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3 years ago

7

Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e

ach molecule at the center of a small cube. Part A What is the length L of an edge of each small cube if adjacent cubes touch but don't overlap

1 answer:

My name is Ann [436]3 years ago

7 0

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

$PV =NkT$

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

$N = 1$

$k = 1.38*10^{-23}J/K$

$T = 27\°C = 27\°C + 273 = 300K$

$P = 1atm = 101325Pa$

Rearrange the equation to find V we have,

$V = \frac{NkT}{P}$

$V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}$

$V = 4.0858*10^{-26}m^3$

We know that length of a cube is given by

$V = L^3$

Therefore the Length would be given as,

$L = V^{1/3}$

$L = (4.0858*10^{-26})^{1/3}$

$L = 3.445*10^{-9}m$

Therefore each length of the cube is 3.44nm

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A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then trav

bija089 [108]

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

$t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h$

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

$d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km$

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

$v_{average}=\frac{155km}{2.25h}=68.9km/h$

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3 years ago

The apparent backward motion of a planet along the celestial sphere is called ____________ motion.

motikmotik

Answer:

Retrograde

Explanation:

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3 years ago

The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?

maria [59]

F=v/wavelength f=1.2/60=50Hz.

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3 years ago

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Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

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3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

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3 years ago