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Lady_Fox [76]
3 years ago
9

Define condensation point

Physics
1 answer:
ella [17]3 years ago
7 0

Answer:

This is a physical point where a substance changes from its gas state to a liquid state at a constant temperature.

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A college student is working on her physics homework in her dorm room. her room contains a total of 6.0×1026 gas molecules. as s
ella [17]
<span>6.6 degrees C Let's model the student as a 125 w furnace that's been operating for 11 minutes. So 125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule So the average kinetic energy increase of each gas molecule is 82500 J / 6.0x10^26 = 1.38x10^-22 J Now the equation that relates kinetic energy to temperature is: E = (3/2)Kb*Tk E = average kinetic energy of the gas particles Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K) Tk = Kinetic temperature in Kelvins Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by. So let's substitute the known values and solve for Tk E = (3/2)Kb*Tk 1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk 1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk 6.64 K = Tk Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.</span>
7 0
3 years ago
A bug of mass 2.2 g is sitting at the edge of a cd of radius 3.0 cm. if the cd is spinning at 280 rpm, what is the angular momen
m_a_m_a [10]
M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.

The angular speed is
ω = 280 rpm
    = (280 rev/min)*(2π rad/rev)*(1/60 min/s)
    = 29.3215 rad/s

The moment of inertia of the bug is
I = mr²
  = (2.2 x 10⁻³ kg)*(0.03 m)²
  = 1.98 x 10⁻⁶ kg-m²

Calculate the angular momentum of the bug.
J = Iω
  = (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
  = 5.806 x 10⁻⁵ (kg-m²)/s

Answer: 5.806 x 10⁻⁵ (kg-m²)/s

5 0
4 years ago
How is modeling useful in studying the structure of the atom?
Mandarinka [93]

Modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B

<h3>What is modeling?</h3>

A model is a representation of reality. We know that a model could help us to recreate reality in a manner that we could be able to relate fully with it. A model could be used also a means of explanation.

The atomic models that we have usually help us to understand more abut the atom. Therefore, modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B

Learn more about modelling the atom:brainly.com/question/1596638

#SPJ1

6 0
2 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
shtirl [24]

Answer:

\lambda=1.39\times 10^{-4}\ s^{-1}

Explanation:

Given that,

The half-life of Barium-139 is 4.96\times 10^3

A sample contains 3.21\times 10^{17} nuclei.

We need to find the decay constant for this decay. The formula for half life is given by :

T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}

Put all the values,

\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}

So, the decay constant is 1.39\times 10^{-4}\ s^{-1}.

4 0
3 years ago
Calculate the magnetic field at 2m from a straight wire carrying a current of 5 A. (K = 2 x 10 ^-7)
blagie [28]

Answer:

B=5\times 10^{-7}\ T

Explanation:

Given that,

Current in a wire, I = 5 A

We need to find the magnetic field at 2m from a straight wire carrying a current of 5 A.

The magnetic field due to a wire is given by :

B=\dfrac{\mu_0I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 5}{2\pi \times 2}\\\\B=5\times 10^{-7}\ T

So, the required magnetic field is 5\times 10^{-7}\ T.

6 0
3 years ago
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