Read the article planet. You will use information from this article to answer the following question. Make a list of the distances from each planet to the sun written in scientific notation. If there is a shortest and longest distance use the average distance from the sun

Answer 1

It seems I do not have access to external articles or documents, including the one you mentioned regarding planets and their distances from the sun. However, I can provide you with average distances from the sun for each planet in our solar system in scientific notation:

1. Mercury: \( 5.79 \times 10^{10} \) meters (or about 0.39 AU)
2. Venus: \( 1.08 \times 10^{11} \) meters (or about 0.72 AU)
3. Earth: \( 1.50 \times 10^{11} \) meters (or about 1.00 AU)
4. Mars: \( 2.28 \times 10^{11} \) meters (or about 1.52 AU)
5. Jupiter: \( 7.78 \times 10^{11} \) meters (or about 5.20 AU)
6. Saturn: \( 1.43 \times 10^{12} \) meters (or about 9.58 AU)
7. Uranus: \( 2.87 \times 10^{12} \) meters (or about 19.22 AU)
8. Neptune: \( 4.50 \times 10^{12} \) meters (or about 30.07 AU)

If you have the specific values from the article and there is a need to consider the shortest and longest distances, please provide those distances or any additional information you want to include, and I can help you compute the average or perform other calculations as needed.

Answer 2

Choose two of the distances of planets from the sun and multiply them. Show your work

Answer 3

Let's choose the distances of Earth and Mars from the Sun to perform the multiplication.

1. Distance from the Sun:

   • Earth: \( 1.50 \times 10^{11} \) meters
   • Mars: \( 2.28 \times 10^{11} \) meters

2. Multiplication: We will use the property of multiplying numbers in scientific notation, which is:

   \[ (a \times 10^m) \times (b \times 10^n) = (a \times b) \times 10^{m+n} \]

   Applying this to our distances:

   \[ (1.50 \times 10^{11}) \times (2.28 \times 10^{11}) = (1.50 \times 2.28) \times 10^{11 + 11} \]

   First, we multiply the coefficients (1.50 and 2.28):

   \[ 1.50 \times 2.28 = 3.42 \]

   Now, add the exponents (11 + 11):

   \[ 11 + 11 = 22 \]

   Putting it all together:

   \[ 3.42 \times 10^{22} \]

Thus, the product of the distances from the Sun for Earth and Mars is:

\[ 3.42 \times 10^{22} \text{ meters}^2 \]

Answer 4

Choose two of the distances of planets from the sun and divide them. Show your work.

Answer 5

Let's choose the distances of Jupiter and Earth from the Sun to perform the division.

1. Distance from the Sun:

   • Jupiter: \( 7.78 \times 10^{11} \) meters
   • Earth: \( 1.50 \times 10^{11} \) meters

2. Division: We will use the property of dividing numbers in scientific notation, which is:

   \[ \frac{(a \times 10^m)}{(b \times 10^n)} = \left(\frac{a}{b}\right) \times 10^{m-n} \]

   Applying this to our distances:

   \[ \frac{(7.78 \times 10^{11})}{(1.50 \times 10^{11})} = \left(\frac{7.78}{1.50}\right) \times 10^{11 - 11} \]

   First, we divide the coefficients (7.78 by 1.50):

   \[ \frac{7.78}{1.50} \approx 5.1867 \quad (\text{rounded to four decimal places}) \]

   Next, we subtract the exponents (11 - 11):

   \[ 11 - 11 = 0 \]

   Putting it all together:

   \[ 5.1867 \times 10^{0} \]

Since \(10^0 = 1\), we can simplify this to:

\[ 5.1867 \]

Thus, the result of dividing the distance from the Sun for Jupiter by the distance for Earth is approximately \(5.19\), indicating that Jupiter is about 5.19 times farther from the Sun than Earth.

Answer 6

Choose different planets

Answer 7

Let's choose the distances of Saturn and Mars from the Sun to perform the division.

1. Distance from the Sun:

   • Saturn: \( 1.43 \times 10^{12} \) meters
   • Mars: \( 2.28 \times 10^{11} \) meters

2. Division: We will again use the property of dividing numbers in scientific notation:

   \[ \frac{(a \times 10^m)}{(b \times 10^n)} = \left(\frac{a}{b}\right) \times 10^{m-n} \]

   Applying this to our distances:

   \[ \frac{(1.43 \times 10^{12})}{(2.28 \times 10^{11})} = \left(\frac{1.43}{2.28}\right) \times 10^{12 - 11} \]

   First, we divide the coefficients (1.43 by 2.28):

   \[ \frac{1.43}{2.28} \approx 0.6272 \quad (\text{rounded to four decimal places}) \]

   Next, we subtract the exponents (12 - 11):

   \[ 12 - 11 = 1 \]

   Putting it all together:

   \[ 0.6272 \times 10^{1} = 6.272 \]

Thus, the result of dividing the distance from the Sun for Saturn by the distance for Mars is approximately \(6.27\). This indicates that Saturn is about 6.27 times farther from the Sun than Mars.

Answer 8

Can u do Uranus and Neptune

Answer 9

Sure! Let's divide the distances of Uranus and Neptune from the Sun.

1. Distance from the Sun:

   • Uranus: \( 2.87 \times 10^{12} \) meters
   • Neptune: \( 4.50 \times 10^{12} \) meters

2. Division: We will use the property of dividing numbers in scientific notation:

   \[ \frac{(a \times 10^m)}{(b \times 10^n)} = \left(\frac{a}{b}\right) \times 10^{m-n} \]

   Applying this to our distances:

   \[ \frac{(2.87 \times 10^{12})}{(4.50 \times 10^{12})} = \left(\frac{2.87}{4.50}\right) \times 10^{12 - 12} \]

   First, we divide the coefficients (2.87 by 4.50):

   \[ \frac{2.87}{4.50} \approx 0.6389 \quad (\text{rounded to four decimal places}) \]

   Next, we subtract the exponents (12 - 12):

   \[ 12 - 12 = 0 \]

   Putting it all together:

   \[ 0.6389 \times 10^{0} = 0.6389 \]

Thus, the result of dividing the distance from the Sun for Uranus by the distance for Neptune is approximately \(0.639\). This indicates that Uranus is about 63.9% the distance of Neptune from the Sun, or conversely, Neptune is farther from the Sun than Uranus.

Answer 10

Simplify it

Answer 11

Certainly! To simplify the result of the division of the distances of Uranus and Neptune, we calculated:

\[ \frac{(2.87 \times 10^{12})}{(4.50 \times 10^{12})} \approx 0.6389 \]

Since \(10^0 = 1\), we can express that more simply. The simplified result indicates that:

\[ \frac{2.87}{4.50} \approx 0.639 \]

This means that Uranus is approximately 0.639 times the distance from the Sun compared to Neptune.

To express it as a fraction for further clarity, we can say:

\[ \frac{Uranus}{Neptune} \approx \frac{0.639}{1} \quad \text{or} \quad Uranus \approx 63.9% \text{ of Neptune’s distance from the Sun.} \]

Thus, the final simplified result is:

\[ \text{Uranus is approximately } 0.639 \text{ times the distance from the Sun as Neptune.} \]