Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
1 answer:
Answer:
B. About 12 degrees
Explanation:
The orbital period is calculated using the following expression:
T = 2π*()
Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.
Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.
Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:
=
Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be years, or 31.62 years.
We find the amount of degrees it moves in 1 year:
or about 12 degrees.
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Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
Answer:
Explanation:
-The chemical formula for Molybdenum (V) Dichromate is
-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate
-We apply Avogadro's constant to find the number of atoms of oxygen:
Hence, there are
Answer:
The length at the final temperature is 11.7 cm.
Explanation:
We need to use the thermal expansion equation:
Where:
- L(0) is the initial length
- ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
- ΔL is the final length minus the initial length (L(f)-L(0))
- α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)
So, we have:
Therefore, the length at the final temperature is 11.7 cm.
I hope it helps you!