Hello I would like to know if there is a course or a book in which to learn quantum mechanics is easier, I am currently studying quantum mechanics, but it is a little complicated because I have some mathematical doubts, but nevertheless I would like to know if there is a course or book that explains and develops the problems so that they are understandable. and if there is no book or course I would appreciate an honest and complete answer, thank you very much and greetings.Sorry I know that this is not a problem but what better to ask the experts do not believe.

Answers

Answer 1

Answer:

Answer:

YES THERE ARE VERY MANY BOOKS THAT YOU CAN BUY OR BORROW FROM YOUR LOCAL LIBRARY, HOPE THIS HELPS! HAVE A GREAT DAY!

Explanation:

Related Questions

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What type of circuit measurement is made by placing a meters test leads in parallel with a deenergized component
You suspect that a power supply is faulty, but you use a power supply tester to measure its voltage output and find it to be acceptable. Why is it still possible that the power supply may be faulty?

The temperature of a solution will be estimated by taking n independent readings and averaging them. Each reading is unbiased, with a standard deviation of σ = 0.5°C. How many readings must be taken so that the probability is 0.90 that the average is within ±0.1◦C of the actual temperature? Round the answer to the next largest whole number.

Answers

Answer:

68 readings.

Explanation:

We need to take this problem as a statistic problem where the normal distribution table help us.

We can start considerating that X is the temperature of the solution, then

$0.9 = P(|\bar{x}-\mu|<0.1)$

$0.9 = P(\frac{|\bar{x}-\mu|}{(\sigma)/(√(n))}<(0.1)/((\sigma)/(√(n))))$

$0.9 = P(|Z|<(0.1)/((\sigma)/(√(n))))$

For a confidence level of 90% our $Z_(critic)$ is 1.645

Therefore,

$(0.1)/((\sigma)/(√(n))) = 1.645$

Substituting for $\sigma = 5$ and re-arrange for n, we have that n is equal to

$n=((1.645\sigma)/(0.1))^2$

$n=((1.645)^2(0.5)^2)/(0.1^2)$

$n=67.65$

$n=68$

We need to make 68 readings for have a probability of 90% and our average is within $0.1\°\frac$

A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.Determine the following:

a. The total amount of energy transfer by work (kJ)
b. The total amount of energy transfer by heat (kJ)

Answers

Answer:

Part a: The total amount of energy transfer by the work done is 54.81 kJ.

Part b: The total amount of energy transfer by the heat is 54.81 kJ

Explanation:

Mass of Carbon Dioxide is given as m1=3 kg

Pressure is given as P1=3 bar =300 kPA

Volume is given as V1=0.5 m^3

Pressure in tank 2 is given as P2=2 bar=200 kPa

T=290 K

Now the Molecular weight of $CO_2$ is given as

M=44 kg/kmol

the gas constant is given as

$R=\frac{\bar{R}}{M}\nR=(8.314)/(44)\nR=0.189 kJ/kg.K$

Volume of the tank is given as

$V=(mRT)/(P_1)\nV=(3 * 0.189 * 290)/(300 )\nV=0.5481 m^3$

Final mass is given as

$m_2=(P_2V)/(RT)\nm_2=(200* 0.5481)/(0.189* 290)\nm_2=2 kg$

Mass of the CO2 moved to the cylinder

$m=m_1-m_3\nm=3-2=1 kg$

The initial mass in the cylinder is given as

$m_((cyl)_1)=(P_((cyl)_1)V_1)/(RT)\nm_((cyl)_1)=(200* 0.5)/(0.189 * 290)\nm_((cyl)_1)=1.82 kg$

The mass after the process is

$m_((cyl)_2)=m_((cyl)_1)+m\nm_((cyl)_2)=1.82+1\nm_((cyl)_2)=2.82\n$

Now the volume 2 of the cylinder is given as

$V_((cyl)_2)=(m_((cyl)_2)RT)/(P_2)\nm_((cyl)_2)=(2.82* 0.189* 290)/(200)\nm_((cyl)_1)=0.774 m^3$

Part a:

So the Work done is given as

$W=P(V_2-V_1)\nW=200(0.774-0.5)\nW=54.81 kJ$

The total amount of energy transfer by the work done is 54.81 kJ.

Part b:

The total energy transfer by heat is given as

$Q=\Delta U+W\nQ=0+W\nQ=54.81 kJ$

As the temperature is constant thus change in internal energy is 0.

The total amount of energy transfer by the heat is 54.81 kJ

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.

Answers

Final answer:

The intensity of sounds is dependent on the square of the amplitude, not the frequency. Therefore, the intensity of sound remains the same when frequency is altered but the amplitude is constant. When the amplitude is quadrupled, the intensity of the sound becomes sixteen times greater.

Explanation:

In the field of physics, the intensity of a sound wave is defined as the power per unit area carried by the wave. This question involves calculating the change in sound wave intensity when the frequency and displacement amplitude of the source are altered.

(a) When the frequency is increased to 2.20 kHz while keeping the displacement amplitude constant, the intensity does not change, as the intensity in this case is not dependent on the frequency but on the square of the amplitude. Therefore, the intensity remains 0.750 W/m2.

(b) When the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled, the intensity changes. Since the intensity of a sound wave is proportional to the square of the amplitude, by quadrupling the amplitude, the intensity will become 16 times greater (since 4 squared is 16). Hence, the new intensity will be 16 * 0.750 = 12 W/m2.

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Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answers

Answer:

a). ΔP1=-2.4 $x10^(3) (D*m)/(s)$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^(3) (D*m)/(s)$

Explanation:

Initial momentum

$P_(1)=m_(1)*v_(i1)$

Final momentum

$P_(1f)=m_(1)*v_(f1)=-m_(1)*v_(i1)$

The change of momentum m1 is:

a).

ΔP1= $P_(1f)-P_(1)$

ΔP1= $-m_(1)*v_(i1)-m_(1)*v_(i1)$

ΔP1= $-2*m_(1)*v_(i1)$

ΔP1= $-2*6 D*200(m)/(s)$

ΔP1= $-2.4x10^(3)(D*m)/(s)$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=- $-2.4x10^(3)(D*m)/(s)$

ΔP2= $2.4x10^(3)(D*m)/(s)$

Final answer:

The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.

Explanation:

A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.

B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.

C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.

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Which one defines force?

Answers

Answer:

Explanation:

a push or a pull that occurs when an object interacts with another object or field.

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