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Vilka [71]
3 years ago
15

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

molecule at the top of the container (assuming the potential energy is zero at the bottom) with the average kinetic energy of the molecules. Is it reasonable to neglect the potential energy?
Physics
1 answer:
Sergio039 [100]3 years ago
8 0

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

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A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
Llana [10]

Answer:

a) C = 40.138\,pF, b) q = 16.056\,nC, c) U = 3.212\,\mu J

Explanation:

a) The capacitance of two parallel plates capacitor with dielectric is given by the following expression:

C = K\cdot \epsilon_{o}\cdot \frac{A}{d}

Where:

K - Dielectric constant.

\epsilon_{o} - Vaccum permitivity.

A - Plate area.

d - Distance between plates.

Hence, the capacitance of the system is:

C = (4.00)\cdot (8.854\times 10^{-12}\,\frac{F}{m} )\cdot \left(\frac{17\times 10^{-4}\,m^{2}}{0.150\times 10^{-2}\,m}\right)

C = 4.014\cdot 10^{-11}\,F

C = 40.138\,pF

b) The charge can be found by using the definition of capacitance:

q = C\cdot V_{batt}

q = (4.014\times 10^{-11}\,F)\cdot (400\,V)

q = 1.606\times 10^{-8}\,C

q = 16.056\,nC

c) The energy stored in the charged capacitor is:

U=\frac{1}{2}\cdot Q\cdot V_{batt}

U=\frac{1}{2}\cdot (1.606\times 10^{-8}\,C)\cdot (400\,V)

U = 3.212\times 10^{-6}\,J

U = 3.212\,\mu J

3 0
2 years ago
Read 2 more answers
Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración
guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0
3 years ago
What level of intensity is playing golf while pulling or carrying a set of clubs?
Nuetrik [128]

Answer:

It would be considered Moderate intensity.

5 0
3 years ago
758 j of heat are added to 0.750 kg of copper. how much does its temperature change?(unit=degrees c) PLEASE HELPPPPPPP MEEEEEE
DiKsa [7]

Answer:

2.6^{\circ}C

Explanation:

When a substance is supplied with a certain amount of heat energy, the temperature of the substance increases according to the equation

Q=mC\Delta T

where

m is the mass of the substance

Q is the amount of energy supplied

C is the specific heat of the substance

\Delta T is the temperature change

In this problem:

Q = 758 J is the energy supplied

m = 0.750 kg is the mass of the sample

C=385 J/kg^{\circ}C is the specific heat of copper

Re-arranging the equation, we can find the increase in temperature:

\Delta T=\frac{Q}{mC}=\frac{758}{(0.750)(385)}=2.6^{\circ}C

5 0
3 years ago
Homework
yarga [219]

Answer:

Please reframe your questions

4 0
2 years ago
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