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2 years ago

8

Explicitly solve the Heisenberg equations of motion to find the time–dependent raising and lowering (creation and annihilation)

operators for a one-dimensional oscillator in the Heisenberg picture. Show these operators are consistent with the time–dependent position and momentum operators previously derived in Lecture

1 answer:

Sedaia [141]2 years ago

3 0

Answer:

see detailed solution attached.

Explanation:

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A car travels 30 miles in 1 hour on a winding mountain road. Which of the following is a true statement?

siniylev [52]

Answer:

The true statement is:

"(C) The magnitude of the average velocity is equal to 30 m.p.h."

Explanation:

Given that a car travels 30 miles in 1 hour on a winding mountain road.

Let' check all the statements one by one:

(A) The magnitude of the total displacement is larger than the distance traveled.

Since the entire motion of the car is not exactly given in the question, so it is not possible to tell whether the magnitude of the total displacement is larger than the distance traveled or not.

Thus, this statement is not true.

(B) The magnitude of the average velocity is greater than 30 m.p.h.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Total distance covered by the car = 30 miles.

Total time taken by the car to cover this distance = 1 hour.

Therefore, the average velocity of the car for this time interval = $\rm \dfrac{30\ miles}{1\ hour }= 30\ m.p.h.$

Thus, this statement is also not true.

(C) The magnitude of the average velocity is equal to 30 m.p.h.

As is cleared in part (B) section above, the average velocity of the car in the given time interval is 30 m.p.h.

Thus, this statement is true.

(D)The magnitude of the average velocity is less than to 30 m.p.h.

Since. the average velocity of the car is 30 m.p.h.

Thus, this statement is not true.

(E)The car traveled with a constant speed of 30 m.p.h.

The motion of the car on the mountain road is not thoroughly given in the question, so again it is not possible to tell whether the car traveled with a constant speed of 30 m.p.h. or not.

Thus, this statement is also not true.

4 0

2 years ago

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Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Plank’s constant (6.63E-34

Nostrana [21]

To help you I need to assume a wavelength and then calculate the momentum.

The momentum equation for photons is:

p = h / λ , this is the division of Plank's constant by the wavelength.

Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:

p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s

p = 1.01067 * 10^ - 27 kg*m/s which must be rounded to three significant figures.

With that, p = 1.01 * 10 ^ -27 kg*m/s

The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s

So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.

7 0

2 years ago

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A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

8 0

2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

$a = \frac{F}{m}$

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

4 0

2 years ago

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is dire

DedPeter [7]

<h2>Answer: at an angle $\theta$ below the inclined plane. </h2>

If we draw the <u>Free Body Diagram</u> for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight $W$ of the block, which is directly proportional to the gravity acceleration $g$ :

$W=m.g$

This force is directed vertically at an angle $\theta$ below the inclined plane, this means it has an X-component and a Y-component:

$W=W_{X}+W_{Y}$

$W_{X}=m.g.cos(\theta)$

$W_{Y}=m.g.sin(\theta)$

Therefore the correct option is c

6 0

2 years ago

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