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2 years ago

If a force of 24 N is applied to an object with a mass of 5 kg, the object will accelerate at _____ m/s2

Physics

1 answer:

Luba_88 [7]2 years ago

7 0

4.8m/s2 according to the formula F=MA

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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

Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is

Talja [164]

Randall has unconscious assumption that attractive people are more competent

4 0

2 years ago

Who published the first journal of experimental psychology?

raketka [301]

Answer:

Who published the first journal of experimental psychology?

American Psychological Association published the first journal of experimental psychology.

xXxAnimexXx

Happy labor day!

6 0

2 years ago

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PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!

adelina 88 [10]

A line on a graph has a downward slope

8 0

2 years ago

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0

2 years ago