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

Answer:

1.18266×10⁻²⁰ J

Explanation:

Applying,

E = (3/2)kT.................... Equation 1

Where E = kinetic energy of the gas molecule, k = Boltzmann's constant, T = Temperature

But,

PV = nRT.............. Equation 2

Where P = pressure, V = Volume, n = number of moles, R = Universal gas constant.

make T the subject of the equation

T = PV/nR............. Equation 3

Substitute equation 3 into equation 1

E = (3/2)k(PV/nR)

E = 3kPV/2nR............... Equation 4

Given: k = 1.38066×10⁻²³ J/K, V = 6.6 L = 0.0066 m³, P = 7.1 atm = (101325×7.1) N/m² = 719407.5 N/m², n = 2.9 mol, R = 8.31451 J/K.mol

Substitute into equation 4

E = (3×1.38066×10⁻²³×0.0066×719407.5)/(2×8.31451)

E = 1182.66×10⁻²³ J

E = 1.18266×10⁻²⁰ J

Answer:

Possible options:

A. Moment of inertia

B. Angular acceleration

C. Center of mass

D. Torque

Answer is B. Angular acceleration

Explanation:

Angular acceleration is the time rate of change of angular velocity. In three dimensions, it is a pseudovector. In SI units, it is measured in radians per second squared (rad/s2) and is usually denoted by the Greek letter alpha (α).

The angular acceleration should be with respect to the shorter pencil and it should be larger.

Angular acceleration refers to the time that shows the rate of change with respect to the angular velocity. In terms of three dimensions, it represents the pseudovector. Also, in SI units, it should be measured in radians per second squared and should be denoted by the  Greek letter alpha (α).

Therefore, The angular acceleration should be with respect to the shorter pencil and it should be larger.

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Answer:

distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is  145.32 m

Explanation:

Given that :

The radius of the wheel  R =  28 cm

Initial angular speed of the wheel [tex]\theta_o = 16 rev/s[/tex]

= [tex]16*\frac{2}{1} *\frac{\pi}{1}rad/s[/tex]

= [tex]32 \pi\ rad/s[/tex]

= 100.5 rad/s

At time t = 9 s ;  the angle  rotated by wheel [tex]\theta_o = \omega_o t[/tex]

= 32 π × 9

= 905 rad

At time [tex]t_1 = 10 \ s[/tex]; the angular speed is definitely the same and the initial velocity [tex]\omega_1[/tex] is [tex]\omega_o = 32 \pi \ rads[/tex]

However ; after time  [tex]t_1 = 10 \ s[/tex] ; the angular acceleration of the wheel ∝ = 1.3 rad/s²

At time [tex]t_2 = 15\ s[/tex] ; angular speed of wheel [tex]\omega_2 = \omega_1 + \alpha \delta \ t[/tex]

[tex]\omega_2 = 32 \pi + 1.3( 15 - 10)[/tex]

[tex]\omega_2 =107 \ rad/s[/tex]

Now; the angle rotated by the wheel during time  [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is expressed as:

[tex]\theta_1 = \omega_1(t_2-t_1)+ \frac{1}{2} \alpha (t_2-t_1)^2\\\\\theta_1 = 32 \pi (15-10) + \frac{1}{2}*1.3*(15-10)^2\\\\\theta_1 = 519 \ rad[/tex]

From the question; we are being told that :

This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R.

i.e [tex]v_{cm} = R \omega[/tex]

then we can say :

acceleration [tex]a_{cm} = R \alpha[/tex]    and

linear distance [tex]s = R \theta[/tex]    indicating   how far did the center of the wheel move

So; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is ;

[tex]s = R \theta[/tex]

= 0.28 × 519

s = 145.32 m

Therefore; distance moved by the center of the wheel during time [tex]t_1 = 10 \ s[/tex]  to   [tex]t_2 = 15\ s[/tex]  is  145.32 m

The center of the wheel moved a distance of 0.418 meters between 10s and 15s.

In rolling motion without slipping, the instantaneous velocity of the atoms in contact with the ground is zero. This implies that the linear speed of the center of the wheel is equal to the product of the angular speed and the radius. To calculate the distance the center of the wheel moved between 10s and 15s, we need to find the average linear speed during that time period.

Therefore, the center of the wheel moved a distance of 0.418 meters between 10s and 15s.

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Answer:

Option D

Explanation:

In order to maintain homeostasis, a function is to be performed that leads to functional normality of the body and its organs.  

Here, the death of cells is affecting the homeostasis hence in order to restore homeostasis the remaining cells must divide at higher pace so that they can replenish the lost cells and their functions.

Hence, option D is correct

Final answer:

To maintain homeostasis after cell death in an organ, the remaining cells will typically reproduce to replace the ones that have died, restoring organ function and balance within the body.

Explanation:

When cells within an organ die, homeostasis may be disrupted. Homeostasis is the complex process by which organisms maintain a stable internal environment despite external changes. To maintain homeostasis after cell death, the remaining cells will attempt to compensate for the loss. Typically, these cells will reproduce and divide to replace the dead cells, thereby restoring the organ's function and maintaining the necessary balance within the body. This is a form of homeostatic regulation, which involves continuous adjustments in cellular activity to sustain a set point or a normal level of function.

complete question

attached

Answer:

188.0 decibels

Explanation:

Substitute I = 6.3 x 10^6 into the given formula  

D(6.3 x 10^6) = 10log(10^12 x 6.3 x 10^6) = 188.0 to one decimal place.  

This is greater than 160 decibels, so the sound could rupture the human eardrum.  

Substitute I = 6.3 x 10^6 into the given formula  

The sound produced by the animal has a decibel level of 174 dB, which is intense enough to rupture a human eardrum at close range. However, the actual harm it might cause depends on other factors such as the frequency of the sound.

The intensity of the animal's sound is given as 2.6 × 10^6 watts/meter squared. We can calculate the decibel level by substituting it into the given formula D = 10 log (10^12* I). Solving this equation gives us D = 174 dB.

Decibel levels of 160 and above can rupture the human eardrum. Since the sound produced by this animal is 174 dB, it's intense enough at close range to rupture the human eardrum.

However, it's worth noting that loudness is not only determined by intensity, but by frequency as well, which can affect how we perceive the loudness of a sound. Therefore, frequency might play a role in whether or not this sound would actually cause harm to a human ear.

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Answer:

Student b

Explanation:

Because, Battery pumps the charges around the circuit. Here, charges are the free electrons. And this charges are already present in the conductor. When we connect this conductor with the battery - It set up an electric field and The diffrence between the potential between the ends of the wire pushes this free charges to move in perticular direction. (This is called electric energy)

This electric energy is converted by chemical reactions in the battery. When the battery is dead - It means that the chemical energy producing chemicals has been consumed. thereafter there will be no more chemical reactions takes place to produce electric energy. Hence we say that the battery is dead.

Answer:

Student B is correct

Explanation:

The battery acts like a pump to drive the electric charge. The electric charge comes from the conducting material but the battery provides a potential difference or a gradient of flow for this charge which is necessary for the movement of the charge along the conductor.

Answer:

λ = 4.97 x 10⁻⁷ m = 497 nm

Explanation:

The formula for the distance of a bright fringe from center in Young's double slit experiment, is given by:

y = mλL/d

where,

y = distance of bright fring from center = 3.22 mm = 3.22 x 10⁻³ m

m = No. of bright fringe = 1

L = Distance between slits and screen = 3.24 m

d = slit separation = 0.5 mm = 0.5 x 10⁻³ m

λ = wavelength of light = ?

3.22 x 10⁻³ m = (1)(λ)(3.24 m)/(0.5 x 10⁻³ m)

(3.22 x 10⁻³ m)(0.5 x 10⁻³ m)/(3.24 m) = λ

λ = 4.97 x 10⁻⁷ m = 497 nm

Answer:

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle, [tex]q=6.41\ \mu C=6.41\times 10^{-6}\ C[/tex]

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force, [tex]F=6.4\times 10^{-3}\ N[/tex]

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

[tex]F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s[/tex]

So, the particle's velocity is 212.15 m/s.

Answer:

The group of light rays is reflected back towards  the focal point thereby producing a magnifying effect.

Explanation:

Answer:

The charge flows through a point in the circuit during the change is 0.044 C.

Explanation:

Given that,

Number of turns in the copper wire, N = 200

Area of cross section, [tex]A=1.2\times 10^{-3}\ m^2[/tex]

Resistance of the circuit, R = 118 ohms

If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.

We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

Using Ohm's law :

[tex]\epsilon=IR[/tex]

[tex]IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}[/tex]

Electric current is equal to the rate of change of electric charge. So,

[tex]dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C[/tex]

So, the charge flows through a point in the circuit during the change is 0.044 C.

Answer:

(A) Acceleration will be [tex]1.145rad/sec^2[/tex]

(B) Coefficient of static friction will be 0.116

Explanation:

We have given angular speed

[tex]\omega =33rpm=\frac{2\pi \times 33}{60}=3.454rad/sec[/tex]

Distance from the axis r = 9.6 cm = 0.096 m

(a) Acceleration is equal to

[tex]a_c=\omega ^2r=3.454^2\times 0.096=1.145rad/sec^2[/tex]

(b) For seed is not to slip

[tex]ma=\mu mg[/tex]

[tex]\mu =\frac{a}{g}[/tex]

[tex]\mu =\frac{1.145}{9.8}=0.116[/tex]

So coefficient of static friction will be 0.116

Answer:

35.71%

Explanation:

See attached file for calculation

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

The gas pressure inside the box is [tex]\( 1.018 \times 10^5 \, \text{Pa} \)[/tex] to three significant figures.

To find the gas pressure [tex]\( p_{\text{gas}} \)[/tex] inside the box using the information provided, we'll follow the hydrostatic equilibrium principles as outlined.

The density of mercury [tex](\( \rho \)) = \( 1.36 \times 10^4 \, \text{kg/m}^3 \)[/tex]

The height of the mercury column in the left arm [tex](\( h_1 \)) = 10.0 cm = 0.10 m[/tex]

The height of the mercury column in the right arm [tex](\( h_2 \)) = 6.00 cm = 0.06 m[/tex]

Atmospheric pressure [tex](\( p_{\text{atm}} \)) = 1.00 atm = \( 1.013 \times 10^5 \, \text{Pa} \)[/tex]

Acceleration due to gravity [tex](\( g \)) = \( 9.81 \, \text{m/s}^2 \)[/tex]

1. Convert the heights to meters:

[tex]\[ h_1 = 0.10 \, \text{m} \] \[ h_2 = 0.06 \, \text{m} \][/tex]

2. Draw a schematic:

The left arm of the U-tube is connected to the box containing the gas. The right arm is open to the atmosphere. Mercury levels in the U-tube indicate that the pressure on the left side (due to the gas) is higher than the pressure on the right side (atmospheric pressure).

3. Pressure at the interface between mercury and gas (left side):

[tex]\[ p_{\text{left}} = p_{\text{gas}} \][/tex]

4. Pressure at the interface on the open side (right side):

[tex]\[ p_{\text{right}} = p_{\text{atm}} \][/tex]

5. Difference in mercury column height:

[tex]\[ \Delta h = h_1 - h_2 = 0.10 \, \text{m} - 0.06 \, \text{m} = 0.04 \, \text{m} \][/tex]

6. Hydrostatic pressure difference:

  The difference in height of the mercury columns corresponds to the pressure difference due to the gas in the box and the atmospheric pressure:

[tex]\[ \Delta p = \rho g \Delta h \] \[ \Delta p = (1.36 \times 10^4 \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (0.04 \, \text{m}) \] \[ \Delta p = 1.36 \times 10^4 \times 9.81 \times 0.04 \] \[ \Delta p = 532.224 \, \text{Pa} \][/tex]

7. Calculate the gas pressure: Since the pressure in the gas must account for the atmospheric pressure plus the pressure due to the height difference of the mercury:

[tex]\[ p_{\text{gas}} = p_{\text{atm}} + \Delta p \] \[ p_{\text{gas}} = 1.013 \times 10^5 \, \text{Pa} + 532.224 \, \text{Pa} \] \[ p_{\text{gas}} = 1.01832 \times 10^5 \, \text{Pa} \][/tex]

[tex]\[p_{\text{gas}} = 1.018 \times 10^5 \, \text{Pa}\][/tex]

Complete question:- To practice Tactics Box 13.1 Hydrostatics.

In problems about liquids in hydrostatic equilibrium, you often need to find the pressure at some point in the liquid. This Tactics Box outlines a set of rules for thinking about such hydrostatic problems.

Answer:

Assume patmos=1.00atmpatmos=1.00atm. What is the gas pressure?

Express your answer in pascals to three significant figures.

Pgas=_____ Pa

EDIT (Needed Information):

Draw a picture. Show open surfaces, pistons, boundaries, and other features that affect pressure. Include height and area measurements and fluid densities. Identify the points at which you need to find the pressure. These objects make up the system; the environment is everything else.

Determine the pressure p0p0 at surfaces.

Surface open to the air: p0=patmosp0=patmos, usually 1 atmatm.

Surface in contact with a gas: p0=pgasp0=pgas.

Closed surface: p0=F/Ap0=F/A, where FFF is the force the surface, such as a piston, exerts on the fluid and AAA is the area of the surface.

Use horizontal lines. The pressure in a connected fluid (of one kind) is the same at any point along a horizontal line.

Allow for gauge pressure. Pressure gauges read pg=p−1atmpg=p−1atm.

Use the hydrostatic pressure equation: p=p0+ρghp=p0+ρgh, where ρρrho is the density of the fluid, ggg is the acceleration due to gravity, and hhh is the height of the fluid.

Use these rules to work out the following problem: A U-shaped tube is connected to a box at one end and open to the air at the other end. The box is full of gas at pressure pgaspgasp_gas, and the tube is filled with mercury of density 1.36×104 kg/m3kg/m3 . When the liquid in the tube reaches static equilibrium, the mercury column is h1h1h_1 = 10.0 cmcm high in the left arm and h2h2h_2 = 6.00 cmcm high in the right arm, as shown in the figure.(Figure 1) What is the gas pressure pgaspgasp_gas inside the box?

The moment of inertia of the given cube, which is the rotational equivalent of mass in linear motion, is 0.000075 kg.m^2. This is calculated by applying the formula I = (M*s^2)/6 with given parameters. The parallel axis theorem helps in finding the moment of inertia about any axis parallel to and a distance away from an object's center of mass.

The moment of inertia of a cube about an axis normal to and through the center of one of its faces is calculated using the formula I = (M*s^2)/6, where M is mass, and s is the side length of the cube. In this case, M = 0.500 kg and s = 0.030 m. Therefore, the moment of inertia I = (0.500 kg * (0.030 m)^2)/6 = 0.000075 kg.m^2. Moment of inertia can be thought of as the rotational equivalent of mass for linear motion.

The moment of inertia of a cube, or any object, about any axis through its center of mass can be calculated using the parallel axis theorem. The theorem links the moment of inertia about an axis passing through the object's center of mass to the moment of inertia about any other parallel axis. It states that the moment of inertia about any axis parallel to and a distance d away from the axis through the center of mass is equal to the moment of inertia of the object about the axis through the center of mass plus the mass of the object times the square of the distance between the axes (I=I_cm + md^2). Therefore, the moment of inertia is also related to the distribution of mass within an object.

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Complete Question

A thin film of cooking oil (n=1.43) is spread on a puddle of water (n=1.34). What is the minimum thickness Dmin of the oil that will strongly reflect blue light having a wavelength in air of 451 nm, at normal incidence? Dmin= nm What are the next three thicknesses that will also strongly reflect blue light of the same wavelength, at normal incidence?

a)   158 nm, 237 nm, and 315 nm

b)   237 nm, 394 nm, and 552 nm

c)   473 nm, 788 nm, and 1100 nm

d)  361 nm, 602 nm, and 842 nm

e)  315 nm, 473 nm, and 631 nm

Answer:

The minimum thickness is   [tex]t= 78.8nm[/tex]

The correct option is B

Explanation:

  From the question we are told that

      The refractive index of cooking oil is [tex]n _c = 1.43[/tex]

       The refractive index of water is [tex]n_w = 1.34[/tex]

      The  wavelength of reflection is is  [tex]\lambda _ r = 451nm[/tex]

The formula for the thickness of the oil film is mathematically represented as

        [tex]2 n t = (m + \frac{1}{2} ) \lambda[/tex]

Where n is the refractive index of oil

            m is the integer number of fringe

            t is the thickness

   for a minimum reflection  m= 0

Now making t the subject of the formula

         [tex]t = \frac{(m + \frac{1}{2} \lambda ) }{2 n}[/tex]

Substituting value

        [tex]t = \frac{(0 + 0.5) * 451 *10^{-9}}{2 * 1.43}[/tex]

          [tex]t= 78.8nm[/tex]

For the next thickness m = 1

   so we have

        [tex]t_1 = \frac{(1 + 0.5 ) * 481}{2 * 1.43}[/tex]

            [tex]= 237nm[/tex]

For the next thickness m = 2

     so we have that  

        [tex]t_2 = \frac{(2 +0.5) *451 *10^{-9}}{2 * 1.43}[/tex]

             [tex]= 394nm[/tex]

For the next thickness m = 3

     so we have that

         [tex]t_2 = \frac{(3 +0.5) *451 *10^{-9}}{2 * 1.43}[/tex]

             [tex]= 552nm[/tex]

According to the question,

(a)

→              [tex]2nt = (m+\frac{1}{2} ) \lambda[/tex]

By substituting the values, we get

→ [tex]2\times 1.43\times t = (0+\frac{1}{2} )481[/tex]

→                  [tex]t = 84.1 \ nm[/tex]

(b)

The next 3 thicknesses are:

→ [tex]2\times 1.43\times t = (1+\frac{1}{2} ) 481[/tex]

                   [tex]t = 252 \ nm[/tex]

→ [tex]2\times 1.43\times t = (2+\frac{1}{2} )481[/tex]

                   [tex]t = 420 \ nm[/tex]

→ [tex]2\times 43\times t = (3+\frac{1}{2} ) 481[/tex]

                [tex]t = 588 \ nm[/tex]

Thus the above approach is correct.

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Answer:

See attached file pls

Explanation:

Answer:voltage=22.4volts

Explanation:

current=3.2A

Resistance=7ohms

Voltage =current x resistance

Voltage=3.2 x 7

Voltage=22.4volts

Answer:

Explanation:

Both skaters are initially at rest

Then,

Ux = Uy = 0m/s

It is assumed that it is a frictionless surface

Mass of skater X is less than Y

Mx < My

My > Mx

This shows that,

My = a•Mx,

Such that a > 1

Skater X pushes Y such that it moves with velocity.

Vy = 2Vo to the right, I.e positive x axis

Vy = 2Vo •i

We want to find direction of the velocity of skater X

Using conservation of momentum

Initial momentum =Final momentum

Mx•Ux + My•Uy = Mx•Vx + My• Vy

Ux = Uy = 0,

also My = aMx. And Vy = 2Vo •i

0 + 0 = Mx•Vx + a•Mx•2Vo •i

0 = Mx•Vx + 2a•Mx•Vo •i

Mx•Vx = —2a•Mx•Vo •i

Divide through by Mx

Vx = —2a Vo •i

Therefore, since a is always positive

Then, Vx is in the negative direction or opposite direction to Vy

Now, center of Mass

The center of mass calculated using

Vcm = 1 / M Σ Mi•Vi

Where

M is sum of all the masses

M = Mx + My = Mx + aMx = (a+1) Mx

Then,

Vcm = (Σ Mi•Vi ) / M

Vcm = (MxVx+MyVy) / (a+1)Mx

Vcm=Mx•(-2aVo) +aMx(2Vo)/(a+1)Mx

Vcm = -2a•Mx•Vo + 2a•MxVo / (a+1)Mx

Vcm = 0 / (a+1) Mx

Vcm = 0

So, conclusion

Skater X direction is to the left and the centre of mass is 0.

This options are not written well

Check attachment for correct option

Answer:

Explanation:

The speed of the astronaut can be found with the help of law of conservation of momentum .

mv = MV , M is mass of astronaut , m is mass of object thrown , v is velocity of object thrown and V is velocity of  astronaut.

Putting the values

77.5 x V = .94 x 12

V = .14554 m /s

This will be the uniform velocity of astronaut.

Distance to be covered = 37.3 m

time taken = distance / velocity

= 37.3 / .14554

= 256.28 s

= 4.27 minutes.

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

The electroscope initially becomes positively charged, but is neutralized when touched. Once a negatively charged object is brought near, the leaves become negatively charged and repel each other.

When a positively charged object is brought near an uncharged gold leaf electroscope, the free electrons in the electroscope are attracted towards the positive charge, leaving the leaves positively charged and causing them to repel each other.

When you touch the electroscope, you provided a grounding path which allows electrons to flow from the ground to the positively charged electroscope to neutralize it again, causing the leaves to fall back together.

After the positively charged object is removed, the electroscope remains neutral until a negatively charged object is brought near it. The electrons in the electroscope will be repelled towards the leaves, giving them a negative charge and causing them to repel again.

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Answer:

5

Explanation:

i dont have one

Answer:

5

Explanation:

good luck :)

Answer:

Explanation:

angular acceleration α = 5 rad /s ²

θ = 1/2 α t² ,     θ is angle of rotation , t is time .

= .5 x 5 x 4²

= 40 rad .

θ = l / r , θ is angle formed , l is length unwound , r is radius o wheel .

l = θ x r

= 40 x .1

= 4 m .

Answer:

Let's analyse the definition and applications of angular momentum, and its relation with torque.

First of all, it's important to consider that the angular momentum is a property of rotational dynamics. Also, it's the analogue of the linear momentum.

Mathematically, the angular momentum is defined as

[tex]L=r \times p[/tex]

Where [tex]L[/tex] represents the angular momentum vector, [tex]r[/tex] represents is the position vector and [tex]p[/tex] is the linear momentum vector.

Notice that the angular momentum is also a vector, which is the cross product  of two vectorial magnitudes. In other words, the direction of the resulting vector (linear momentum) follows the right hand rule, which means that the resulting direction is according to the rotation direction, also means that the cross product is not commutative, which is a common assumption students make.

Now, the realtion between angular momentum and torque is that the change of the angular momentum with respect to time is equivalent to its torque:

[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] =\tau[/tex]

Remember that torque is defined as [tex]\sum \tau = r \times \sum F[/tex], and the derivative of the cross product is

[tex]\frac{d}{dt}[L]=\frac{d}{dt}[r \times p] = \frac{dr}{dt} \times p + r \times \frac{dp}{dt}[/tex]

Then,

[tex]\frac{d}{dt}[L] =(v \times mv)+r \times \frac{dp}{dt}[/tex]

But, [tex]v \times mv = 0[/tex], because those vector are parallel.

So, [tex]\frac{d}{dt}[L] = r\times \sum F = \tau[/tex]

At this point, we demonstrate it the relation between torque and rotational momentum.

In words, the net torque on a particle is equal to the rate of change of the angular momentum with respect to time.

Now, the application of angular momentum can be seen in skating spins, notice that when the skater puts his arms closer to its body, he'll rotate faster. The reason of this phenomenon is because arms represents rotating mass and the axis is the body, so the postion of this arm mass changes to zero distance to the rotational axis, that will increase the angular momentum, making higher. If the angular momentum is higher, the torque will be also higher, that's way the skater increses its rotational velocity.

Angular momentum is the rotational equivalent of linear momentum, defined as L = Iω for extended bodies. It is conserved when net external torque is zero, similar to how linear momentum is conserved in the absence of external forces. Torque and angular momentum are related by net τ = ΔL/Δt.

In physics, the concept of angular momentum is the rotational equivalent of linear momentum. It is defined as L = r × p for a single particle, where r is the position vector and p is the linear momentum vector. For an extended body rotating about an axis of symmetry, angular momentum (L) is given by the product of the moment of inertia (I) and angular velocity (ω): L = Iω.

Just as linear momentum is conserved in the absence of external forces, angular momentum is conserved when the net external torque is zero. This principle is crucial in various applications, such as planetary motion and rotating machinery.

The relationship between torque (τ) and angular momentum is given by the equation: net τ = ΔL/Δt, meaning the net torque acting on a system is equal to the rate of change of angular momentum.

Answer:C

Explanation:

Radar waves will be useful to examine the surface of the planet.

Radio waves are the shorter wavelength microwaves ranging approximately 1 cm.

Short wavelength causes large reflection from large objects.  These waves are useful to detect smaller objects like water drops in the cloud.

They can penetrate clouds and that is why they are used in weather forecasting, air traffic control, etc.

The best wavelength of electromagnetic radiation to use when trying to examine a planet covered by a thick layer of clouds is radar waves, as they can penetrate the clouds and provide information about the surface beneath.

If you want to examine the surface of a planet that is completely covered by a thick layer of clouds all the time, the smartest wavelength of electromagnetic radiation to use would be C. Radar waves. The reason is that radar waves, which are a type of radio wave, can penetrate through clouds and provide information about the surface beneath. Other types of electromagnetic radiation such as visible light, x-rays, and ultra-violet would be absorbed or scattered by the clouds, preventing you from seeing the surface.

For example, radar is used here on Earth for weather prediction because it can 'see' through clouds, providing details about storm structures. Similarly, space probes such as the Magellan mission to Venus used radar wavelengths to map the surface of Venus, a planet notoriously covered in thick clouds.

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Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr  = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final  velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

Answer:

155 N

Explanation:

We are given following data for a uniform, solid, horizontal disk:  

r = 1.5 m

m= 110 kg

t=2 s

w = 0.6  

Torque is given by:  

т = F*r

  =I*α

Solving it for the force exerted on the rope:  

F = I*α/r

  = (1/2*m*r^2)*(2π*w/t )/r

  = (1/2*110*1.5^2)*(2π*0.6/2 )/1.5

  = 155 N

Answer:

For M84:

M = 590.7 * 10³⁶ kg

For M87:

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc  = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

[tex]v = \sqrt{\frac{GM}{r} }[/tex]

making m the subject of the formula in the equation above to calculate the mass of the black hole:

[tex]M = \frac{v^{2} r}{G}[/tex].............(1)

For M84:

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

[tex]M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }[/tex]

M = 590.7 * 10³⁶ kg

For M87:

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

[tex]M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }[/tex]

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

Answer:

The approximate number of decays  this represent  is  [tex]N= 23*10^{10}[/tex]  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is [tex]I_a = 2.28 \ mSv[/tex]

     The source of the radiation is [tex]S = 5.49 MeV \ alpha \ particle[/tex]

 Generally

            [tex]1 \ J/kg = 1000 mSv[/tex]

   Therefore  [tex]2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg[/tex]

Also  [tex]1eV = 1.602 *10^{-19}J[/tex]

  Therefore  [tex]2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg} * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg[/tex]

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        [tex]1 kg ------> 1.423*10^{16}ev \\88.7kg --------> x[/tex]

Cross-multiplying and making x the subject

           [tex]x = 88.7 * 1.423*10^{16} eV[/tex]

              [tex]x = 126.2*10^{16}eV[/tex]

Therefore the total  energy  deposited is [tex]x = 126.2*10^{16}eV[/tex]

The approximate number of decays  this represent  is mathematically evaluated as

            N = [tex]\frac{x}{S}[/tex]

Where n is the approximate number of decay

   Substituting values

                             [tex]N = \frac{126 .2*10^{16}}{5.49*10^6}[/tex]  

                                  [tex]N= 23*10^{10}[/tex]  

The energy deposited in the body from radon exposure and the number of decays it represents. It also discusses the significance of radon-222 as a source of radiation exposure.

The energy deposited in your body each year from radon can be calculated by converting the dose equivalent to energy. Given that the average American receives 2.28 mSv annually and all of it comes from a 5.49 MeV alpha particle emitted by Rn-222, the energy deposited would be 5.49 MeV.

To determine how many decays this represents, you divide the total energy deposited by the energy per decay. In this case, 5.49 MeV divided by 5.49 MeV per decay gives you 1 decay annually.

Radon-222 is a significant source of radiation exposure due to its radioactivity and ability to seep into homes from the ground. It highlights the importance of understanding the impact of radioactive elements on our environment.

Answer:

Nuclear engineer

Explanation:

Because it requires a lot of movements during that career.