or no and why?
zheka24 [161]

Answer:

yes

Explanation:

because it will help find who did the crime and it can also open jobs and opportunities for the people who likes sicnenae

8 0
3 years ago
Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i
Arte-miy333 [17]
I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
R_x=0.96m
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
R_y=-0.84m
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}
v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}
Total velocity is:
v=\sqrt{v_x^2+v_y^2}
Once we calculate velocities we get:
v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0
3 years ago
Someone please help its a simple power problem.
SOVA2 [1]
Well 200 doubled or (x2)=400 if that’s what it means
7 0
3 years ago
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