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BlackZzzverrR [31]
2 years ago
14

Which force stops the car from moving?

Physics
2 answers:
Alex73 [517]2 years ago
8 0

Answer:

The force of friction.

Explanation:

Gravity keeps the car on the ground.

Motion Allows the car to move.

The force of speed doesnt make sense.

Friction would cause the car to stop moving.

yawa3891 [41]2 years ago
3 0
Yes, friction is the force enabling the car to move along the road.
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Explain what kind of heat transfer occurs when you burn yourself on a hot car seat in the summer.
liubo4ka [24]
Conduction is a mode of transfer of heat there
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3 years ago
White blood cells are most closely associated with which two body’s systems
Vlad1618 [11]

Answer:

Explanation:

circulatory and digestive. ... An immune response is primarily due to the body's white blood cells recognizing.

8 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
Choose all the answers that apply.
Serhud [2]
<span>According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

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8 0
3 years ago
Evaluate u+xy where U=3 X=4 Y=7
Reptile [31]

Answer:

31

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

31

6 0
3 years ago
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