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3 years ago

Match the correct term with the definition

Physics

1 answer:

klio [65]3 years ago

7 0

-Positive particle in the nucleus is a (proton)

-Question two..(isotope)

-Number of protons that an atom has..(atomic number)

-protons + neutrons in the nucleus..(atomic mass)

-neutrally charged element in the nucleus..(neutron)

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9.96 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the c

Archy [21]

Answer:

At 300 kPa: The temperature is 0.529 ºC and the total enthalpy, 2016.77kJ.

At 600 kPa: The temperature is 21.45 ºC and the total enthalpy, 2317.00 kJ.

Explanation:

1. Get the container volume in cubic meters: $1m^{3}=1000L$

$V=14L*\frac{1m^{3}}{1000L} =0.014m^{3}$

2. Find the specific volume of R-134a:

$v=\frac{V}{m}=\frac{0.014m^{3}}{9.96kg}=0.001416\frac{m^{3}}{kg}$

3. Find the phase of R-134a, for this, look at the steam tables (Note: I am using the table B.5.1 from van Wylen 6th Edition.) Look for a pressure of 300 kPa at the table; I found this values:

T(ºC) P(kPa) vf( $\frac{m^{3}}{kg}$ vg( $\frac{m^{3}}{kg}$

0 294.0 0.000773 0.06919

5 350.9 0.000783 0.05833

It is possible to predict that the properties of saturated R134-a at 300 kPa would be so closely to that of 294 kPa. From that data we find that the specific volume of our R134-a is between vf and vg, so we conclude that it is a vapor liquid mixture.

4. Find the quality (x)

From a linear interpolation for the pressure, it is possible to know the saturation data for 300 KPa:

T(ºC) P(kPa) vf( $\frac{m^{3}}{kg}$ vg( $\frac{m^{3}}{kg}$

0.529 300 0.0007405 0.06796

Applying the quality relation for specific volume:

$v=v_{f}+xv_{fg}\\x=\frac{v-v_{f}}{v_{fg}}\\x=0.009989$

The total enthalpy is calculated in the same way:

$h=h_{f}+xh_{fg}\\h=200.71+0.00898*197.96=202.48 \frac{kJ}{kg}$

$H=202.48*9.96=2016.77kJ$

The temperature is 0.529 ºC and the total enthalpy, 2016.77kJ.

When the substance is heated, the volume remains constant because the container is rigid, so the calculation requires to do the same process again with 600 kPa.

T(ºC) P(kPa) vf( $\frac{m^{3}}{kg}$ vg( $\frac{m^{3}}{kg}$

21.45 600 0.000820 0.03458

hf( $\frac{kJ}{kg}$ hg( $\frac{kJ}{kg}$

229.56 406.13

$x=0.0173$

$h=232.63 \frac{kJ}{kg}$

$H=m*h=2317.00kJ$

The temperature is 21.45 ºC and the total enthalpy, 2317.00 kJ.

3 0

4 years ago

How do you calculate the radius of a planets orbit with mass and time ?

Advocard [28]

Answer:

By observing the time between transits, we know the orbital period. Kepler's Third law can be used to determine the orbital radius of the planet if the mass of the orbiting star is known (R3=T2−Mstar/Msun, the radius is in AU and the period is in earth years).

Explanation:

hope this helps!

6 0

3 years ago

A small object and a medium object are both charged. I want to calculate the magnitude of the electric force on the small object

ArbitrLikvidat [17]

I think the answer is D

4 0

3 years ago

The overall charge of the nucleus is _____<br> a. positive <br> b. negative <br> c. neutral

hoa [83]

The overall charge of the nucleus is POSITIVE. This is because the nucleus of an atom is made up of protons and neutrons. Protons are positively charged particles while neutrons are neutrally charged particles which means it has no charge and cannot effect the protons electric charge in any way. Therefore, since the only effective charge is positive, we can conclude that the nucleus is also positive.

Let me know if you need anything else. :)

~ Dotz

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4 years ago

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What component of fitness should you work on every time you exercise?

Pepsi [2]

Answer:

1.body composition, (2) flexibility, (3) muscular strength, (4) muscular endurance, and (5) cardio respiratory endurance.

Explanation:

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3 years ago

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