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3 years ago

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A thin rod of length 0.79 m and mass 130 g is suspended freely from one end. It is pulled to one side and then allowed to swing

like a pendulum, passing through its lowest position with angular speed 3.61 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

a) 0.3965 j

b) 0.3112 m

Explanation:

The picture attached explains it all. Thank you

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A wire runs left to right and carries a current in the direction shown. What is the direction of the magnetic field at point Z?

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The correct option is out of the screen.

As the motion of positive charge is the direction of current in the wire. From the right-hand curl rule, the magnetic field direction will be outside the paper or the screen. As the <span>wire runs left to right and carries a current in the direction from left to right, the magnetic field lines will be outside the screen.</span>

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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.

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<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

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Answer:

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It has many applications,

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Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/s through a horizontal tube with a 30mm diameter. wh

MariettaO [177]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>

We have the Poiseuille's formula,

$Q=\frac{\pi Pr^4}{8\beta l}$

where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, $\beta$ is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.

It is given that,

$Q=8*10^{-6}m^3/s\\diameter=30mm, thus,\\r=15mm\\l=100m\\\beta =0.95$

Thus, the pressure drop will be,

$P=\frac{8Q\beta l}{\pi r^4} =\frac{8*8*10^{-6}*0.95*100}{3.14*(15*10^{-3})^4} \\\\P=3.824*10^4Pascals.$

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4 Pascals.

Learn more about the Poiseuille's formula here:

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