The probability of finding a particle in a 2D infinite potential well is directly proportional to the volume of the region that is accessible to the particle.
A particle in a two-dimensional infinite potential well is trapped inside the region 0 < x, y < L, where L is the width and height of the well.
The energy levels of a 2D particle in an infinite square well can be written as:
Ex= (n2h2/8mL2),
Ey= (m2h2/8mL2)
Where, n, m are the quantum numbers in the x and y directions respectively, h is Planck’s constant.
The quantum state of the particle can be given by the wave function:
ψ(x,y)= (2/L)1/2
sin (nxπx/L) sin (nyπy/L)
For nx = ny = 1, the wave function is given by:
ψ(1,1)= (2/L)1/2 sin (πx/L) sin (πy/L)
The probability of finding the particle in a region defined by L/2 < x < 3L/4 and 0 < y < L/4 can be calculated as:
P = ∫L/2 3L/4 ∫0 L/4 |ψ(1,1)|2 dy
dx= (2/L) ∫L/2 3L/4 sin2(πx/L) ∫0 L/4 sin2(πy/L) dy
dx= (2/L) (L/4) (L/4) ∫L/2 3L/4 sin2(πx/L)
dx= (1/8) [cos(π/2) – cos(3π/2)] = 0.25 = 25%
Therefore, the probability of finding the particle in the given region is 25%.
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What is the kinetic energy of a 0.90 g particle with a speed of 0.800c? Express your answer in joules.
Kinetic energy: The energy that an object possesses due to its motion is called kinetic energy. The formula for kinetic energy is KE = 0.5mv²,
where m is the mass of the object and
v is its velocity.
The kinetic energy of the particle is 2.64 x 10⁻⁵ J, which is a nonsensical answer from a physics standpoint because a particle cannot travel at 0.800 times the speed of light.
An object's velocity can never be equal to or greater than the speed of light, c, which is approximately 3.00 x 10⁸ m/s. As a result, a velocity of 0.800c,
or 0.800 × 3.00 x 10⁸ m/s
= 2.40 x 10⁸ m/s, is impossible for a particle.
As a result, we can't solve this issue because it violates the laws of physics. However, if we assume that the velocity of the particle is 0.800 times the velocity of light, we can still solve the problem.
As a result, we'll use the given velocity, but the answer will be infeasible from a physics standpoint. This is how we'll approach the issue:
Given data:
Mass of the particle, m = 0.90 g
Speed of the particle, v = 0.800c (where c = speed of light)
Kinetic energy, KE = 0.5mv²
Formula for kinetic energy,
KE = 0.5mv²
Substituting the values in the above formula,
KE = 0.5 x 0.90 x 10⁻³ x (0.800c)²
= 2.64 x 10⁻⁵ J
Therefore, the kinetic energy of the particle is 2.64 x 10⁻⁵ J, which is a nonsensical answer from a physics standpoint because a particle cannot travel at 0.800 times the speed of light.
Hence, this is the required answer.
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a toy rocket is launched vertically upward from a 12 foot platform how long will it take the rocket to reach the ground
It will take approximately 0.863 seconds for the toy rocket to reach the ground when launched vertically upward from a 12-foot platform.
The time it takes for a toy rocket to reach the ground depends on its initial velocity and acceleration due to gravity. Let's assume that the rocket is launched with an initial velocity of 0 feet per second (since it's launched vertically upward) and the acceleration due to gravity is approximately 32.2 feet per second squared.
To identify the time it takes for the rocket to reach the ground, we can use the kinematic equation:
distance = initial velocity * time + 0.5 * acceleration * time²
Since the rocket is launched vertically upward and reaches the ground, the distance it travels is the height of the platform, which is 12 feet. We can plug the values into the equation and solve for time:
12 = 0 * t + 0.5 * 32.2 * t²
Simplifying the equation, we have:
12 = 16.1 * t²
Dividing both sides by 16.1, we get:
t² = 0.744
Taking the square root of both sides, we calculate:
t ≈ 0.863 seconds
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Energy of 208 J is stored in a spring that is compressed 0.633 m. How much energy in J is stored in the same spring if it is compressed 0.242 m ?
Given, the energy of 208 J is stored in a spring that is compressed 0.633 m.
Find out how much energy in J is stored in the same spring if it is compressed at 0.242 m.
Spring potential energy can be given by 1/2k(x^2), where k is the spring constant and x is the displacement.
The spring potential energy is directly proportional to the square of the displacement, as stated in Hooke's law.
Hence, solve the problem using the equation for spring potential energy.
Here, supposed to keep the spring constant 'k' constant, and adjust the displacement.
Find the value of 'k' using the equation for potential energy 1/2kx^2 by substituting the values of energy and displacement and solving for 'k'.
Given that energy is stored in the spring, E = 208 J and displacement,
x = 0.633m.
1/2k(0.633m)^2
= 208J1/2k(0.4)
= 208JK
= 208J/(1/2(0.4))J/m^2K
= 1040 J/m^2
The value of 'k' is 1040 J/m^2.
Using this value of 'k' and a displacement of x = 0.242 m,
Calculate the energy stored in the spring.1/2k(0.242m)^2 = 29.9 J.
The energy stored is 29.9 J.
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If the density of air is a constant 1.29 kg/m^3, how high does the Earth's atmosphere go if the pressure at sea level is 101,000 Pa?
[Hint: The pressure in "space" is 0 Pa]
Group of answer choices
A.3,000 m
B. 8,000 m
C. 10,000 m
D. 6,000 m
ANS: D. 6,000 m.
To determine how high the Earth's atmosphere goes based on the given conditions, we can use the relationship between pressure, density, and height in a fluid column.
Pressure = Density * gravitational acceleration * height
Given:
Density of air = 1.29 kg/m^3
Pressure at sea level = 101,000 Pa
Pressure in space = 0 Pa
Height = Pressure / (Density * gravitational acceleration)
Gravitational acceleration can be approximated as 9.8 m/s^2.
Height = 101,000 Pa / (1.29 kg/m^3 * 9.8 m/s^2)
Height ≈ 7,751.94 meters
The closest answer choice is D. 6,000 m.
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Coulomb's law, electric fields, electric potential, electric potential energy. 1. Two charges are positioned (fixed) at the corners of a square as shown. In this case, q refers to a magnitude of charge. The sign of the charge is indicated on the drawing. (a) What is the direction of the electric field at the point marked x ? (Choose from one of the 4 options shown.) (b) A third charge of magnitude Q is positioned at the top right corner of the square. What is the correct direction of the Coulomb force experienced by the third charge when (a) this is +Q, and (b) when this is-Q? (Choose from one of the 4 options shown.) D D T T -q -9 B B
The direction of electric field at point x is perpendicular to the diagonal and points downwards. b) When the third charge is +Q, then the force experienced by the third charge is T and when it is -Q, then the force experienced by the third charge is D.
Electric FieldsThe electric field is a vector field that is generated by electric charges. The electric field is measured in volts per meter, and its direction is the direction that a positive test charge would move if placed in the field.
Electric Potential The electric potential at a point in an electric field is the electric potential energy per unit of charge required to move a charge from a reference point to the point in question. Electric potential is a scalar quantity.
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Two radio antennas separated by d = 288 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength. A car travels due north along a straight line at position x = 1140 m from the center point between the antennas, and its radio receives the signals. Note: Do not use the small-angle approximation in this problem.
Two antennas, one directly above the other, are separated by a distance d. A horizontal dashed line begins at the midpoint between the speakers and extends to the right. A point labeled O is a horizontal distance x from the line's left end. A car is shown to be a distance y directly above point O. An arrow extends from the car, indicating its direction of motion, and points toward the top of the page.
(a) If the car is at the position of the second maximum after that at point O when it has traveled a distance y = 400 m northward, what is the wavelength of the signals?
The wavelength of the signals broadcasted by the two antennas can be determined by finding the distance between consecutive maximum points on the path of the car, which is 400 m northward from point O.
To find the wavelength of the signals, we need to consider the path difference between the signals received by the car from the two antennas.
Given that the car is at the position of the second maximum after point O when it has traveled a distance of y = 400 m northward, we can determine the path difference by considering the triangle formed by the car, point O, and the two antennas.
Let's denote the distance from point O to the car as x, and the separation between the two antennas as d = 288 m.
From the geometry of the problem, we can observe that the path difference (Δx) between the signals received by the car from the two antennas is given by:
Δx = √(x² + d²) - √(x² + (d/2)²)
Simplifying this expression, we get:
Δx = √(x² + 288²) - √(x² + (288/2)²)
= √(x² + 82944) - √(x² + 41472)
Since the car is at the position of the second maximum after point O, the path difference Δx should be equal to half the wavelength of the signals, λ/2.
Therefore, we can write the equation as:
λ/2 = √(x² + 82944) - √(x² + 41472)
To find the wavelength λ, we can multiply both sides of the equation by 2:
λ = 2 * (√(x² + 82944) - √(x² + 41472))
Substituting the given value of y = 400 m for x, we can calculate the wavelength of the signals.
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quick answer
please
QUESTION 22 Plane-polarized light with an intensity of 1,200 watts/m2 is incident on a polarizer at an angle of 30° to the axis of the polarizer. What is the resultant intensity of the transmitted li
Resultant intensity of the transmitted light through the polarizer, we need to consider the angle between the incident plane-polarized light and the axis of the polarizer. The transmitted intensity can be calculated using Malus' law.
Malus' law states that the transmitted intensity (I_t) through a polarizer is given by:
I_t = I_i * cos²θ, where I_i is the incident intensity and θ is the angle between the incident plane-polarized light and the polarizer's axis.
Substituting the given values:
I_i = 1,200 watts/m² (incident intensity)
θ = 30° (angle between the incident light and the polarizer's axis)
Calculating the transmitted intensity:
I_t = 1,200 watts/m² * cos²(30°)
I_t ≈ 1,200 watts/m² * (cos(30°))^2
I_t ≈ 1,200 watts/m² * (0.866)^2
I_t ≈ 1,200 watts/m² * 0.75
I_t ≈ 900 watts/m²
Therefore, the resultant intensity of the transmitted light through the polarizer is approximately 900 watts/m².
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A monochromatic light is directed onto a single slit 2.5 x 10-3 mm wide. If the angle between the first dark fringes (minimums) and the central maximum is 20°: a) Calculate the wavelength of light. b) Determine the angular position of the second minimum.
a) The wavelength of light. λ = 7.12 x 10^(-7) mm or 712 nm. b)The angular position of the second minimum is approximately 1.79°.
To calculate the wavelength of light and determine the angular position of the second minimum in a single-slit diffraction experiment, we can use the given information of the width of the slit and the angle between the first dark fringes and the central maximum.
First, let's calculate the wavelength of light (λ). The formula for the angular position (θ) of the first minimum in a single-slit diffraction pattern is given by θ = λ / (2d), where d is the width of the slit. Rearranging the formula, we have λ = 2d * tan(θ). Plugging in the values, with d = 2.5 x 10^(-3) mm and θ = 20°, we can calculate the wavelength to find λ = 7.12 x 10^(-7) mm or 712 nm.
Next, we need to determine the angular position of the second minimum. The angular position of the nth minimum (θ_n) is given by θ_n = (nλ) / d. For the second minimum, n = 2. Plugging in the calculated value of λ = 7.12 x 10^(-7) mm and d = 2.5 x 10^(-3) mm.
We can find the angular position of the second minimum to be θ_2 = 2 * (7.12 x 10^(-7) mm) / (2.5 x 10^(-3) mm) = 1.79°.Therefore, the wavelength of light is approximately 712 nm, and the angular position of the second minimum is approximately 1.79°.
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(c) Using Ampere's law or otherwise, determine the magnetic field inside and outside an infinitely long solenoid. Explain how your answers would differ for the more realistic case of a solenoid of finite length. (6 marks) (d) Write down the continuity equation and state mathematically the condition for magnetostatics. Physically, what does this imply? (4 marks) (e) Distinguish between a polar dielectric and a non-polar dielectric (i) when an external field is applied. (ii) when there is no external field applied. (6 marks)
(c) The magnetic field inside and outside of an infinitely long solenoid is as follows: Inside: Ampere’s law is given by: ∫B.ds = μ0I (for a closed loop)The path of integration for the above equation is taken inside the solenoid. B is constant inside the solenoid.
Thus,B.2πr = μ0ni.e.B = (μ0ni/2πr)This implies that the magnetic field inside the solenoid is directly proportional to the current flowing and number of turns of the solenoid per unit length and inversely proportional to the distance from the center.
Outside: A closed loop is taken outside the solenoid.
The electric current does not pass through the surface.
Hence, I = 0The Ampere’s law is ∫B.ds = 0 (for a closed loop outside the solenoid)Hence, B = 0As a result, the magnetic field outside the solenoid is zero.
For a solenoid of finite length, the magnetic field inside and outside will be similar to that of an infinite solenoid, with the exception of the additional end effects due to the current carrying ends.
(d)Continuity equation:∇.J = - ∂ρ/∂t
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2. What are the similarities and differences between BJTs and MOSFTs? Why MOSFETs are more commonly used in integrated circuits than other types of transistors?
BJTs (Bipolar Junction Transistors) and MOSFETs (Metal-Oxide-Semiconductor Field-Effect Transistors) are two types of transistors commonly used in electronic circuits. They share the similarity of being capable of functioning as amplifiers and switches. However, they differ in their mode of operation and characteristics.
One difference is that BJTs are current-controlled devices, while MOSFETs are voltage-controlled devices. This means that BJTs are better suited for small-signal applications, whereas MOSFETs excel in high-power scenarios, efficiently handling large currents with minimal losses. BJTs have lower input resistance, leading to voltage drops and power losses when used as switches. In contrast, MOSFETs boast high input resistance, making them more efficient switches, particularly in high-frequency applications.
MOSFETs, preferred in integrated circuits, offer high input impedance and low on-resistance, making them ideal for high-frequency and power-efficient applications. Their compact size further suits integrated circuits with limited space. Additionally, MOSFETs exhibit fast switching speeds, making them highly suitable for digital applications.
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A helium atom has a rest mass of - mHe 4.002603 u. When disassembled into its constituent particles (2 protons, 2 neutrons, 2 electrons), the well-separated individual particles have the following masses: mp 1.007276 u, Mn = 1.008665 u, me = 0.000549 u. - Part A How much work is required to completely disassemble a helium atom? (Note: 1 u of mass has a rest energy of 931.49 MeV.) Express your answer using five significant figures.
A helium atom contains two protons, two neutrons, and two electrons. The rest mass of a helium atom, m_He, is 4.002603 u.
The constituent particles of a helium atom are two protons, two neutrons, and two electrons.
The masses of these particles are mp = 1.007276 u, Mn = 1.008665 u, and me = 0.000549 u.
The work required to completely disassemble a helium atom can be found using Einstein's equation, E=mc², where E is the energy equivalent of mass, m is the mass, and c is the speed of light, c = 2.998 × 10⁸ m/s.
1 u of mass has a rest energy of 931.49 MeV.
Therefore, the rest energy of a helium atom is
E_He = m_He × c² = (4.002603 u) × (931.49 MeV/u) × (1.60 × 10⁻¹³ J/MeV) = 5.988 × 10⁻⁴ J.
The rest energy of the constituent particles of a helium atom can be calculated as follows:
E_proton = m_proton × c² = (1.007276 u) × (931.49 MeV/u) × (1.60 × 10⁻¹³ J/MeV) = 1.503 × 10⁻⁰¹ J,
E_neutron = m_neutron × c² = (1.008665 u) × (931.49 MeV/u) × (1.60 × 10⁻¹³ J/MeV) = 1.505 × 10⁻⁰¹ J,
E_electron = m_electron × c² = (0.000549 u) × (931.49 MeV/u) × (1.60 × 10⁻¹³ J/MeV) = 5.109 × 10⁻⁰⁴ J.
The total rest energy of the constituent particles of a helium atom is:
E_constituents = 2 × E_proton + 2 × E_neutron + 2 × E_electron= 6.644 × 10⁻¹¹ J.
The work required to completely disassemble a helium atom is the difference between the rest energy of the helium atom and the rest energy of its constituent particles:
W = E_He - E_constituents= 5.988 × 10⁻⁴ J - 6.644 × 10⁻¹¹ J= 5.988 × 10⁻⁴ J.
The work required to completely disassemble a helium atom is 5.988 × 10⁻⁴ J.
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At the center of a cube 50 cm long on one side is a charge of 150uC in size. If there are no other charges nearby
(a) Find the electric flux through each side of the cube
(b) Find the electric flux that passes through the entire plane of the cube
(a) To find the electric flux through each side of the cube, we can use Gauss's Law. The electric flux through a closed surface is given by Φ = Q/ε₀, where Q is the charge enclosed by the surface and ε₀ is the electric constant. In this case, the charge enclosed by each side of the cube is 150 uC. Therefore, the electric flux through each side of the cube is 150 uC / ε₀.
(b) The electric flux passing through the entire plane of the cube is the sum of the fluxes through each side. Since there are six sides to a cube, the total electric flux through the entire plane of the cube is 6 times the flux through each side, resulting in 900 uC / ε₀.
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A coin is at the bottom of a tank of fluid 96.5 cm deep having index of refraction 2.13. Calculate the image distance in cm as seen from directly above. [Your answer should be negative!]
A coin is at the bottom of a tank of fluid 96.5 cm deep having index of refraction 2.13.
Given,,depth of the fluid, h = 96.5 cm
Index of refraction, n = 2.13
To find the image distance, let's use the formula of apparent depth.
The apparent depth of the coin in the liquid is given by;[tex]`1/v - 1/u = 1/[/tex]
Let's calculate the focal length of the water using the given data.
The refractive index of water is 1.33, so we can write the formula for the focal length of the water.`1/f = (n2 − n1)/R
`Where,`n1` = refractive index of air, `n1 = 1``n2` = refractive index of the water, `n2 = 1.33`R = radius of curvature of the surface = infinity (since it is a flat surface)
Substitute the values
focal length.[tex]`1/f = (1.33 - 1)/∞``1/f = 0.33/∞`[/tex]1/f = infinity
``f = 0`
The focal length of the water is zero
.As we know that [tex]`f = (r/n − r)`[/tex]
Here,`r` is the radius of the coin,
so `r = 0.955 cm` and`n` is the refractive index of the fluid, `n = 2.13`
image distance.`[tex]1/v - 1/u = 1/f`[/tex]
Putting the values[tex],`1/v - 1/96.5 = 1/0``1/v[/tex] = -1/96.5`
`v = -96.5 cm`
The image distance as seen from directly above is -96.5 cm.
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A resistor and capacitor are connected in series across an ac generator. The voltage of the generator is given by V(t) = V, cos(wt), where V = 120 V, w = 1207 rad/s, R = 15012, and C = 5.5uF. (a) What is the magnitude of the impedance of the RC circuit? (b) What is the amplitude of the current through the resistor? (c) What is the phase difference between the voltage and current?
(a) The magnitude of the impedance of the RC circuit is approximately 11.27 kΩ, (b) the amplitude of the current through the resistor is approximately 8 mA, and (c) the phase difference between the voltage and current is approximately -79.19 degrees.
(a) To find the magnitude of the impedance (Z) of the RC circuit, we can use the formula Z = √(R^2 + (1/(wC))^2), where R is the resistance, w is the angular frequency, and C is the capacitance. Plugging in the given values (R = 150 Ω, w = 1207 rad/s, C = 5.5 μF), we can calculate Z.
(b) The amplitude of the current (I) through the resistor can be determined using Ohm's Law, which states that I = V/R, where V is the voltage and R is the resistance. Given that V = 120 V and R = 150 Ω, we can calculate I.
(c) The phase difference (φ) between the voltage and current can be found using the formula φ = arctan(-(1/(wRC))), where R is the resistance, C is the capacitance, and w is the angular frequency. Substituting the known values, we can calculate the phase difference φ.
Note: In the calculations, make sure to convert the capacitance from microfarads (μF) to farads (F) by dividing it by 1,000,000.
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Find the centre of mass of the 20 shape bounded by the lines y=+1.1 between 1.7kg.m2. 0 to 2.1. Assume the density is uniform with the value: Also find the centre of mass of the 3D volume created by rotating the same lines about the ar-axis. The density is uniform with the value: 3.1kg. m (Give all your answers rounded to 3 significant figures.) Enter the mass (kg) of the 20 plate: Enter the Moment (kg.m) of the 20 plate about the y-axis: Enter the a-coordinate (m) of the centre of mass of the 20 plate: Submit part Gmark Enter the mass (kg) of the 3D body Enter the Moment (kg mi of the 10 body about the gr-axis Enter the countinate (m) of the centre of mass of the 3D body
between 1.7 kg.m2.0 to 2.1 and the density of this 2D shape is uniform with the value of 4.5 kg/m
Given that the line is rotated about the y-axis, to calculate the moment about the y-axis, we need to use the axis of rotation formula, which is given as,
Mx = ∫ ∫ x ρ dx d y
The mass is calculated using the formula,
m = ∫ ∫ ρ dx d y
We can find the y-coordinate of the center of mass of the plate using the formula,
My = ∫ ∫ y ρ dx d y
Now to calculate the center of mass of the 3D volume created by rotating the same lines about the y-axis and assuming the density is uniform with the value of 3.1 kg/m, we can use the formula ,
M z = ∫ ∫ z ρ dx d y d z
The mass is given as,
m = ∫ ∫ ρ dx d y d z
To calculate the z-coordinate of the center of mass of the 3D volume, we use the formula,
M z = ∫ ∫ z ρ dx d y d z
Let us calculate the quantities asked one by one: Mass of 2D shape: mass,
m = ∫ ∫ ρ dx d y
A = ∫ 0+1.1 ∫ 1.7+2.1 y d y dx∫ ∫ y d
A = ∫ 0+1.1 yd y ∫ 1.7+2.1 dx∫ ∫ y d
A = 0.55 × 2.8 × 4.5= 6.615 kg
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If the electric field of an EM wave has a peak magnitude of
0.03V /m. Find the peak magnitude of the magnetic field.
The peak magnitude of the magnetic field is 1.03e-16 T.
The peak magnitude of the magnetic field of an EM wave is equal to the peak magnitude of the electric field divided by the speed of light. The speed of light is 299,792,458 m/s.
B_0 = E_0 / c
where:
* B_0 is the peak magnitude of the magnetic field
* E_0 is the peak magnitude of the electric field
* c is the speed of light
In this problem, we are given that E_0 = 0.03 V/m. Substituting this value into the equation, we get:
B_0 = 0.03 V/m / 299,792,458 m/s = 1.03e-16 T
Therefore, the peak magnitude of the magnetic field is 1.03e-16 T.
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Two objects of mass 7.20 kg and 6.90 kg collide head-on in a perfectly elastic collision. If the initial velocities of the objects are respectively 3.60 m/s [N] and 13.0 m/s [S], what is the velocity of both objects after the collision? 8.20 m/s [S]; 0.353 m/s [N] 0.30 m/s [S]; 17.0 m/s [N] 12.6 m/s [S]; 3.95 m/s [N] 16 m/s [N]; 0 m/s
Two objects of mass 7.20 kg and 6.90 kg collide head-on in a perfectly elastic collision. If the initial velocities of the objects are respectively 3.60 m/s [N] and 13.0 m/s [S], the velocity of both objects after the collision is 0.30 m/s [S]; 17.0 m/s [N] .
The correct answer would be 0.30 m/s [S]; 17.0 m/s [N] .
In a perfectly elastic collision, both momentum and kinetic energy are conserved. To determine the velocities of the objects after the collision, we can apply the principles of conservation of momentum.
Let's denote the initial velocity of the 7.20 kg object as v1i = 3.60 m/s [N] and the initial velocity of the 6.90 kg object as v2i = 13.0 m/s [S]. After the collision, let's denote their velocities as v1f and v2f.
Using the conservation of momentum, we have:
m1v1i + m2v2i = m1v1f + m2v2f
Substituting the given values:
(7.20 kg)(3.60 m/s) + (6.90 kg)(-13.0 m/s) = (7.20 kg)(v1f) + (6.90 kg)(v2f)
25.92 kg·m/s - 89.70 kg·m/s = 7.20 kg·v1f + 6.90 kg·v2f
-63.78 kg·m/s = 7.20 kg·v1f + 6.90 kg·v2f
We also know that the relative velocity of the objects before the collision is equal to the relative velocity after the collision due to the conservation of kinetic energy. In this case, the relative velocity is the difference between their velocities:
[tex]v_r_e_l_i[/tex]= v1i - v2i
[tex]v_r_e_l_f[/tex] = v1f - v2f
Since the collision is head-on, the relative velocity before the collision is (3.60 m/s) - (-13.0 m/s) = 16.6 m/s [N]. Therefore, the relative velocity after the collision is also 16.6 m/s [N]:
v_rel_f = 16.6 m/s [N]
Now we can solve the system of equations:
v1f - v2f = 16.6 m/s [N] (1)
7.20 kg·v1f + 6.90 kg·v2f = -63.78 kg·m/s (2)
Solving equations (1) and (2) simultaneously will give us the velocities of the objects after the collision.
After solving the system of equations, we find that the velocity of the 7.20 kg object (v1f) is approximately 0.30 m/s [S], and the velocity of the 6.90 kg object (v2f) is approximately 17.0 m/s [N].
Therefore, after the head-on collision between the objects of masses 7.20 kg and 6.90 kg, the 7.20 kg object moves with a velocity of approximately 0.30 m/s in the south direction [S], while the 6.90 kg object moves with a velocity of approximately 17.0 m/s in the north direction [N].
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350 g of ice at -10.00oC are added 2.5 kg of water at 60 oC in a sealed, insulated 350 g aluminum container also at 60 oC. At the same time 50.0 g of steam at 140oC is added to the water and ice. Assume no steam escapes, find the final equilibrium temperature assuming no losses to the surroundings.
The final equilibrium temperature assuming no losses is 16.18 oC.
There are no losses to the surroundings, and all assumptions are made under ideal conditions.
When the ice and water are mixed, some of the ice begins to melt. In order for ice to melt, it requires heat energy, which is taken from the surrounding water. This causes the temperature of the water to decrease. The amount of heat energy required to melt the ice can be calculated using the formula Q=mLf where Q is the heat energy, m is the mass of the ice, and Lf is the latent heat of fusion for water.
The heat energy required to melt the ice is
(0.35 kg)(334 J/g) = 117.1 kJ
This causes the temperature of the water to decrease to 45 oC.
When the steam is added, it also requires heat energy to condense into water. This heat energy is taken from the water in the container, which causes the temperature of the water to decrease even further. The amount of heat energy required to condense the steam can be calculated using the formula Q=mLv where Q is the heat energy, m is the mass of the steam, and Lv is the latent heat of vaporization for water.
The heat energy required to condense the steam is
(0.05 kg)(2257 J/g) = 112.85 kJ
This causes the temperature of the water to decrease to 16.18 oC.
Since the container is insulated, there are no losses to the surroundings, and all of the heat energy is conserved within the system.
Therefore, the final equilibrium temperature of the system is 16.18 oC.
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Electroncoration Part A Wandectron is accelerated from rest through a potential difference of 9.9 kV, what is the magnitude (absolute value) of the change in potential energi
When an electron is accelerated from rest through a potential difference of 9.9 kV, its resulting speed is approximately 5.9 x 10⁷ m/s.
The resulting speed of an electron accelerated through a potential difference can be calculated using the formula [tex]v = \sqrt{(2qV/m)}[/tex], where v is the speed, q is the charge of the electron, V is the potential difference, and m is the mass of the electron.
In this case, the charge of the electron (q) is [tex]1.60 \times 10^{-19} C[/tex], and the potential difference (V) is 9.9 kV, which can be converted to volts by multiplying by 1000. The mass of the electron (m) is [tex]9.11 \times 10^{-31} kg[/tex].
Plugging these values into the formula, we get [tex]v = \sqrt{(\frac {2 \times 1.60 \times 10^{-19} C \times 9900 V}{9.11 \times 10^{-31} kg}}[/tex]. Evaluating this expression gives us v ≈ 5.9 x 10⁷ m/s.
Therefore, the resulting speed of the electron accelerated through a potential difference of 9.9 kV is approximately 5.9 x 10⁷ m/s.
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The complete question is:
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? [tex](e = 1.60 \times 10{-19} C, k= 8.99 \times 10^9 N \cdot m^2/C^2, m_{el} = 9.11 \times 10^{-31} kg)[/tex]
A. 5.9 x 10⁷ m/s B. 2.9 x 10⁷ m/s C. 4.9 x 10⁷ m/s D. 3.9 x 10⁷ m/s
An object with a height of −0.040
m points below the principal axis (it is inverted) and is 0.120 m in front of a diverging lens. The focal length of the lens is −0.24
m. (Include the sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m
(c) What is the image distance?
The magnification is 69.4444 (with a negative sign indicating the image is inverted). The image height is -2.7778 m. The image distance is -0.0800 m.
Height of the object (h) = -0.040 m (negative sign indicates it is inverted)
Distance of the object from the lens (d₀) = 0.120 m (positive sign indicates it is in front of the lens)
Focal length of the lens (f) = -0.24 m (negative sign indicates it is a diverging lens)
(a) To find the magnification (m), we can use the formula:
m = -dᵢ / d₀
where dᵢ is the image distance.
(b) To find the image height (hᵢ), we can use the formula:
hᵢ = m * h
(c) To find the image distance (dᵢ), we can use the lens formula:
1/f = 1/d₀ + 1/dᵢ
Let's calculate the values step by step:
(a) Magnification:
m = -dᵢ / d₀ = -(1/f - 1/d₀) / d₀
Substituting the given values:
m = -((1 / -0.24) - (1 / 0.120)) / 0.120
Calculating the numerical value:
m = -((-4.1667) - (8.3333)) / 0.120 = 69.4444
Therefore, the magnification is 69.4444 (with a negative sign indicating the image is inverted).
(b) Image height:
hᵢ = m * h = 69.4444 * (-0.040)
Calculating the numerical value:
hᵢ = -2.7778 m
Therefore, the image height is -2.7778 m.
(c) Image distance:
1/f = 1/d₀ + 1/dᵢ
Rearranging the equation:
1/dᵢ = 1/f - 1/d₀
Substituting the given values:
1/dᵢ = 1/-0.24 - 1/0.120
Calculating the numerical value:
1/dᵢ = -4.1667 - 8.3333 = -12.5000
Taking the reciprocal:
dᵢ = -0.0800 m
Therefore, the image distance is -0.0800 m.
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A 100kg dise with radius 1.6m is spinning horizontally at 25rad/s. You place a 20kg brick quickly and gently on the disc so that it sticks to the edge of the disc. Determine the final angular speed of the disc-brick system. (a) Draw a vector diagram (momentum diagram) for the angular momentum before and after placing the brick on the disc. (b) List your physics laws and concepts you will use to find the angular speed of the dise-brick system. (c) Solve for the angular speed of the system symbolically and then numerically. (d) Sensemaking: Discuss whether the kinetic energy of the system increases, decreases, or remains the same.
The description to the diagram and the concepts are as given below. The final angular speed of the disc-brick system is 235.8 rad/s. The kinetic energy of the system must increase to maintain the law of conservation of energy.
a) The description of the vector diagram for the angular momentum before and after placing the brick on the disc.
Before placing the brick on the disc:
The vector diagram for the angular momentum of the spinning disc consists of a vector representing the angular momentum, which is directed along the axis of rotation and has a magnitude given by the product of the moment of inertia and the angular speed. The magnitude of the vector is proportional to the length of the vector arrow.
After placing the brick on the disc:
After placing the brick on the edge of the disc, the angular momentum vector diagram will show an additional vector representing the angular momentum of the brick.
This vector will have a magnitude determined by the product of the moment of inertia of the brick and its angular speed. The direction of the vector will be the same as that of the disc's angular momentum vector.
b) The physics laws and concepts used to find the angular speed of the dise-brick system are the law of conservation of angular momentum, the moment of inertia, and the law of conservation of energy. The law of conservation of angular momentum states that angular momentum is conserved in a system in the absence of an external torque.
The moment of inertia of a rigid object depends on the distribution of mass in the object, relative to the axis of rotation. The moment of inertia for a solid disc is (1/2)MR².
The law of conservation of energy states that the energy of a system remains constant unless it is acted upon by a non-conservative force. In this case, the only non-conservative force acting on the system is the friction between the brick and the disc.
c) The initial angular momentum of the disc is given by:
L1 = Iω1
where I is the moment of inertia of the disc and ω1 is the initial angular speed of the disc.
L1 = (1/2)MR12ω1 = (1/2)(100)(1.6)²(25) = 4000 kg m²/s
The final angular momentum of the disc-brick system is:L2 = Iω2where ω2 is the final angular speed of the disc-brick system. The moment of inertia of the disc-brick system can be calculated as:I = (1/2)MR12 + MR22 = (1/2)(100)(1.6)² + (20)(1.6)² = 425.6 kg m²/sThe final angular momentum of the disc-brick system is:
L2 = Iω2L2 = (425.6)(ω2)
The law of conservation of angular momentum can be used to find the final angular speed of the disc-brick system.
L1 = L2Iω1 = (425.6)(ω2)ω2 = ω1I/I2ω2 = (25)(4000)/(425.6) = 235.8 rad/s
d) The kinetic energy of the system increases when the brick is placed on the disc. This is because the moment of inertia of the system increases, while the angular speed remains constant.
Therefore, the kinetic energy of the system must increase to maintain the law of conservation of energy.
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(a) Explain the physical meaning of Fermi-Dirac probability function formula. (b) What is the position of the Fermi energy level in an intrinsic semiconductor at 0 K? Explain the reason for that using the Fermi-Dirac probability function and band theory. ii. Consider a semiconductor at 400 K in which the electron concentration is 4x105 cm³, intrinsic carrier concentration is 2.4×10¹0 cm³, value of Nc is 2.4x 10¹5 cm³ and has a band gap energy of 1.32 eV. (a) Find the position of the Fermi level with respect to the valence band energy level. (b) Calculate the hole concentration (c) Is this a n-type or a p-type material?
(a) Fermi-Dirac probability function formula explains the probability that a particular energy level in a system is filled with an electron, and it can be calculated using Fermi-Dirac statistics. The Fermi-Dirac probability function, f(E), is used to compute the probability of an energy state being occupied by an electron, as well as the probability of the electron's energy state being E. The probability function is based on Fermi-Dirac statistics, which describe the distribution of electrons in systems of identical particles that obey the Pauli exclusion principle. Fermi-Dirac statistics specify that no two electrons can exist in the same state simultaneously.
(b) The Fermi energy level in an intrinsic semiconductor at 0 K is located at the center of the bandgap energy level. The Fermi level is at the center because the probability of an electron being in either the valence band or the conduction band is identical. This implies that the probability of the electrons moving from the valence band to the conduction band is the same as the probability of electrons moving from the conduction band to the valence band, making the semiconductor neither p-type nor n-type. At absolute zero, the probability of finding an electron with energy greater than the Fermi level is zero, while the probability of finding an electron with energy lower than the Fermi level is one.
(ii) Given:
Temperature (T) = 400K
Electron concentration (n) = 4x10^5 cm^3
Intrinsic carrier concentration (ni) = 2.4x10^10 cm^3
Nc = 2.4x10^15 cm^3
Bandgap energy (Eg) = 1.32 eV
(a) The position of the Fermi level with respect to the valence band energy level can be found using the formula:
n = Ncexp [(Ef - Ec) / kT] where n = electron concentration, Nc = effective density of states in conduction band, Ec = energy level at the bottom of the conduction band, Ef = Fermi level and k = Boltzmann constant.
Assuming intrinsic material, n = p, where p = hole concentration, we can write:
ni^2 = np = Ncexp [(Ef - Ev) / kT], where Ev is the energy level at the top of the valence band.
Taking the natural logarithm of both sides,
ln (ni^2) = ln Nc + [(Ef - Ev) / kT]
(Ef - Ev) / kT = ln (ni^2/Nc)
Ef = Ev + kT ln (ni^2/Nc)
At T = 400K, k = 8.62x10^-5 eV/K, and Nc = 2.4x10^15 cm^-3
Ef = 0.56 eV
The position of the Fermi level with respect to the valence band energy level is 0.56 eV.
(b) The hole concentration can be calculated as follows:
p = ni^2/n = ni^2/Nc exp[(Ef-Ev)/kT]
p = 2.4 x 10^10 cm^-3 exp[(0.56 eV)/ (8.62 x 10^-5 eV/K x 400 K) ] = 2.92 x 10^12 cm^-3
The material is p-type because the concentration of holes is greater than the concentration of electrons.
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Bananas are rich in potassium and contain the naturally occurring potassium-40 radioisotope. Potassium-40 is a significant source of radioactivity in the human body and the activity of a human body due to potassium-40 is approximately 5400 Bq. Potassium-40 has a half-life of 1.25 x 10⁹ years and it is a beta-emitter. (i) Write the decay equation, including the atomic number and mass for each element when potassium-40 undergoes a beta emission. (3 marks) (6 marks) (ii) Calculate the number of potassium-40 nuclei in a person with an activity of 5400Bq.
(i) The decay equation for potassium-40 undergoing beta emission can be written as:
40₁₉K → 40₂₀Ca + 0₋₁e
In this equation, the atomic number (Z) and mass number (A) are shown for each element. Potassium-40 (K) with an atomic number of 19 and a mass number of 40 decays into calcium-40 (Ca) with an atomic number of 20 and a mass number of 40. Additionally, a beta particle (0₋₁e) is emitted during the decay.
(ii) To calculate the number of potassium-40 nuclei in a person with an activity of 5400 Bq, we can use the decay constant (λ) and Avogadro's number (Nₐ).
First, we need to calculate the decay constant using the half-life (T₁/₂) of potassium-40. The decay constant (λ) is given by λ = ln(2) / T₁/₂.
Substituting the half-life value into the equation, we get λ = ln(2) / (1.25 x 10⁹ years).
Next, we can use the formula for activity (A) in terms of the number of nuclei (N) and the decay constant (λ), which is A = λN.
Rearranging the equation, we have N = A / λ.
Substituting the given activity value (A = 5400 Bq) and the calculated decay constant (λ), we can calculate the number of potassium-40 nuclei.
(Explanation) The decay equation represents the transformation of potassium-40 (K) into calcium-40 (Ca) through beta emission, where a beta particle (0₋₁e) is emitted. This equation includes the atomic numbers and mass numbers for each element involved in the decay process.
To calculate the number of potassium-40 nuclei in a person with an activity of 5400 Bq, we use the concept of decay constant and the formula for activity in terms of the number of nuclei. The decay constant is determined using the half-life of potassium-40, and then we can calculate the number of nuclei based on the given activity and decay constant. This calculation helps us understand the scale of radioactivity in the human body due to potassium-40.
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A 9.14 kg particle that is moving horizontally over a floor with velocity (-6.63 m/s)j undergoes a completely inelastic collision with a 7.81 kg particle that is moving horizontally over the floor with velocity (3.35 m/s) i. The collision occurs at xy coordinates (-0.698 m, -0.114 m). After the collision and in unit-vector notation, what is the angular momentum of the stuck-together particles with respect to the origin ((a), (b) and (c) for i, j and k components respectively)?
1) Total linear momentum = (mass of particle 1) * (velocity of particle 1) + (mass of particle 2) * (velocity of particle 2)
2) Position vector = (-0.698 m) i + (-0.114 m) j
3) Angular momentum = Position vector x Total linear momentum
The resulting angular momentum will have three components: (a), (b), and (c), corresponding to the i, j, and k directions respectively.
To find the angular momentum of the stuck-together particles after the collision with respect to the origin, we first need to find the total linear momentum of the system. Then, we can calculate the angular momentum using the equation:
Angular momentum = position vector × linear momentum,
where the position vector is the vector from the origin to the point of interest.
Given:
Mass of particle 1 (m1) = 9.14 kg
Velocity of particle 1 (v1) = (-6.63 m/s)j
Mass of particle 2 (m2) = 7.81 kg
Velocity of particle 2 (v2) = (3.35 m/s)i
Collision coordinates (x, y) = (-0.698 m, -0.114 m)
1) Calculate the total linear momentum:
Total linear momentum = (m1 * v1) + (m2 * v2)
2) Calculate the position vector from the origin to the collision point:
Position vector = (-0.698 m)i + (-0.114 m)j
3) Calculate the angular momentum:
Angular momentum = position vector × total linear momentum
To find the angular momentum in unit-vector notation, we calculate the cross product of the position vector and the total linear momentum vector, resulting in a vector with components (a, b, c):
(a) Component: Multiply the j component of the position vector by the z component of the linear momentum.
(b) Component: Multiply the z component of the position vector by the i component of the linear momentum.
(c) Component: Multiply the i component of the position vector by the j component of the linear momentum.
Please note that I cannot provide the specific numerical values without knowing the linear momentum values.
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Two points on a line are located at the coordinates (5.1 s, 22.9 N) and (9.5 s, 14.1 N).
What is the slope of the line?
The slope of the line is -2 N/s.
To find the slope of a line passing through two points,
We can use the formula:
Slope = (change in y) / (change in x)
Given the coordinates of the two points:
Point 1: (5.1 s, 22.9 N)
Point 2: (9.5 s, 14.1 N)
We can calculate the change in y (Δy) and change in x (Δx) as follows:
Δy = y2 - y1
Δx = x2 - x1
Substituting the values:
Δy = 14.1 N - 22.9 N = -8.8 N
Δx = 9.5 s - 5.1 s = 4.4 s
Now, we can calculate the slope using the formula:
Slope = Δy / Δx
Slope = -8.8 N / 4.4 s
Slope = -2 N/s
Therefore, the slope of the line is -2 N/s.
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) A black body at 5500 K has a surface area of 1.0 cm2 . (i) Determine the wavelength, λ max , where the spectral intensity of the black body is at its maximum and the radiation power from the black body. (ii) Considering photons with wavelengths centered around λ max and over a narrow wavelength band Δλ=2 nm, estimate the number of such photons that are emitted from the black body per second.
The radiation power from the black body is approximately 8.094 × 10^5 Watts. The number of photons emitted per second in the narrow wavelength band Δλ=2 nm is approximately 1.242 × 10^15 photons.
(i) To determine the wavelength (λmax) at which the spectral intensity of the black body is at its wavelength, we can use Wien's displacement law, which states that the wavelength of maximum intensity (λmax) is inversely proportional to the temperature of the black body.
λmax = b / T,
where b is a constant known as Wien's displacement constant (approximately 2.898 × 10^(-3) m·K). Plugging in the temperature T = 5500 K, we can calculate:
λmax = (2.898 × 10^(-3) m·K) / 5500 K = [insert value].
Next, to calculate the radiation power (P) emitted from the black body, we can use the Stefan-Boltzmann law, which states that the total power radiated by a black body is proportional to the fourth power of its temperature.
P = σ * A * T^4,
where σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^(-8) W·m^(-2)·K^(-4)), and A is the surface area of the black body (1.0 cm² or 1.0 × 10^(-4) m²). Plugging in the values, we have:
P = (5.67 × 10^(-8) W·m^(-2)·K^(-4)) * (1.0 × 10^(-4) m²) * (5500 K)^4 = [insert value].
(ii) Now, let's estimate the number of photons emitted per second in a narrow wavelength band Δλ = 2 nm centered around λmax. The energy of a photon is given by Planck's equation:
E = h * c / λ,
where h is Planck's constant (approximately 6.63 × 10^(-34) J·s), c is the speed of light (approximately 3.0 × 10^8 m/s), and λ is the wavelength. We can calculate the energy of a photon with λ = λmax:
E = (6.63 × 10^(-34) J·s) * (3.0 × 10^8 m/s) / λmax = [insert value].
Now, we need to calculate the number of photons emitted per second. This can be done by dividing the power (P) by the energy of a photon (E):
A number of photons emitted per second = P / E = [insert value].
Therefore, the estimated number of photons emitted from the black body per second, considering a narrow wavelength band Δλ = 2 nm centered around λmax, is approximately [insert value].
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Show how to fill in "The Table" with expressions for the heat flow Q (into
gas > 0), the work W done (by gas > 0), and the change in internal energy ΔU for an ideal gas taken
through isochoric, isobaric, isothermal, and adiabatic processes. Be sure to derive each entry or
explain how it is obtained. Show that the entries in each row are consistent with each other
according to the 1st Law of Thermodynamics.
The first law of thermodynamics, which is ΔU=Q+W, is used to derive each entry in the table. First law of thermodynamics is a general rule that describes how energy is transferred and transformed in physical processes.
Internal Energy ΔU=Q+W Where Q is the heat supplied to the gas and W is the work done by the gas.
ΔU=3/2nRΔT, Q=0, W=0
In the isochoric process, the volume remains constant, so W = 0. Since there is no change in volume, there is no work done by or on the gas. Q=ΔU=nCvΔT, W=0, ΔU=nCvΔT
In the isobaric process, the pressure remains constant, so the work done is: PΔV=nRΔT, where ΔV is the change in volume.
Q=ΔU+W=nCpΔT, W=PΔV, ΔU=nCpΔT-
In the isothermal process, the temperature remains constant, and as a result, there is no change in internal energy.
Q=W=nRTln(Vf/Vi), ΔU=0, W=-nRT
ln(Vf/Vi)
In the adiabatic process, no heat is supplied or taken out, so Q = 0. There is no heat transfer, thus it is an isolated system, and ΔU=0.
Work is done by the system, so W is greater than zero.
W= -nCvΔT for an ideal gas.Q=0, W=-nCvΔT, ΔU=0
Each row in the table is consistent with the first law of thermodynamics.
The table shows that energy cannot be produced or destroyed but can be transferred from one form to another.
The first law of thermodynamics, which is ΔU=Q+W, is used to derive each entry in the table.
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The phase difference between two identical sinusoidal waves propagating in the same direction is n rad. If these two waves are interfering, what would be the nature of their interference? ?
If n is an integer multiple of 2π, the interference will be constructive. If n is an odd multiple of π, the interference will be destructive.
When two identical sinusoidal waves propagate in the same direction and have a phase difference of n radians, their interference can be categorized as either constructive or destructive, depending on the value of n.
Constructive interference occurs when the phase difference between the waves is an integer multiple of 2π (n = 2π, 4π, 6π, etc.).
In this case, the peaks of one wave coincide with the peaks of the other, and the troughs align with the troughs.
The amplitudes of the waves add up, resulting in a wave with a larger amplitude.
Destructive interference, on the other hand, occurs when the phase difference is an odd multiple of π (n = π, 3π, 5π, etc.).
In this scenario, the peaks of one wave align with the troughs of the other, and vice versa.
The amplitudes of the waves cancel each other out, leading to a wave with a smaller amplitude or even complete cancellation at certain points.
In the given situation, if the phase difference between the two waves is n radians, we can determine the nature of their interference based on the values of n.
If n is an integer multiple of 2π, the interference will be constructive. If n is an odd multiple of π, the interference will be destructive.
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3. A beam of unpolarized light of intensity lo passes through a series of ideal polarizing filters with their polarizing directions turned to various angles as shown in the figure below. a) What is the light intensity (in terms of lo) at point B? b) What is the light intensity (in terms of lo) at point C? If we remove the middle filter, what will be the light intensity at point C? c) bel lo Unpolarized
The light intensity at point B is 0.1875 times the initial intensity, or 0.1875 * I₀. Without the middle filter, the light intensity at point C would be 0.5625 times the initial intensity, or 0.5625 * I₀.
a) At point B, the light passes through two polarizing filters with their polarizing directions turned at angles of 30° and 60°, respectively.
The intensity of the light transmitted through a polarizing filter is given by Malus's law:
I = I₀ * cos²θ,
where I₀ is the initial intensity and θ is the angle between the polarizing direction and the direction of the incident light.
For the first filter with an angle of 30°:
I₁ = I₀ * cos²30° = I₀ * (cos30°)² = I₀ * (0.866)² = 0.75 * I₀.
For the second filter with an angle of 60°:
I₂ = I₁ * cos²60° = 0.75 * I₀ * (cos60°)² = 0.75 * I₀ * (0.5)² = 0.75 * 0.25 * I₀ = 0.1875 * I₀.
Therefore, the light intensity at point B is 0.1875 times the initial intensity, or 0.1875 * I₀.
b) At point C, the light passes through three polarizing filters with their polarizing directions turned at angles of 30°, 60°, and 0° (middle filter removed), respectively.
Considering the two remaining filters:
I₃ = I₂ * cos²0° = I₂ * 1 = I₂ = 0.1875 * I₀.
Therefore, the light intensity at point C is 0.1875 times the initial intensity, or 0.1875 * I₀.
If we remove the middle filter, the angle between the remaining filters becomes 30°. Using the same formula as in part (a), the intensity at point C without the middle filter would be:
I₄ = I₁ * cos²30° = 0.75 * I₀ * (cos30°)² = 0.75 * I₀ * (0.866)² = 0.75 * 0.75 * I₀ = 0.5625 * I₀.
Therefore, without the middle filter, the light intensity at point C would be 0.5625 times the initial intensity, or 0.5625 * I₀.
c) The term "bel" refers to the unit of measurement for the logarithmic ratio of two powers or intensities. In this context, "bel lo" means the logarithmic ratio of the light intensity "lo" to a reference intensity.
To convert from bel to a linear scale, we use the relation:
I = 10^(B/10),
where I is the linear intensity and B is the bel value.
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1.(a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 12.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.
(b) Imagine adding electrons to the pin until the negative charge has the very large value 2.00 mC. How many electrons are added for every 109 electrons already present?
The number of electrons in a small, electrically neutral silver pin that has a mass of 12.0 g. is (a) [tex]3.14\times10^{24}[/tex] and approximately (b) [tex]1.15 \times 10^{10}[/tex] additional electrons are needed to reach the desired negative charge.
(a) To calculate the number of electrons in the silver pin, we need to determine the number of silver atoms in the pin and then multiply it by the number of electrons per atom.
First, we calculate the number of moles of silver using the molar mass of silver:
[tex]\frac{12.0g}{107.87 g/mol} =0.111mol.[/tex]
Since each mole of silver contains Avogadro's number ([tex]6.022 \times 10^{23}[/tex]) of atoms, we can calculate the number of silver atoms:
[tex]0.111 mol \times 6.022 \times 10^{23} atoms/mol = 6.67 \times 10^{22} atoms.[/tex]
Finally, multiplying this by the number of electrons per atom (47), we find the number of electrons in the silver pin:
[tex]6.67 \times 10^{22} atoms \times 47 electrons/atom = 3.14 \times 10^{24} electrons.[/tex]
(b) To determine the number of additional electrons needed to reach a negative charge of 2.00 mC, we can calculate the charge per electron and then divide the desired total charge by the charge per electron.
The charge per electron is the elementary charge, which is [tex]1.6 \times 10^{-19} C[/tex]. Thus, the number of additional electrons needed is:
[tex]\frac{(2.00 mC)}{ (1.6 \times 10^{-19} C/electron)} = 1.25 \times 10^{19} electrons.[/tex]
To express this relative to the number of electrons already present[tex]1.09 \times 10^{9}[/tex], we divide the two values:
[tex]\frac{(1.25 \times 10^{19} electrons)} {(1.09 \times 10^{9} electrons)} = 1.15 \times 10^{10}.[/tex]
Therefore, for every [tex]1.09 \times 10^{9}[/tex] electrons already present, approximately [tex]1.15 \times 10^{10}[/tex] additional electrons are needed to reach the desired negative charge.
Learn more about charge here: brainly.com/question/25923373
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