Ask question

Ask question

All categories

3 years ago

8

The average distance of the earth from the sun is about 1.5 × 10^8 km. Assume that the earth’s orbit around the sun is circular

and that the sun is at the origin of your coordinate system. (a) Estimate the speed of the earth as it moves in its orbit around the sun. Express your answer in miles per hour with the appropriate number of significant figures. (b) Estimate the angle θ between the position vector of the earth now and what it will be in 4 months. (c) Calculate the distance between these two positions.

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

Explanation:

Distance between sun and earth will act as radius of circular path

R = 1.5 X 10¹¹ m

angular velocity of the earth

ω = 2π / T

= 2 X 3.14 / ( 365 X 24 X 60 X 60 )

= 1.99 X 10⁻⁷ radian / s

velocity v = ω R

= 1.99 X 10⁻⁷ X 1.5 X 10¹¹

= 2.985 X 10⁴ m /s

= 29.85 km/s

29.85 x 60 x 60 km/h

= 107460 km/h

107460/1.6 mile / h

= 67162.5 mile /h

b ) Angle moved in 12 months

= 360 degree

angle moved in 4 months

= 360 / 3

= 120 degree

c ) Circumference of the orbit

= 2 x 3.14 x 1.5 x 10⁸ km

9.42 x 10⁸ km

required distance will be 1/3 rd of circumference

= 9.42 / 3 x 10⁸ km

= 3.14 x 10⁸ km

You might be interested in

What's the Genotype for these? If your answer is something off-topic I ShAlL rEpOrT iT u.U

RideAnS [48]

Pick any free letter for each trait. For example, "A" for the dominant allele and "a" for the recessive allele for the number of arms. Assuming complete dominance, the genotype for the dominant trait will always have at least one "capital" allele, so a creature with 2 arms would have genotype either AA (homzygous dominant) or Aa (heterozygous), while a creature with the recessive trait of having 4 arms will always be homzygous recessive with genotype aa.

4 0

2 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

<u>For process 2:</u>

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

<u>For process 3:</u>

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

<u>For process 4:</u>

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

<u>For process 5:</u>

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

What is a force that attracts all matter to each other?

Temka [501]

The answer is gravity. I hope this helps.

5 0

3 years ago

Read 2 more answers

How do theories differ from Laws and Principles?

g100num [7]

A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.

8 0

3 years ago

Summarize: Based on what you have learned, how will the sound that the observer hears

Reil [10]

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

Explanation: Let assume that the observer is at a stationary position. The wavelength of the sound from the airplane reduces and the frequency increases as the plane is moving toward the observer. As the airplane passes by, that is, moving away from the observer, the frequency starts to reduce while the wavelength of the sound starts to increase.

The sound that the observer hears will change base on the illustration above.

3 0

2 years ago