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3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

om Saturn to the sun and then calculate it.

Physics

1 answer:

soldi70 [24.7K]3 years ago

3 0

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

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At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

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]

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

Explanation:

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:

For process 1:

We are given:

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

Putting values in equation 1, we get:

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

For process 2:

We are given:

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

Putting values in equation 2, we get:

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

For process 3:

We are given:

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

Putting values in equation 1, we get:

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

For process 4:

We are given:

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

Putting values in equation 2, we get:

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

For process 5:

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

?????Help please?????

Sindrei [870]

The correct selections are C, C, B, D, A, B, and A .

6 0

3 years ago

How is the temperature of a gas related to the kinetic energy of its particles

love history [14]

The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move. All gases at a given temperature have the same average kinetic energy. Lighter gas molecules move faster than heavier molecules.

7 0

3 years ago

Match the lithification processes.

Varvara68 [4.7K]

Answer:

ReCrystallization

Contact pressure causing grains to "grow" together.

It's a metamorphic process used to rearrange the atoms of the minerals in order to modify them in to the required form under the specific temperature and pressure. For example conversion of limestone into marble.

Cementation:

Precipitation of bonding agents between grains.

Cementation is the process in which the minerals in the form of fluid fills in spaces between the grains and binds them together upon crystallizing.

Compaction:

Increase in density due to weight of overburden.

Compaction as it's name indicate is the compaction of the sediments due to the heavy weight of the rocks which reduce their size and increase their density.

5 0

3 years ago