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2 years ago

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur

face of the earth. In this problem, use 9.81m/s2 for the acceleration due to gravity on earth.Part AWhat is its mass on the earth's surface?Part BWhat is its mass on the surface of Io?Part CWhat is its weight on the surface of Io?

1 answer:

kvasek [131]2 years ago

6 0

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

W = mg

W = 5.91 x 9.8

= 10.697N.

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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

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$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

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$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

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Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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