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natulia [17]
3 years ago
6

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

lculate its orbital period). A. About I degree B. About 12 degrees C About 30 degrees D. About 90 degrees E. 360 degroes
Physics
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

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:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

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Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là
natka813 [3]

Answer:

28 cm and 32 cm

Explanation:

1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of

a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.

As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.

So, the minimum length is 28 cm and the maximum length is 32 cm.

8 0
3 years ago
Which type of energy is released when a bond between atoms is broken
scoray [572]

Answer:

D

Explanation:

7 0
3 years ago
Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

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.

6 0
3 years ago
How many total oxygen atoms are in the compound Molybdenum (V) Dichromate?
alekssr [168]

Answer:

1.2646\times10^{25}\ atoms

Explanation:

-The chemical formula for Molybdenum (V) Dichromate is Mo(Cr_2O_7)_3

-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:

Avogadro's \ Constant=6.022\times 10^{23} \ mol_1\\\\No\ of \ Atoms=Moles\times Avogadro's \ Constant\\\\=21\times 6.022\times 10^{23} \\\\=1.2646\times10^{25}\ atoms

Hence, there are 1.2646\times10^{25} \ atoms

3 0
3 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

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:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0
3 years ago
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