Each length unit on the speedway diagram represents one-tenth of a mile. What is the area of the sector associated with turn A i
n units of square miles? Use the value 3.142, and round your answer to four decimal places.
1 answer:
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1. The distance between the perihelion and the aphelion is 116 million miles
2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles
3. The equation of the elliptical orbit of Mercury is
4. The eccentricity of the ellipse is 0.207 to the nearest thousandth
5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle
Step-by-step explanation:
Let us revise the equation of the ellipse is
, where the major axis is parallel to the x-axis
- The length of the major axis is 2a
- The coordinates of the vertices are (± a , 0)
- The coordinates of the foci are (± c , 0) , where c² = a² - b²
∵ The Sun is located at a focus of the ellipse
∴ The sun located ate c
∵ The perihelion is the point in a planet’s orbit that is closest to the
Sun ( it is the endpoint of the major axis that is closest to the Sun )
∴ The perihelion is located at the vertex (a , 0)
∵ The closest Mercury comes to the Sun is about 46 million miles
∴ The distance between a and c is 46 million miles
∵ The aphelion is the point in the planet’s orbit that is furthest from
the Sun ( it is the endpoint of the major axis that is furthest from
the Sun )
∴ The aphelion is located at the vertex (-a , 0)
∵ The farthest Mercury travels from the Sun is about 70 million miles
∴ The distance from -a to c is 70 million miles
∴ The distance between the perihelion and the aphelion =
70 + 46 = 116 million miles
1. The distance between the perihelion and the aphelion is 116 million miles
∵ The distance between the perihelion and the aphelion is the
length of the major axis of the ellipse
∵ The length of the major axis is 2 a
∴ 2 a = 116
- Divide both sides by 2
∴ a = 58
∴ The distance from the center of Mercury’s elliptical orbit to the
closest end point to the sun is 58 million miles
∵ The distance between the sun and the closest endpoint is
46 million miles
∴ The distance from the center of Mercury’s elliptical orbit and
the Sun = 58 - 46 = 12 million miles
2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles
∵ The major axis runs horizontally
∴ The equation is
∵ a = 58
∵ c is the distance from the center to the focus of the ellipse
∴ c = 12
∵ c² = a² - b²
∴ (12)² = (58)² - b²
- Add b² to both sides
∴ (12)² + b² = (58)²
- Subtract (12)² from both sides
∴ b² = (58)² - (12)² = 3220
- Substitute these values in the equation
∴
3. The equation of the elliptical orbit of Mercury is
The eccentricity (e) of an ellipse is the ratio of the distance from the
center to the foci (c) and the distance from the center to the
vertices (a) ⇒
∵ c = 12
∵ a = 58
∴ = 0.207
4. The eccentricity of the ellipse is 0.207 to the nearest thousandth
If the eccentricity is zero, it is not squashed at all and so remains a circle.
If it is 1, it is completely squashed and looks like a line
∵ The eccentricity of the ellipse is 0.207
∵ This number is closed to zero than 1
∴ The shape of the ellipse is near to the shape of the circle
5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle
Learn more:
You can learn more about conics section in brainly.com/question/4054269
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Answer:
or 2.738
Step-by-step explanation:
Let’s just look at the triangle on the top with the on the top and x on the bottom. (Basically the top half to the equilateral triangle)
There is a small square in the bottom right corner, which indicates that this triangle is a right triangle. This means that we can use the Pythagorean Theorem:
We know that \sqrt{10} is our hypotenuse, and therefore our c in our equation. Let’s say that x=a in our equation. Therefore we are left to find b. However, b is half the length of the side of the original equilateral triangle. An equilateral triangle means that all three sides are the same length. Therefore our side would also be \sqrt{10} units long. However we know that b is half of that value, so b= or
Plugging these values into the equation:
x^2+ (\frac{\sqrt{10} }{2})^{2}=\sqrt{10} ^{2}
This approximately equals 2.738