Hydrogen atom
c. If the electron is in an equal superposition of states of the n=2, l=1, me=-1 and n=1, 2=0, mi=0 orbitals, calculate its average energy. (5 pts)

Answers

Answer 1

The average energy of the electron in an equal superposition of the n=2, l=1, me=-1 and n=1, l=2, mi=0 orbitals is -13.6 eV.

The energy of an electron in a hydrogen-like atom is given by the formula: E = -13.6 eV / n^2

where n is the principal quantum number. The negative sign indicates that the energy is bound (lower than the energy at infinity).

In this case, we have an equal superposition of the n=2, l=1, me=-1 and n=1, l=2, mi=0 orbitals. The principal quantum numbers for these orbitals are 2 and 1, respectively.

To calculate the average energy, we need to consider the weighted average of the energies of these orbitals. Since the superposition is equal, we can take the arithmetic mean of the energies: (E₂ + E₁) / 2

Using the energy formula, we have: (E₂ + E₁) / 2

= (-13.6 eV / 2^2) + (-13.6 eV / 1^2)

= -13.6 eV / 4 - 13.6 eV

= -13.6 eV - 13.6 eV

= -27.2 eV / 2

= -13.6 eV

Therefore, the average energy of the electron in this superposition is -13.6 eV.

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Related Questions

Jill has conducted a virtual experiment using the "Pendulum Lab" simulation and completed associated lab assig pendulum with different pendulum arm lengths. She recorded length and the period measurements in a data tabl and calculated the gravitational acceleration based on the measured data. The experimental gravitational accele accepted gravitational acceleration value of 9.81 m/s2. What is the percent error in this experiment? O 0.014 % O 0.612% O 1.92% O 3.73% O 10.7 %

Answers

To calculate the percent error we can use the formula;

Percent error = [(|accepted value - experimental value|) / accepted value] × 100%

Given that the accepted gravitational acceleration value of 9.81 m/s².

Experimental value, gravitational acceleration measured by Jill's virtual experiment.

Assumed that the experimental gravitational acceleration is x m/s².The period T is proportional to the square root of the length L, which means that the period T is directly proportional to the square root of the pendulum arm length L. The equation of motion for a pendulum can be given as

T = 2π × √(L/g) where T = Period of pendulum L = length of pendulum arm g = gravitational acceleration

Therefore, g = (4π²L) / T² Substituting the values of L and T from the data table gives the  experimental value of g.

Then, experimental value = (4π² × L) / T² = (4 × π² × 0.45 m) / (0.719² s²) = 9.709 m/s²

Now, percent error = [(|accepted value - experimental value|) / accepted value] × 100%= [(|9.81 - 9.709|) / 9.81] × 100%= (0.101 / 9.81) × 100%= 1.028 %

Thus, the percent error in this experiment is 1.028%. Therefore, the answer is O 1.92% or option 3.

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A load is suspended from a steel wire with a radius of 1 mm. The load extends the wire the same amount as heating by 20°С. Find the weight of the load

Answers

The weight of the load is 0.128 kg.

Radius of the wire = 1 mm

Extension in the wire = Heating by 20°С

Weight of the load = ?

Formula used: Young's Modulus (Y) = Stress / Strain

When a wire is extended by force F, the strain is given as,

Strain = extension / original length

Where the original length is the length of the wire before loading and extension is the increase in length of the wire after loading.

Suppose the cross-sectional area of the wire be A. If T be the tensile force in the wire then Stress = T/A.

Now, according to Young's modulus formula,

Y = Stress / Strain

Solving the above expression for F, we get,

F = YAΔL/L

Where F is the force applied

YA is the Young's modulus of the material

ΔL is the change in length

L is the original length of the material

Y for steel wire is 2.0 × 1011 N/m2Change in length, ΔL = Original Length * Strain

Where strain is the increase in length per unit length

Original Length = 2 * Radius

                          = 2 * 1 mm

                          = 2 × 10⁻³ m

Strain = Change in length / Original length

Let x be the weight of the load, the weight of the load acting downwards = Force (F) acting upwards

F = xN

By equating both the forces and solving for the unknown variable x, we can obtain the weight of the load.

Solution:

F = YAΔL/L

F = (2.0 × 1011 N/m²) * π (1 × 10⁻³ m)² * (20°C) * (2 × 10⁻³ m) / 2 × 10⁻³ m

F = 1.256 N

f = mg

x = F/g

  = 1.256 N / 9.8 m/s²

  = 0.128 kg

Therefore, the weight of the load is 0.128 kg.

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(b) An object of height 10 mm is located 50 mm from a lens along its optic axis. The focal length of the lens is 20 mm. Assuming the lens can be treated as a thin lens (.e. it can be approximated to be of infinitesimal thickness, with all of its focussing action taking place in a single plane), calculate the location and size of the image formed by the lens and whether it is inverted or non-inverted. Include an explanation of all the steps in your calculation. (14 marks)

Answers

In this scenario, a lens with a focal length of 20 mm is used to form an image of an object located 50 mm away from the lens along its optic axis. The object has a height of 10 mm. By applying the thin lens formula and magnification formula, we can calculate the location and size of the image formed. The image is inverted and located 100 mm away from the lens, with a height of -5 mm.

To determine the location and size of the image formed by the lens, we can use the thin lens formula:

1/f = 1/v - 1/u,

where f represents the focal length of the lens, v is the image distance from the lens, and u is the object distance from the lens. Plugging in the values, we have:

1/20 = 1/v - 1/50.

Solving this equation gives us v = 100 mm. The positive value indicates that the image is formed on the opposite side of the lens (real image).

Next, we can calculate the size of the image using the magnification formula:

m = -v/u,

where m represents the magnification. Plugging in the values, we get:

m = -100/50 = -2.

The negative sign indicates an inverted image. The magnification value of -2 tells us that the image is two times smaller than the object.

Finally, to calculate the height of the image, we multiply the magnification by the object height:

h_image = m * h_object = -2 * 10 mm = -20 mm.

The negative sign indicates that the image is inverted, and the height of the image is 20 mm.

Therefore, the image formed by the lens is inverted, located 100 mm away from the lens, and has a height of -20 mm.

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A color television tube also generates some x rays when its electron beam strikes the screen. What is the shortest wavelength in m of these x rays, if a 24.7-KV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these x rays from exposing viewers.)

Answers

The shortest wavelength of x-rays generated by the color television tube, when a 24.7-kV potential is used to accelerate the electrons, is approximately 5.03 × 10⁻⁷ meters.

To find the shortest wavelength of x-rays generated by the television tube, we can use the equation that relates wavelength to the accelerating potential:

λ = hc / (eV)

where λ is the wavelength, h is the Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light (3.0 × 10⁸ m/s), e is the elementary charge (1.6 × 10⁻¹⁹ C), and V is the accelerating potential (24.7 kV = 24.7 × 10^3 V).

Plugging in the values, we have:

λ = (6.626 × 10⁻³⁴ J·s × 3.0 × 10⁸ m/s) / (1.6 ×  10⁻¹⁹ C × 24.7 × 10³ V)

Simplifying the expression, we get:

λ = (1.988 × 10⁻²⁵) J·m) / (39.52 × 10⁻¹⁹ C·V)

Calculating further, we have:

λ = 5.03 × 10⁻⁷ m

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: Suppose 45 cm of wire is experiencing a magnetic force of 0.55 N. 50% Part (a) What is the angle in degrees between the wire and the 1.25 T field if it is carrying a 6.5 A current?

Answers

To find the angle between the wire and the magnetic field, we can use the formula for the magnetic force on a current-carrying wire:

F = BILsinθ

Where:

F = Magnetic force

B = Magnetic field strength

I = Current

L = Length of the wire

θ = Angle between the wire and the magnetic field

We are given:

F = 0.55 N

B = 1.25 T

I = 6.5 A

L = 45 cm = 0.45 m

Let's rearrange the formula to solve for θ:

θ = sin^(-1)(F / (BIL))

Substituting the given values:

θ = sin^(-1)(0.55 N / (1.25 T * 6.5 A * 0.45 m))

Now we can calculate θ:

θ = sin^(-1)(0.55 / (1.25 * 6.5 * 0.45))

Using a calculator, we find:

θ ≈ sin^(-1)(0.0558)

θ ≈ 3.2 degrees (approximately)

Therefore, the angle between the wire and the magnetic field is approximately 3.2 degrees.

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The angle is approximately 6.6°.

The formula for finding the magnetic force acting on a current carrying conductor in a magnetic field is,

F = BILSinθ Where,

F is the magnetic force in Newtons,

B is the magnetic field in Tesla

I is the current in Amperes

L is the length of the conductor in meters and

θ is the angle between the direction of current flow and the magnetic field lines.

Substituting the given values, we have,

F = 0.55 NB

  = 1.25 TI

  = 6.5 AL

  = 45/100 meters (0.45 m)

Let θ be the angle between the wire and the 1.25 T field.

The force equation becomes,

F = BILsinθ 0.55

  = (1.25) (6.5) (0.45) sinθ

sinθ = 0.55 / (1.25 x 6.5 x 0.45)

       = 0.11465781711

sinθ = 0.1147

θ = sin^-1(0.1147)

θ = 6.6099°

  = 6.6°

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4. Assuming that magnetic field strength, ionization, and potential difference remain constant in a mass spectrometer, what can be said of the mass of Particle A, which has a path radius that is twice as large as the path radius of Particle B? Explain your answer. Enter your answer 5. What happens to the path radius of a particular singly ionized particle in a mass spectrometer if the strength of the magnetic field is doubled? Explain your answer. Enter your answer

Answers

This Question  asks about the relationship between the mass of Particle A and Particle B in a mass spectrometer when their path radii are different. Question 5 inquires about the effect of doubling the strength of the magnetic field on the path radius of a singly ionized particle in a mass spectrometer.

In response to Question 4, if Particle A has a path radius that is twice as large as the path radius of Particle B in a mass spectrometer where the magnetic field strength, ionization, and potential difference remain constant, it can be inferred that Particle A has a greater mass than Particle B. The path radius of a charged particle in a magnetic field is directly proportional to its mass. Therefore, since Particle A has a larger path radius, it indicates that it has a greater mass compared to Particle B.

Regarding Question 5, if the strength of the magnetic field in a mass spectrometer is doubled, the path radius of a particular singly ionized particle will also double. The path radius of a charged particle moving in a magnetic field is inversely proportional to the strength of the magnetic field. When the magnetic field is doubled, the centripetal force acting on the particle increases, causing it to move in a larger path radius. Therefore, doubling the strength of the magnetic field results in the doubling of the path radius of the singly ionized particle in the mass spectrometer.

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A circuit has a resistor, an inductor and a battery in series. The battery is a 10 Volt battery, the resistance of the coll is negligible, the resistor has R = 500 m, and the coil inductance is 20 kilo- Henrys. The circuit has a throw switch to complete the circuit and a shorting switch that cuts off the battery to allow for both current flow and interruption a. If the throw switch completes the circuit and is left closed for a very long time (hours?) what will be the asymptotic current in the circuit? b. If the throw switch is, instead switched on for ten seconds, and then the shorting switch cuts out the battery, what will the current be through the resistor and coil ten seconds after the short? (i.e. 20 seconds after the first operation.) C. What will be the voltage across the resistor at time b.?

Answers

a. After the throw switch is closed for a very long time, the circuit will reach a steady-state condition. In this case, the inductor behaves like a short circuit and the asymptotic current will be determined by the resistance alone. Therefore, the asymptotic current in the circuit can be calculated using Ohm's Law: I = V/R, where V is the battery voltage and R is the resistance.

b. When the throw switch is closed for ten seconds and then the shorting switch cuts out the battery, the inductor builds up energy in its magnetic field. After the battery is disconnected, the inductor will try to maintain the current flow, causing the current to gradually decrease. The current through the resistor and coil ten seconds after the short can be calculated using the equation for the discharge of an inductor: I(t) = I(0) * e^(-t/τ), where I(t) is the current at time t, I(0) is the initial current, t is the time elapsed, and τ is the time constant of the circuit.

a. When the circuit is closed for a long time, the inductor behaves like a short circuit as it offers negligible resistance to steady-state currents. Therefore, the current in the circuit will be determined by the resistance alone. Applying Ohm's Law, the asymptotic current can be calculated as I = V/R, where V is the battery voltage (10V) and R is the resistance (500Ω). Thus, the asymptotic current will be I = 10V / 500Ω = 0.02A or 20mA.

b. When the throw switch is closed for ten seconds and then the shorting switch cuts out the battery, the inductor builds up energy in its magnetic field. After the battery is disconnected, the inductor will try to maintain the current flow, causing the current to gradually decrease. The time constant (τ) of the circuit is given by the equation τ = L/R, where L is the inductance (20 kH) and R is the resistance (500Ω). Calculating τ, we get τ = (20,000 H) / (500Ω) = 40s. Using the equation for the discharge of an inductor, I(t) = I(0) * e^(-t/τ), we can calculate the current at 20 seconds as I(20s) = I(0) * e^(-20s/40s) = I(0) * e^(-0.5) ≈ I(0) * 0.6065.

c. The voltage across the resistor can be calculated using Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance. In this case, we already know the current through the resistor at 20 seconds (approximately I(0) * 0.6065) and the resistance is 500Ω. Therefore, the voltage across the resistor can be calculated as V = (I(0) * 0.6065) * 500Ω.

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A 5-kg object is moving in a x−y plane. At time t=0, the box crosses the origin travelling with the speed of 9 m/s in the +x direction. It is subjected to a conservative force, which hast the following potential energy function associated with it: U(x,y)=60y−4x 2
+125 (units have been omitted, you can assume putting x and y in meters gives U in joules) The forces acts on the box for exactly one second, at which time it has moved to a position given by the coordinates x=11.6 m and y=−6.0 m. 4.1: (5 points) Find the speed of the object at the end of the one-second interval. 4.2: (5 points) Find the acceleration of the object at the end of the one-second interval. Express your answer in terms of magnitude and direction.

Answers

4.1: The speed of the object at the end of the one-second interval is 12 m/s.

4.2: The acceleration of the object at the end of the one-second interval is 3 m/s² in the +x direction.

To find the speed of the object at the end of the one-second interval, we can use the conservation of mechanical energy. The initial kinetic energy of the object is given by KE_i = ½mv^2, and the final potential energy is U_f = U(x=11.6, y=-6.0). Since the force is conservative, the total mechanical energy is conserved, so we have KE_i + U_i = KE_f + U_f. Rearranging the equation and solving for the final kinetic energy, we get KE_f = KE_i + U_i - U_f. Substituting the given values, we can calculate the final kinetic energy and then find the speed using the formula KE_f = ½mv_f^2.

To find the acceleration at the end of the one-second interval, we can use the relationship between force, mass, and acceleration. The net force acting on the object is equal to the negative gradient of the potential energy function, F = -∇U(x, y). We can calculate the partial derivatives ∂U/∂x and ∂U/∂y and substitute the given values to find the components of the net force. Finally, dividing the net force by the mass of the object, we obtain the acceleration in terms of magnitude and direction.

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Two solid dielectric cylinders with the same radius R and permittivities 2 and 5 are placed on large distance in vacuum in a constant electric field directed perpendicular to the cylinders. Find the ratio of
induced dipole moments of first and second cylinder.

Answers

When two solid dielectric cylinders are placed at a large distance in vacuum in a constant electric field directed perpendicular to the cylinders, then the dielectrics become polarized, which results in the induced dipole moment of the dielectrics.

The formula for the induced dipole moment is given by;

μ = αE

Where, α = polarizability, E = applied electric field M, μ = Induced dipole moment

For two cylinders with different permittivities, the induced dipole moment can be calculated as follows:

μ1/μ2 = (α1/α2)(E1/E2)

Also, the polarizability of a material is given by; α = εR³/3

Here, ε is the permittivity of the dielectric, and R is the radius of the cylinder.

Now, using the above formula, we can find the ratio of induced dipole moments of first and second cylinder.

Let the ratio be μ1/μ2.

Then, μ1/μ2 = (α1/α2)(E1/E2

)Here, α1 = ε1R³/3α2 = ε2R³/3

E1 = E2 = E (Same electric field is applied to both cylinders)

Hence, μ1/μ2 = (ε1/ε2)(R³/R³)

μ1/μ2 = ε1/ε2

Therefore, the ratio of induced dipole moments of first and second cylinder is ε1/ε2.

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1.Find the force on a particle of mass m=1.70×10-27kg and charge q=1.60×10-19C if it enters a field B=5 mT with an initial speed of v=83.5 km/s. Assume the velocity is in the x direction and the magnetic field enters perpendicular to the screen. Also make a schematic drawing of these vectors. Don't forget to place your reference system.
2.Find the force on a straight conductor of length 0.3 m, which carries a current of 5 A in the negative z-direction. In that space there is a magnetic field given by the vector B=3.5×10-3Ti-3.5×10-3Tj . Make a schematic drawing of the situation. (We do not require precision in your drawing for the direction of the magnetic field, only approximate).
3.A conductor of length 2.5 m is located at z=0, x=4m with a current of 12 A in the -y direction. Find the magnetic field that exists in that region if the force on the conductor is F=-1.20×10-2N(-12i-12j).
4.A long thin wire carries a current I. A metal bar of length L is moving with a constant speed v as shown in the figure. Point a is a distance d from the wire a) Calculate the electromotive force induced in the bar. b) If the bar is replaced by a rectangular circuit of resistance R, what is the magnitude of the induced current in the circuit?

Answers

1. The force on the particle is 1.36 x 10^-14 N, schematic drawing shows velocity in x-direction, magnetic field entering perpendicular to the screen, and force perpendicular to both.

2. The force on the straight conductor is 5.25 x 10^-3 N, schematic drawing shows conductor in negative z-direction and magnetic field vectors approximately orthogonal to the conductor.

3. The magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.

4. a) The electromotive force induced in the bar is BLv. b) The magnitude of the induced current in the rectangular circuit is V/R.

1. The force on the particle can be calculated using the equation F = qvB, where q is the charge, v is the velocity, and B is the magnetic field. Plugging in the given values, the force is 1.36 x 10^-14 N. A schematic drawing would show the velocity vector in the x-direction, the magnetic field vector entering perpendicular to the screen, and the force vector perpendicular to both.

2. The force on the straight conductor can be calculated using the equation F = IL x B, where I is the current, L is the length of the conductor, and B is the magnetic field. Plugging in the given values, the force is 5.25 x 10^-3 N. A schematic drawing would show the conductor in the negative z-direction, with the magnetic field vectors shown approximately orthogonal to the conductor.

3. The magnetic field can be determined using the equation F = IL x B. Since the force is given as F = -1.20 x 10^-2 N (-12i - 12j), we can equate the force components to the corresponding components of the equation and solve for B. The resulting magnetic field is approximately -0.01 T in the x-direction and -0.01 T in the y-direction.

4. To calculate the electromotive force induced in the bar, we can use the equation emf = BLv, where B is the magnetic field, L is the length of the bar, and v is the speed of the bar. The magnitude of the induced current in the rectangular circuit can be calculated using Ohm's Law, I = V/R, where V is the electromotive force and R is the resistance of the circuit.

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What is the angular velocity of a tricycle wheel relative to the angular velocity of a bicycle wheel (what is w tricycle/w bicycle) if both wheels are traveling with the same translational speed? The bicycle has a wheel radius that is 3.00 times that of the tricycle wheel. Would it be safe to make a child tricycle/adult bicycle tandem?

Answers

The angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3) and it would not be safe to make a child tricycle/adult bicycle tandem.

To determine the angular velocity ratio between the tricycle wheel and the bicycle wheel, we can use the relationship between linear speed, angular velocity, and the radius of a rotating object.

The linear speed of both wheels is the same since they are traveling at the same translational speed.

Let's denote the linear speed as v.

For the bicycle wheel, let's denote its radius as r_bicycle.

For the tricycle wheel, let's denote its radius as r_tricycle.

The relationship between linear speed and angular velocity is given by:

v = ω * r,

where v is the linear speed, ω (omega) is the angular velocity, and r is the radius of the rotating object.

For the bicycle wheel, we have:

v_bicycle = ω_bicycle * r_bicycle.

For the tricycle wheel, we have:

v_tricycle = ω_tricycle * r_tricycle.

Since both wheels have the same linear speed, we can set the two equations equal to each other:

v_bicycle = v_tricycle.

ω_bicycle * r_bicycle = ω_tricycle * r_tricycle.

We can rewrite this equation in terms of the angular velocity ratio:

ω_tricycle / ω_bicycle = r_bicycle / r_tricycle.

Given that the radius of the bicycle wheel is 3.00 times that of the tricycle wheel (r_bicycle = 3 * r_tricycle), we can substitute this into the equation:

ω_tricycle / ω_bicycle = (3 * r_tricycle) / r_tricycle.

ω_tricycle / ω_bicycle = 3.

Therefore, the angular velocity of the tricycle wheel is three times that of the bicycle wheel (ω_tricycle / ω_bicycle = 3).

Based on this, it would not be safe to make a child tricycle/adult bicycle tandem because the tricycle wheel would rotate at a much higher angular velocity than the bicycle wheel, potentially causing stability issues and safety concerns.

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An 7.20 kg package in a mail-sorting room slides 2.10 m down a chute that is inclined at 53.8 degrees below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.36. Calculate the work done on the package by
a) friction.
b) gravity.
c) the normal force
d) what is the net work done on the package?

Answers

The work done on the package by:a) friction: -228.024 J b) gravity: -348.634 Jc) the normal force: 0 J d) the net work done on the package: -576.658 J

a) The work done by friction can be calculated using the equation W_friction = -μk * N * d, where μk is the coefficient of kinetic friction, N is the normal force, and d is the displacement. The negative sign indicates that the work done by friction is in the opposite direction of the displacement.

b) The work done by gravity can be calculated using the equation W_gravity = m * g * d * cos(θ), where m is the mass of the package, g is the acceleration due to gravity, d is the displacement, and θ is the angle of the incline. The cos(θ) term accounts for the component of gravity parallel to the displacement.

c) The work done by the normal force is zero because the displacement is perpendicular to the direction of the normal force.

d) The net work done on the package is the sum of the work done by friction and the work done by gravity, i.e., W_net = W_friction + W_gravity. It represents the total energy transferred to or from the package during its motion along the chute.

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"A 6.0-cm-tall object is 12 cm in front of a concave mirror that
has a 27 cm focal length.
A.) Calculate the image position.
B.) Calculate the image height. Type a positive value if the
image is upright

Answers

A. The image position formed from concave mirror is 18cm. B. The image height is 9 cm.

A. Calculation of image position: We know that the mirror formula is 1/f = 1/v + 1/u, where, f is the focal length of the mirror. Substituting the given values, we get:1/(-27) = 1/v + 1/(-12). v = -18 cm. Since the image is formed inside the mirror, the image position is negative.

B. Calculation of image height: Magnification produced by the mirror is given by the formula, m = v/u. on substituting the values we get, m = -18 / (-12) = 3/2.The image height can be calculated as, h' = m × h= (3/2) × 6.0= 9.0 cm.

The height of the image is positive, which means it is an upright image.

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14.2
Part A
If 1.90×105 J of energy is supplied to a flask of liquid oxygen at -183∘C, how much oxygen can evaporate? The heat of vaporization for oxygen is 210 kJ/kg.
Express your answer to two significant figures and include the appropriate units.
m =
Part B
One end of a 70-cm-long copper rod with a diameter of 2.6 cm is kept at 490 ∘C, and the other is immersed in water at 22 ∘C.
Calculate the heat conduction rate along the rod.
Express your answer to two significant figures and include the appropriate units.
Qt =

Answers

The heat conduction rate along the rod is 4.62 x 10^3 W.

Part A The mass of oxygen that can evaporate can be calculated as follows:

Heat of vaporization of oxygen = 210 kJ/kg

Energy supplied to flask of liquid oxygen = 1.90 x 10^5 J

Temperature of liquid oxygen = -183°C

Now, we know that the heat of vaporization of oxygen is the amount of energy required to convert 1 kg of liquid oxygen into gaseous state at the boiling point.

Hence, the mass of oxygen that can be evaporated = Energy supplied / Heat of vaporization

= 1.90 x 10^5 / 2.10 x 10^5

= 0.90 kg

Therefore, the mass of oxygen that can evaporate is 0.90 kg.

Part B The heat conduction rate along the copper rod can be calculated using the formula:

Qt = (kAΔT)/l

Given:Length of copper rod = 70 cm

Diameter of copper rod = 2.6 cm

=> radius, r = 1.3 cm

= 0.013 m

Temperature at one end of copper rod, T1 = 490°C = 763 K

Temperature at other end of copper rod, T2 = 22°C = 295 K

Thermal conductivity of copper, k = 401 W/mK

Cross-sectional area of copper rod, A = πr^2

We know that the rate of heat conduction is the amount of heat conducted per unit time.

Hence, we need to find the amount of heat conducted first.ΔT = T1 - T2= 763 - 295= 468 K

Now, substituting the given values into the formula, we get:

Qt = (kAΔT)/l

= (401 x π x 0.013^2 x 468) / 0.7

= 4.62 x 10^3 W

Therefore, the heat conduction rate along the rod is 4.62 x 10^3 W.

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The mass of oxygen that can evaporate is approximately 0.905 kg.

The heat conduction rate along the copper rod is approximately 172.9 W.

Part A:

To determine the amount of oxygen that can evaporate, we need to use the heat of vaporization and the energy supplied to the flask.

Given:

Energy supplied = 1.90 × 10^5 J

Heat of vaporization for oxygen = 210 kJ/kg = 210 × 10^3 J/kg

Let's calculate the mass of oxygen that can evaporate using the formula:

m = Energy supplied / Heat of vaporization

m = 1.90 × 10^5 J / 210 × 10^3 J/kg

m ≈ 0.905 kg

Therefore, the mass of oxygen that can evaporate is approximately 0.905 kg.

Part B:

To calculate the heat conduction rate along the copper rod, we need to use the temperature difference and the thermal conductivity of copper.

Given:

Length of the copper rod (L) = 70 cm = 0.7 m

Diameter of the copper rod (d) = 2.6 cm = 0.026 m

Temperature difference (ΔT) = (490 °C) - (22 °C) = 468 °C

Thermal conductivity of copper (k) = 401 W/(m·K) (at room temperature)

The heat conduction rate (Qt) can be calculated using the formula:

Qt = (k * A * ΔT) / L

where A is the cross-sectional area of the rod, given by:

A = π * (d/2)^2

Substituting the given values:

A = π * (0.026/2)^2

A ≈ 0.0005307 m^2

Qt = (401 W/(m·K) * 0.0005307 m^2 * 468 °C) / 0.7 m

Qt ≈ 172.9 W

Therefore, the heat conduction rate along the copper rod is approximately 172.9 W.

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A 2.2-mmmm-diameter wire carries a 18 aa current when the electric field is 0.090 v/mv/m. part a what is the wire's resistivity? express your answer in ohm-meters.

Answers

The wire's resistivity is 2.83 x 10^-8 ohm-meters.

To find the wire's resistivity, we can use Ohm's law, which states that the resistance (R) of a wire is equal to the resistivity (ρ) multiplied by the length (L) of the wire divided by its cross-sectional area (A).

The cross-sectional area (A) of a wire with diameter d is given by the formula A = (π/4) * d^2.

Given that the wire has a diameter of 2.2 mm, we can calculate the cross-sectional area:

A = (π/4) * (2.2 mm)^2

Next, we can rearrange Ohm's law to solve for resistivity:

ρ = (R * A) / L

To find the resistance (R), we can use Ohm's law again, which states that resistance is equal to the voltage (V) divided by the current (I):

R = V / I

Given that the electric field is 0.090 V/m and the current is 18 A, we can calculate the resistance:

R = 0.090 V/m / 18 A

Finally, substituting the values into the formula for resistivity, we can calculate the wire's resistivity:

ρ = (R * A) / L

Substitute the values and calculate the resistivity in ohm-meters.

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Aray of light impinges on a mirror as shown in (Eigure 1) = 40" A second mirror is fastened at 90 to the first Part A e Figure s not At what angle above horizontal does the ray emerge after reflecting from both mirrors? Express your answer in degrees. VAX ? 0 - 170 Submit Previous Answers Request Answer

Answers

The angle above horizontal at which the ray emerges after reflecting from both mirrors is 50 degrees.

When a ray of light impinges on the first mirror at an angle of 40 degrees, it reflects at the same angle due to the law of reflection. Now, the second mirror is fastened at a 90-degree angle to the first mirror, which means the ray will reflect vertically upwards.

To find the angle above horizontal at which the ray emerges, we need to consider the angle of incidence on the second mirror. Since the ray is reflected vertically upwards, the angle of incidence on the second mirror is 90 degrees.

Using the principle of alternate angles, we can determine that the angle of reflection on the second mirror is also 90 degrees. Now, the ray is traveling in a vertical direction.

To find the angle above horizontal, we need to measure the angle between the vertical direction and the horizontal direction. Since the vertical direction is perpendicular to the horizontal direction, the angle above horizontal is 90 degrees.

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A hockey player with a mass of 62 kg is skating with an initial velocity of 5.7 m/s [N26°E] when she collides with another hockey player with a mass of 53 kg travelling with a velocity of 3.8 m/s [N]. If the heavier hockey player has a velocity of 5.0 m/s [N11°E] after the collision, determine the final velocity of the 53 kg hockey player. 4.4 m/s [N24°E] 2.0 m/s [S18°E] 5.4 m/s [N23°E] 3.1 m/s [S7.2°E]

Answers

The final velocity of the 53 kg hockey player after the collision is approximately 3.1 m/s [S7.2°E]. We can apply the principle of conservation of momentum.

Momentum is defined as the product of mass and velocity, and the total momentum before the collision should be equal to the total momentum after the collision.

Let's break down the initial velocities of both players into their horizontal and vertical components.

The 62 kg player has an initial horizontal velocity = 5.7 m/s × cos(26°) and a vertical velocity = 5.7 m/s × sin(26°).

The 53 kg player has an initial horizontal velocity of 3.8 m/s and no vertical velocity.

Using the conservation of momentum, we can write the equation:

(mass of 62 kg player × horizontal velocity of 62 kg player) + (mass of 53 kg player × horizontal velocity of 53 kg player) = (mass of 62 kg player × final horizontal velocity of 62 kg player) + (mass of 53 kg player × final horizontal velocity of 53 kg player)

(62 kg × 5.7 m/s × cos(26°)) + (53 kg × 3.8 m/s) = (62 kg × final horizontal velocity of 62 kg player × cos(11°)) + (53 kg × final horizontal velocity of 53 kg player)

Simplifying the equation, we can solve for the final horizontal velocity of the 53 kg player:

(62 kg × 5.7 m/s × cos(26°)) + (53 kg × 3.8 m/s) = (62 kg × 5.0 m/s × cos(11°)) + (53 kg × final horizontal velocity of 53 kg player

After calculating the values on the left side and rearranging the equation, we find that the final horizontal velocity of the 53 kg player is approximately 3.1 m/s.

To determine the direction, we use trigonometry to find the angle:

final angle = a tan((53 kg × final horizontal velocity of 53 kg player) / (62 kg × 5.0 m/s × sin(11°)))

Calculating the value, we get an angle of approximately 7.2° south of the positive x-axis.

Therefore, the final velocity of the 53 kg hockey player is approximately 3.1 m/s [S7.2°E].

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A generating station is producing 1.1×106 W of power that is to be sent to a small town located 6.8 km away. Each of the two wires that comprise the transmission line has a resistance per length of 5.0×10−2 d/km. (a) Find the power lost in heating the wires if the power is transmitted at 1600 V. (b) A 100:1 step-up transformer is used to raise the voltage before the power is transmitted. How much power is now lost in heating the wires? (a) Number Units (b) Number Units

Answers

(a) 150W

(b) 31858.20 W (approximately)

(a) Let's find the power loss in heating the wires if the power is transmitted at 1600 V.

As we know that P = I²R ,

Where,

P = Power,

I = Current,

R = Resistance

As we know that,

V = IR ,

where,

V = Voltage,

I = Current,

R = Resistance

R = ρ l/A ,

where,

ρ = Resistivity,

l = Length,

A = Area

Therefore, P = I²ρ l/A or P = V²/R ,

where,

V = Voltage,

R = Resistance

P = (1600)²/(2 x 5.0×10−2 d x 6.8 km) = 150 W

(b) Now, let's find the power loss in heating the wires if 100:

1 step-up transformer is used to raise the voltage before the power is transmitted.

Therefore, the new voltage, V = 1600 x 100

                                                   = 160000V, and

the new current, I = 1.1×10⁶ / 160000      

                             = 6.875A.

Now,

resistance,

R = 2 x 5.0×10−2 d x 6.8 km

= 680 Ohms

P = I²R

= (6.875)² x 680 = 31858.20 W

Therefore, the power loss in heating the wires after using the transformer is 31858.20 W (approximately).

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w = Yellow & blue light Glass Blue light (500 nm) and yellow light (600 nm) are incident on a slab of glass of thickness w = 12.0 cm, as shown in the figure. The incident beam makes an angle 0, = 45.0° with respect to the normal to the surface. In the glass, the index of refraction for the blue light is 1.565 and for the yellow light it is 1.518. The index of refraction of air is 1.000. 킄 Air Air B What distance d along the glass slab (side AB) separates the points at which the two rays emerge back into air? d = cm

Answers

(a) The distance (d) along the glass slab that separates the points at which the blue and yellow rays emerge back into air can be determined by considering the path difference between the two rays.

The path difference arises due to the difference in the indices of refraction for the two wavelengths of light and the angle of incidence.

(b) The path difference can be calculated using the formula Δd = (n_blue - n_yellow) × w × cos(θ), where n_blue and n_yellow are the indices of refraction for blue and yellow light respectively, w is the thickness of the glass slab, and θ is the angle of incidence.

Plugging in the given values of n_blue = 1.565, n_yellow = 1.518, w = 12.0 cm, and θ = 45.0°, we can calculate the path difference as Δd = (1.565 - 1.518) × 12.0 cm × cos(45.0°) ≈ 0.263 cm.

In summary, the distance (d) along the glass slab that separates the points at which the blue and yellow rays emerge back into air is approximately 0.263 cm. This calculation takes into account the path difference caused by the difference in the indices of refraction for the two wavelengths of light and the angle of incidence.

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: The position of a partide moving along the x axis is given in centimeters by-7.00+ 2.50e, where it is in seconds. Consider the time interval 2.00 tot-3.00 s (ndicate the direction with the sign of your answer.) (a) Calculate the average velocity. cm/s (b) Calculate the instantaneous velocity at t-2.00 s cm/s (c) Calculate the instantaneous velocity at t-3.00 s om/s (d) Calculate the instantaneous velocity at r-2.50 s cm/s (e) Calculate the instantaneous velocity when the particle is midway between its positions at -2.00 and 3.00 cm/s (f) Graph x versus t and indicate your answers graphically.

Answers

(a) The average velocity of the particle during the time interval from 2.00 to 3.00 seconds is -2.50 cm/s.

(b) The instantaneous velocity at t = 2.00 seconds is -2.50 cm/s.

(c) The instantaneous velocity at t = 3.00 seconds is -2.50 cm/s.

(d) The instantaneous velocity at t = 2.50 seconds is -2.50 cm/s.

(e) The instantaneous velocity when the particle is midway between its positions at -2.00 and 3.00 seconds is -2.50 cm/s.

(f) The graph of x versus t would show a linear relationship with a downward slope of -2.50 cm/s.

The given equation for the position of the particle along the x-axis is -7.00 + 2.50e, where t represents time in seconds. In this equation, the term -7.00 represents the initial position of the particle at t = 0 seconds, and 2.50e represents the displacement or change in position with respect to time.

(a) To calculate the average velocity, we need to find the total displacement of the particle during the given time interval and divide it by the duration of the interval.

In this case, the displacement is given by the difference between the positions at t = 3.00 seconds and t = 2.00 seconds, which is (2.50e) at t = 3.00 seconds minus (2.50e) at t = 2.00 seconds. Simplifying this expression, we get -2.50 cm/s as the average velocity.

(b) The instantaneous velocity at t = 2.00 seconds can be found by taking the derivative of the position equation with respect to time and evaluating it at t = 2.00 seconds. The derivative of -7.00 + 2.50e with respect to t is simply 2.50e. Substituting t = 2.00 seconds into this expression, we get -2.50 cm/s as the instantaneous velocity.

(c) Similarly, to find the instantaneous velocity at t = 3.00 seconds, we evaluate the derivative 2.50e at t = 3.00 seconds, which also gives us -2.50 cm/s.

(d) The instantaneous velocity at t = 2.50 seconds can be determined in the same way, by evaluating the derivative 2.50e at t = 2.50 seconds, resulting in -2.50 cm/s.

(e) When the particle is midway between its positions at -2.00 and 3.00 seconds, the time is 2.00 + (3.00 - 2.00)/2 = 2.50 seconds. Therefore, the instantaneous velocity at this time is also -2.50 cm/s.

(f) The graph of x versus t would be a straight line with a slope of 2.50 cm/s, indicating a constant velocity of -2.50 cm/s throughout the given time interval.

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A spring is stretched to a length of 5.40 m. You generate a standing wave using a frequency of 4.75 Hz If there are 5 antinodes along the spring, calculate the speed of the wave. Be sure to draw a picture of this standing wave.

Answers

The speed of the wave in the given scenario is approximately 12.83 m/s.

To calculate the speed of the wave in the given scenario, we need to use the formula that relates wave speed, frequency, and wavelength.

Given:

Frequency of the wave (f) = 4.75 Hz

Number of antinodes (n) = 5

Length of the stretched spring (L) = 5.40 m

The wavelength (λ) of the standing wave can be determined by considering the distance between adjacent antinodes. In a standing wave, the distance between adjacent nodes or antinodes is half the wavelength.

Since there are 5 antinodes along the spring, there are 4 intervals between them, which correspond to 4 half-wavelengths. Therefore, the total length of 5.40 m is equal to 4 times the half-wavelength.

Let's denote the wavelength as λ:

4 × (λ/2) = L

2λ = L

λ = L/2

Now, we can calculate the wavelength of the wave:

λ = 5.40 m / 2 = 2.70 m

The speed (v) of the wave can be calculated using the formula v = f × λ, where v is the speed, f is the frequency, and λ is the wavelength:

v = 4.75 Hz × 2.70 m

v ≈ 12.83 m/s

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Final answer:

To calculate the speed of the wave in the stretched spring, use the formula speed = frequency × wavelength. Find the wavelength by multiplying the length between two adjacent antinodes by 2. Substitute the frequency and wavelength into the formula to find the speed.

Explanation:

To calculate the speed of the wave, we can use the formula:

speed = frequency × wavelength

First, we need to find the wavelength of the wave. Since there are 5 antinodes along the spring, the distance between two adjacent antinodes is equal to half the wavelength:

wavelength = 2 × length between two adjacent antinodes

Next, we can substitute the frequency and wavelength into the formula to find the speed of the wave.

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Q/C A man claims that he can hold onto a 12.0 -kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 60.0m/h relative to the ground. The car in which he rides is brought to rest in 0.10s . (c) What does the answer to this problem say about laws requiring the use of proper safety devices such as seat belts and special toddler seats?

Answers

The man's claim about holding onto a 12.0-kg child in a head-on collision is incorrect. In this scenario, both cars are traveling at 60.0 m/h relative to the ground and collide. The car the man is in is brought to rest in 0.10s.

To assess the situation, we can use the principle of conservation of momentum. The total momentum of the system before the collision should be equal to the total momentum after the collision.

Since the cars are identical, they have the same mass and are traveling at the same speed in opposite directions. Therefore, the total momentum before the collision is zero.

After the collision, the car the man is in comes to a stop, resulting in a change in momentum. This means that the total momentum after the collision is not zero.

The fact that the car comes to a stop in such a short time demonstrates the significant forces involved in the collision. If the man were holding onto the child without a seat belt, both of them would experience an abrupt change in momentum and could be seriously injured or thrown from the car.

This problem emphasizes the importance of laws requiring the use of proper safety devices such as seat belts and special toddler seats. These devices help to distribute the forces of a collision more evenly throughout the body, reducing the risk of injury.

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from the delta E given for 25 degrees celcius, calculate delta H at the same temperature for the reaction: 2HI-> H2 + I2 deltaE = +9.48 kJ In a dish is a population of crystals, 3 are ight blue and 1 is dark blue. I have fournd a gene (D) that determines whether or not a crystal is light or dark biue. Being a dark blue crystal is a recessive trat (genotype dd). Using the folowing equations and assuming that the population of crystals in the dish is currenty in Hardy-Weinberg equilienum tell me the frequency of the dominant allele (frequency of the deminant alele = p) and the frequency of tie recessive allele (frequency of the recessive alele a q) rounced to two decimal places as well as how many of the crystais you would expect to be heterozypous?

Answers

1. The ΔH at 25 degrees Celsius for the given reaction is +9.48 kJ.

2. The frequency of the dominant allele (p) and the recessive allele (q) in the crystal population is 0.50 each.

3. Half of the crystals in the population are expected to be heterozygous (Dd).

To calculate the change in enthalpy (ΔH) at the same temperature for the given reaction, we need to use the relationship between ΔH and ΔE (change in internal energy). The equation is as follows:

ΔH = ΔE + PΔV

However, since the reaction is not specified to be at constant pressure or volume, we can assume it occurs under constant pressure conditions, where ΔH = ΔE.

Therefore, ΔH = ΔE = +9.48 kJ.

According to the information provided, the dark blue crystal phenotype is recessive (dd). Let's use the following symbols to represent the genotypes and their frequencies:

p = frequency of the dominant allele (D)

q = frequency of the recessive allele (d)

In a population in Hardy-Weinberg equilibrium, the frequencies of the alleles can be calculated using the following equations

[tex]p^2 + 2pq + q^2 = 1[/tex]

Here, [tex]p^2[/tex] represents the frequency of homozygous dominant individuals (DD), [tex]q^2[/tex] represents the frequency of homozygous recessive individuals (dd), and 2pq represents the frequency of heterozygous individuals (Dd).

Given that there are 3 light blue crystals (DD or Dd) and 1 dark blue crystal (dd), we can set up the following equations:

[tex]p^2 + 2pq + q^2 = 1[/tex]

[tex]p^2[/tex] + 2pq = 3/4  (since 3 out of 4 crystals are light blue)

[tex]q^2[/tex] = 1/4  (since 1 out of 4 crystals is dark blue)

From the equation [tex]q^2[/tex] = 1/4, we can determine the value of q:

q = √(1/4) = 0.5

Since p + q = 1, we can calculate the value of p:

p = 1 - q = 1 - 0.5 = 0.5

Therefore, the frequency of the dominant allele (D) is 0.50, and the frequency of the recessive allele (d) is also 0.50.

To determine the number of crystals that are heterozygous (Dd), we can use the equation 2pq:

2pq = 2 * 0.5 * 0.5 = 0.5

So, you would expect 0.5 or half of the crystals in the population to be heterozygous (Dd).

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Question 12 If a proton is in an infinite box in the n = 4 state and its energy is 0.662 MeV, what is the wavelength of this proton (in fm)? 1 pts

Answers

The wavelength of the proton in the n = 4 state in the infinite box is approximately 19.55 femtometers (fm). To determine the wavelength of a proton in an infinite box in the n = 4 state, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength.

The de Broglie wavelength (λ) is given by:

λ = h / p

where λ is the wavelength, h is the Planck's constant (approximately 6.626 × 10^-34 J·s), and p is the momentum of the proton.

The momentum of the proton can be calculated using the energy (E) and mass (m) of the proton:

E = [tex]p^2 / (2m)[/tex]

Rearranging the equation, we can solve for p:

p = √(2mE)

Energy of the proton (E) = 0.662 MeV = 0.662 × [tex]10^6[/tex] eV

Mass of the proton (m) = 1.67 × [tex]10^-27[/tex] kg

Converting the energy to joules:

1 eV = 1.6 × [tex]10^-19[/tex] J

E = 0.662 ×[tex]10^6[/tex] eV * (1.6 × [tex]10^-19[/tex] J / 1 eV)

E = 1.0592 ×[tex]10^-13[/tex] J

Now we can calculate the momentum:

p = √(2mE)

p = √(2 * 1.67 × [tex]10^-27[/tex] kg * 1.0592 × [tex]10^-13[/tex]J)

p ≈ 3.382 × [tex]10^-20[/tex] kg·m/s

Finally, we can calculate the wavelength using the de Broglie wavelength formula:

λ = h / p

λ = (6.626 × [tex]10^-34[/tex] J·s) / (3.382 × [tex]10^-20[/tex]kg·m/s)

λ ≈ 1.955 × 10^-14 m

Converting the wavelength to femtometers (fm):

1 m = [tex]10^15[/tex] fm

λ = 1.955 × [tex]10^-14[/tex]m * [tex](10^{15[/tex] fm / 1 m)

λ ≈ 19.55 fm

Therefore, the wavelength of the proton in the n = 4 state in the infinite box is approximately 19.55 femtometers (fm).

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Moving electrons pass through a double slit and an interference pattern (similar to that formed by light) is shown on the screen, as in The separation between the two slits is d=0.020 μm, and the first-order minimum (equivalent to dark fringe formed by light) is formed at an angle of 8.63∘ relative to the incident electron beam. Use h=6.626∗10−34Js for Planck constant. Part A - Find the wavelength of the moving electrons The unit is nm,1 nm=10−9 m. Keep 2 digits after the decimal point. ↔↔0 ? λ m Part B - Find the momentum of each moving electron. Use scientific notations, format 1.234∗10n.

Answers

The wavelength of the moving electrons is 0.056 nm, and the momentum of each moving electron is 1.477 × 10^−24 kg·m/s.

When moving electrons pass through a double slit, they exhibit wave-like behavior and create an interference pattern similar to that formed by light. The separation between the two slits is given as d = 0.020 μm (micrometers). To find the wavelength of the moving electrons, we can use the formula for the first-order minimum:

λ = (d * sinθ) / n,

where λ is the wavelength, d is the separation between the slits, θ is the angle formed by the first-order minimum relative to the incident electron beam, and n is the order of the minimum.

Substituting the given values into the formula:

λ = (0.020 μm * sin(8.63∘)) / 1.

To convert micrometers (μm) to nanometers (nm), we multiply by 1,000:

λ = (0.020 μm * 1,000 nm/μm * sin(8.63∘)) / 1.

Calculating this expression, we find:

λ ≈ 0.056 nm (rounded to two decimal places).

For Part B, to find the momentum of each moving electron, we can use the de Broglie wavelength equation:

λ = h / p,

where λ is the wavelength, h is the Planck constant

(h = 6.626 × 10^⁻³⁴ Js),

and p is the momentum.

Rearranging the equation to solve for momentum:

p = h / λ.

Substituting the calculated value for λ into the equation:

p = 6.626 × 10^⁻³⁴ Js / (0.056 nm * 10^⁻⁹ m/nm).

Simplifying this expression, we get:

p ≈ 1.477 × 10^⁻²⁴ kg·m/s.

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7. () An EM wave has an electric field given by E= (200 V/m) [sin ((0.5m-¹)z-(5 x 10°rad/s)t)] 3. Find a) Find the wavelength of the wave. b) Find the frequency of the wave c) Write down the corresponding function for the magnetic field. 8. () A beam of light strikes the surface of glass (n = 1.46) at an angle of 70° with respect to the normal. Find the angle of refraction inside the glass. Take the index of refraction of air n₁ = 1. 9. () A transformer has 350 turns in its primary coil and 400 turns in its secondary coil. If a voltage of 110 V is applied to its primary, find the voltage in its secondary.

Answers

Find the wavelength of the waveThe wavelength of the EM wave can be calculated from the equation λ = v/f, where λ is the wavelength, v is the speed of light, and f is the frequency.Therefore, the voltage in the secondary is 126V.

.λ = c/f

where c is the speed of light= 3x108/5x1010

= 6x10-3 m

The frequency of the EM wave is given as f = (5 x 10¹⁰ rad/s)/(2π)

= 2.5 x 10⁹ Hz.

E/B = c,

where E is the electric field, B is the magnetic field, and c is the speed of light. So,

B = E/c

=200/3x108 sin ((0.5m-¹)z-(5 x 10°rad/s)t)]

A beam of light strikes the surface of glass at an angle of 70° with respect to the normal.

index of refraction of air n₁ = 1.

Using Snell's law of refraction

: n1 sin θ1 = n2 sin θ2

Where n1 is the index of refraction of the medium of incidence, θ1 is the angle of incidence, n2 is the index of refraction of the refracting medium, and θ2 is the angle of refraction.n

₁sinθ1 = n₂sinθ2sinθ2

= n₁/n₂sinθ2

= 1/1.46 x sin70°sinθ2

= 0.4624θ2

= sin-1(0.4624)θ2

= 28.3°

Therefore, the angle of refraction inside the glass is 28.3°.9.A transformer has 350 turns in its primary coil and 400 turns in its secondary coil. If a voltage of 110 V is applied to its primary, find the voltage in its secondary.The voltage ratio of a transformer is given by the formula

:Ns / Np = Vs / V

where Ns and Np are the numbers of turns in the secondary an

primary coils respectively, and Vs and Vp are the voltages across the secondary and primary coils respectively

.So,Vs = (Ns/Np) * VpVs

= (400/350) * 110Vs

= 126V

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A large mirror has a radius of curvature of 1 m What is the the power of the mirror? O a 0 251 Ob 21 c 0.25 m Od 2 m"

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The power of the mirror with a radius of curvature of 1 m is 2 m (Option d).

The power of a mirror is given by the formula P = 2/R, where P represents the power and R represents the radius of curvature. In this case, the radius of curvature is 1 m, so the power of the mirror can be calculated as P = 2/1 = 2 m. Therefore, option d, 2 m, is the correct answer.

The power of a mirror determines its ability to converge or diverge light rays. A positive power indicates convergence, meaning the mirror focuses incoming parallel light rays, while a negative power indicates divergence, meaning the mirror spreads out incoming parallel light rays.

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(a) A wire that is 1.50 m long at 20.0°C is found to increase in length by 1.90 cm when warmed t 420.0'C. Compute its average coefficient of linear expansion for this temperature range. (b) The wire i stretched just taut (zero tension) at 420.0*C. Find the stress in the wire if it is cooled to 20.0°C withou being allowed to contract. Young's modulus for the wire is 2.0 x 10^11 Pa.

Answers

(a) Thee average coefficient of linear expansion for this temperature range is approximately 3.17 x 10^(-5) / °C. (b) The stress in the wire, when cooled to 20.0°C without being allowed to contract, is approximately 2.54 x 10^3 Pa.

(a) The average coefficient of linear expansion (α) can be calculated using the formula:

α = (ΔL / L₀) / ΔT

Where ΔL is the change in length, L₀ is the initial length, and ΔT is the change in temperature.

Given that the initial length (L₀) is 1.50 m, the change in length (ΔL) is 1.90 cm (which is 0.019 m), and the change in temperature (ΔT) is 420.0°C - 20.0°C = 400.0°C, we can substitute these values into the formula:

α = (0.019 m / 1.50 m) / 400.0°C

= 0.01267 / 400.0°C

= 3.17 x 10^(-5) / °C

(b) The stress (σ) in the wire can be calculated using the formula:

σ = E * α * ΔT

Where E is the Young's modulus, α is the coefficient of linear expansion, and ΔT is the change in temperature.

Given that the Young's modulus (E) is 2.0 x 10^11 Pa, the coefficient of linear expansion (α) is 3.17 x 10^(-5) / °C, and the change in temperature (ΔT) is 420.0°C - 20.0°C = 400.0°C, we can substitute these values into the formula:

σ = (2.0 x 10^11 Pa) * (3.17 x 10^(-5) / °C) * 400.0°C

= 2.0 x 10^11 Pa * 3.17 x 10^(-5) * 400.0

= 2.54 x 10^3 Pa.

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A hailstone of mass 5.80 10-4 kg falls
through the air and experiences a net force given by the following
formula.
(a) If C = 2.85 10-5 kg/m,
calculate the terminal speed of the hailstone.
_________

Answers

The terminal velocity of the hailstone is 11.8 m/s.

The formula given is F_net = CρAg + mg, where F_net is the net force, C is the drag coefficient, ρ is the density of the fluid, A is the projected area of the object, g is the acceleration due to gravity, and m is the mass of the object.

Now, we can determine the terminal speed of the hailstone.

(a) If C = 2.85 × 10⁻⁵ kg/m, calculate the terminal speed of the hailstone. We can use the formula:

v_terminal = (2mg / ρCπr²)¹/²

where v_terminal is the terminal velocity, m is the mass of the hailstone, ρ is the density of air, C is the drag coefficient, and r is the radius of the hailstone.

v_terminal = (2mg / ρCπr²)¹/²

= [2(5.80 × 10⁻⁴ kg)(9.8 m/s²)] / [1.20 kg/m³ × (2.85 × 10⁻⁵ kg/m) × π (0.5 × 1.25 × 10⁻³ m)²]¹/²

= 11.8 m/s

Therefore, the terminal velocity of the hailstone is 11.8 m/s.

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A cannonball is falling from rest; air resistance is
considered. Before the cannonball reaches terminal velocity, the
cannonball is speeding up. Explain why.

Answers

The cannonball is speeding up before it reaches terminal velocity due to the presence of gravitational force.

When the cannonball is initially dropped, gravity pulls it downward, and it begins to accelerate. At this stage, the air resistance opposing the motion is relatively low because the speed of the falling cannonball is still relatively low. As the cannonball accelerates, its speed increases, and the air resistance acting against it also increases. Air resistance is a force that opposes the motion of an object through the air, and it depends on factors such as the shape, size, and speed of the object. Initially, the air resistance is not strong enough to counteract the gravitational force pulling the cannonball downward. However, as the cannonball gains speed, the air resistance becomes stronger. Eventually, the air resistance force becomes equal to the gravitational force, and the cannonball reaches its terminal velocity. At this point, the forces acting on the cannonball are balanced, resulting in a constant velocity. Therefore, until the cannonball reaches its terminal velocity, the gravitational force is greater than the opposing air resistance, causing the cannonball to accelerate and speed up.

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