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2 years ago

ASAP pls answer right and I will mark brainiest

Physics

1 answer:

Ronch [10]2 years ago

6 0

1. D:Asteroids have geological activity
2.A:Short-period comets circle around the sun in 200 years or less and originate in the Kuiper Belt
3.D:All of the above
4.A:It’s gravity is too weak to clear it’s orbit
5.B:Ceres
6.C:Haumea
7.A:Is bigger than Pluto

I hope this helps

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Usually, when the temperature is increased, what will happen to the rate of dissolving?

ArbitrLikvidat [17]

It will decrease

When the temperature increased, the rate will decrease

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3 years ago

A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor

tankabanditka [31]

100 J

Please mark me brainliest it would be greatly appreciated haha

5 0

3 years ago

Read 2 more answers

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

According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme

Bas_tet [7]

The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.

8 0

2 years ago

The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i

Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

$\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170$

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0

3 years ago