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3 years ago

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Physics

1 answer:

kolbaska11 [484]3 years ago

4 0

Answer: 2.2x10^4

Explanation: in a adiabatic process pV^y = constant.

So V2=3V1 and p2 = p1 / 3^1.67 = 2.2x10^4 Pa

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A quantum system has three states, with energies 0, 1.6 × 10-21, and 1.6 × 10-21, in Joules. It is coupled to an environment wit

xenn [34]

To develop the problem it is necessary to apply two concepts, the first is related to the calculation of average data and the second is the Boltzmann distribution.

Boltzmann distribution is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of that state's energy and the temperature of the system. It is given by

$z = \sum\limit_i e^{-\frac{\epsilon_i}{K_0T}}$

Where,

$\epsilon_i =$ energy of that state

k = Boltzmann's constant

T = Temperature

With our values we have that

T= 250K

$k = 1.381*10^{23} m^2 kg s^{-2} K^{-1}$

$\epsilon_1=0J$

$\epsilon_2=1.6*10^{-21}J$

$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

$\beta = \frac{1}{(1.381*10^{23} m^2)(250)}$

$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

$\bar{\epsilon} =\frac{\sum \epsilon_i e^{-\beta \epsilon_i}}{\sum e^{-\beta \epsilon_i}}$

Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$

$\bar{\epsilon} = 0.9*10^{-22}J$

Therefore the average internal energy is $\bar{\epsilon} = 0.9*10^{-22}J$

3 0

3 years ago

Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) High and low pressures in kilopascals:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures in pounds per square inch:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures in meter water column in meters water column:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0

3 years ago

I want friends comment your email and ill hit you up

eimsori [14]

Answer:

ddddbdhxhhxhhbb

Explanation:

6 0

3 years ago

Read 2 more answers

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

3 years ago

4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data

melisa1 [442]

Answer:

heat is the transfer of thermal energy from a system to its surroundings or from ... It is very important to know that, in science, heat and temperature are not the same thing. ... Have you noticed that when you put a cold, metal teaspoon into your hot cup of ... AIM: To investigate which materials are the best conductors of heat.

Explanation:

5 0

3 years ago