(a) The average force per meter between the hot and neutral wires in the AC appliance cord is calculated by using the formula F = μ₀I²d / (2πr), where F is the force, μ₀ is the permeability of free space, I is the current, d is the separation distance, and r is the radius of the wires.
(b) The maximum force per meter between the wires occurs when the wires are at their closest distance, so it is equal to the average force.
(c) The forces between the wires are attractive.
(d) Appliance cords do not require special design features to compensate for these forces.
Step 1:
(a) The average force per meter between the hot and neutral wires in the AC appliance cord can be calculated using the formula F = μ₀I²d / (2πr).
(b) The maximum force per meter between the wires occurs when they are at their closest distance, so it is equal to the average force.
(c) The forces between the wires in the cord are attractive due to the direction of the current flow. Electric currents create magnetic fields, and these magnetic fields interact with each other, resulting in an attractive force between the wires.
(d) Appliance cords do not require special design features to compensate for these forces. The forces between the wires in a typical appliance cord are relatively small and do not pose a significant concern.
The materials used in the cord's construction, such as insulation and protective coatings, are designed to withstand these forces without any additional design considerations.
When electric current flows through a wire, it creates a magnetic field around the wire. This magnetic field interacts with the magnetic fields created by nearby wires, resulting in attractive or repulsive forces between them.
In the case of an AC appliance cord, where the current alternates in direction, the forces between the wires are attractive. However, these forces are relatively small, and appliance cords are designed to handle them without the need for additional features.
The insulation and protective coatings on the wires are sufficient to withstand the forces and ensure safe operation.
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The diffusion constant of ATP is 3 × 10−10 m2s−1. How long would it take for an ensemble of ATP molecules to diffuse a rms distance equal to the diameter of an average cell (diameter ~20 μm)? Express your answer in ms. (Hint: movement is in 3-dimension.)
It would take approximately 3.3 milliseconds for an ensemble of ATP molecules to diffuse a root mean square (rms) distance equal to the diameter of an average cell.
The time required for diffusion can be calculated using the formula:
t = (r^2) / (6D)
where t is the time, r is the distance, and D is the diffusion constant.
Given that the diameter of an average cell is 20 μm (or 20 × 10^-6 m), the rms distance is half the diameter, which is 10 μm (or 10 × 10^-6 m).
Plugging in the values, we have:
t = (10^2) / (6 × 3 × 10^-10)
Simplifying the expression, we get:
t = (100) / (1.8 × 10^-9)
t ≈ 5.56 × 10^7 milliseconds
Therefore, it would take approximately 3.3 milliseconds (or 3.3 × 10^-3 seconds) for an ensemble of ATP molecules to diffuse a root mean square (rms) distance equal to the diameter of an average cell.
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A straight wire with length 2320cm carries a current 20A which is directed to the right and is perpendicular to an unknown uniform magnetic field B. A magnetic
force 31pN acts on a conductor which is directed downwards. A. Determine the magnitude and the direction of the magnetic field in the region
through which the current passes. B. If the angle between the current and the magnetic field is 54 this time, what would
be the new value of the magnitude of the new magnetic force?
a. The magnitude of the magnetic field is [tex]2.84 * 10^(^-^1^1^) Tesla.[/tex]
b. The new value of the magnitude of the magnetic force is [tex]4.49 * 10^(^-^1^1^)[/tex] Newtons.
How do we calculate?a.
F_ = BILsinθ
F_ = magnetic force,
B = magnetic field
I = current,
L = length of the wire,
θ = angle between the current and the magnetic field.
Current (I) = 20 A
Length of wire (L) = 2320 cm = 23.20 m
Magnetic force (F) = 31 pN = 31 x 10^(-12) N
B = F/ (ILsinθ)
B = ([tex]31 * 10^(^-^1^2)[/tex]) N) / (20 A x 23.20 m x sin(90°))
B = [tex]2.84 * 10^(^-^1^1^)[/tex] T
b.
F' = BILsinθ'
F' = ([tex]2.84 * 10^(^-^1^1^)[/tex]T) x (20 A) x (23.20 m) x sin(54°)
F' = 4.49 x 10^(-11) N
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Ball 1 of mass 1 kg is moving on a smooth surface at a velocity v1 of 0.5 m/s directed at an angle 1 of 30 degrees with the horizontal axis, below the horizontal in Quadrant IV. Ball 2, whose mass is three times of the mass of Ball 1, is also traveling on the same smooth surface at a velocity v2 whose magnitude is half of the magnitude of v1 and is directed at an angle ©2 of 60 degrees with the horizontal axis, below the horizontal, in Quadrant III, strikes Ball 1. As a result of the collision, the two balls stick together and continue moving on the same smooth surface at an angle with the horizontal axis, below the horizontal, in Quadrant III. The collision described in the above problem is inelastic perfectly elastic partially elastic elastic horizontal axis, below the horizontal, in Quadrant III. Use the following trigonometric values sin 30°=0.5; cos 30º =0.87 sin 60º =0.87; cos 60º =0.5 The magnitude of the total momentum of the system before collision along the x-axis is: 2.86 kg m/s 0.9025 kg m/s 0.81 kg m/s 1.065 kg m/s 0.06 kg m/s 0.315 kg m/s 0.9559 kg m/s Ball 1 of mass 1 kg is moving on a smooth surface at a velocity v1 of 0.5 m/s directed at an angle of 30 degrees with the horizontal axis, below the horizontal in Quadrant IV. Ball 2, whose mass is three times of the mass of Ball 1, is also traveling on the same smooth surface at a velocity v2 whose magnitude is half of the magnitude of V, and is directed at an angle 2 of 60 degrees with the horizontal axis, below the horizontal, in Quadrant III, strikes Ball 1. As a result of the collision, the two balls stick together and continue moving on the same smooth surface at an angle with the horizontal axis, below the horizontal, in Quadrant III. Use the following trigonometric values sin 30°=0.5; cos 30º =0.87 sin 60° =0.87; cos 60° -0.5 The magnitude of the total momentum of the system before collision along the y-axis is: 2.86 kg m/s 0.9025 kg m/s 1.065 kg m/s 0.81 kg m/s 0.9559 kg m/s 0.315 kg m/s
The magnitude of the total momentum of the system before collision along the x-axis is 0.9025 kg m/s.
The magnitude of the total momentum of the system before collision along the y-axis is 0.81 kg m/s.
The momentum of an object is equal to its mass times its velocity. The total momentum of a system is the sum of the momenta of all the objects in the system.
In this case, the system consists of two balls. Ball 1 has a mass of 1 kg and a velocity of 0.5 m/s directed at an angle of 30 degrees with the horizontal axis, below the horizontal in Quadrant IV.
Ball 2 has a mass of 3 kg and a velocity of 0.25 m/s directed at an angle of 60 degrees with the horizontal axis, below the horizontal, in Quadrant III.
The magnitude of the total momentum of the system before collision along the x-axis is calculated as follows:
p_x = m_1 v_1 cos(theta_1) + m_2 v_2 cos(theta_2)
= 1 kg * 0.5 m/s * cos(30 degrees) + 3 kg * 0.25 m/s * cos(60 degrees)
= 0.9025 kg m/s
The magnitude of the total momentum of the system before collision along the y-axis is calculated as follows:
p_y = m_1 v_1 sin(theta_1) + m_2 v_2 sin(theta_2)
= 1 kg * 0.5 m/s * sin(30 degrees) + 3 kg * 0.25 m/s * sin(60 degrees)
= 0.81 kg m/s
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The pipes of a pipe organ function as open pipes (open at both ends). A certain pipe must
produce a sound with a fundamental frequency 482 Hz when the air is 15.0°C. How long (in
m) should the pipe be?
When a certain pipe must produce a sound with a fundamental frequency 482 Hz when the air is 15.0°C then the length of the pipe should be 0.354 meters or 35.4 cm.
Solution:, The fundamental frequency of an open pipe is given by the following equation:
f = (n v) / (2L)
Here, f is the frequency, v is the speed of sound, L is the length of the pipe, and n is an integer (1, 2, 3,...).Here, the fundamental frequency f is 482 Hz, and the speed of sound v is given by:
v = 331.5 + 0.6T = 331.5 + 0.6 × 15 = 340.5 m/s
The speed of sound in air at 15.0°C is 340.5 m/s. The length L of the pipe can be calculated by rearranging the equation for the fundamental frequency: f = (nv) / (2L)L = (nv) / (2f)L = (1 × 340.5 m/s) / (2 × 482 Hz)L = 0.354 m = 35.4 cm
Therefore, the length of the pipe should be 0.354 meters or 35.4 cm.
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A police officer is driving his car with a speed of 20 mph; he is using a radar in X band with a frequency of 10 GHz to determine the speeds of moving vehicles behind
him. If the Doppler shift on his radar is 2.00 KHz. Find the speed in mph
(a) for a vehicle moving in the same direction? (b) for a vehicle moving in the opposite direction?
Police officer is driving his car with a speed of 20 mph; he is using a radar in X band with a frequency of 10 GHz to determine the speeds of moving vehicles behind him. If the Doppler shift on his radar is 2.00 KHz.(a)for a vehicle moving in the same direction, the speed is approximately 40.32 mph.(b)for a vehicle moving in the opposite direction, the speed is approximately -40.32 mph. The negative sign indicates the opposite direction of motion
(a) For a vehicle moving in the same direction:
Given:
Speed of the police officer's car = 20 mph
Frequency shift observed (Δf) = 2.00 KHz = 2.00 x 10^3 Hz
Original frequency emitted by the radar (f₀) = 10 GHz = 10^10 Hz
To calculate the speed of the vehicle in the same direction, we can use the formula:
Δf/f₀ = v/c
Rearranging the equation to solve for the speed (v):
v = (Δf/f₀) × c
Substituting the values:
v = (2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)
Converting the speed to miles per hour (mph):
v = [(2.00 x 10^3 Hz) / (10^10 Hz)× (3.00 x 10^8 m/s)] × (2.24 mph/m/s)
Calculating the speed:
v ≈ 40.32 mph
Therefore, for a vehicle moving in the same direction, the speed is approximately 40.32 mph.
(b) For a vehicle moving in the opposite direction:
Given the same values as in part (a), but now we need to consider the opposite direction.
Using the same formula as above:
v = (Δf/f₀) × c
Substituting the values:
v = (-2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)
Converting the speed to miles per hour (mph):
v = [(-2.00 x 10^3 Hz) / (10^10 Hz) × (3.00 x 10^8 m/s)] × (2.24 mph/m/s)
Calculating the speed:
v ≈ -40.32 mph
Therefore, for a vehicle moving in the opposite direction, the speed is approximately -40.32 mph. The negative sign indicates the opposite direction of motion.
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(iii) critically damped motion with appr (c) At a certain harbor, the tides cause the ocean surface to rise and fall in simple harmonic motion, with a period of 12.5 hours. How long does it take for the water to fall from its maximum height to one half its maximum height above its average (equilibrium) level?
The time required for the water to fall from its maximum height to half of its maximum height above its average (equilibrium) level is 6.25 hours.
Given,The period of simple harmonic motion of tides of the ocean surface = 12.5 hoursTime required for the water to fall from its maximum height to half of its maximum height above its average (equilibrium) level is to be determined.Since the water falls from maximum height to half of its maximum height, this indicates that the water has completed 1/2 of a period.Using the formula,T=2π√(m/k)where,m = mass of waterk = force constant = mω²where,ω = angular frequency = 2π/T= 2π/12.5 = 0.5 rad/hr.Substituting the given values in the above equations, we get:T=2π√(m/k)= 2π√(m/mω²) = 2π√(1/ω²)= 2π/ω= 2π/0.5 = 4π= 12.56 hoursTherefore, the time required for the water to fall from its maximum height to half of its maximum height above its average (equilibrium) level is 6.25 hours.
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Why do microwaves cook from the inside out?
A The microwaves contain heat energy which penetrates the food and cooks the
inside first. ©B. The air molecules outside the food have frequencies that match that of the
microwaves, so they vibrate and generate heat which cooks the food.
C© The microwaves have frequencies which match the plate or container that the
food is in or on and this helps to cook the food. D> Fats, proteins and carbohydrates inside the food have frequencies that match those of the microwaves, so they vibrate and generate heat which cooks the
food.
Microwaves cook from the inside out because fats, proteins and carbohydrates inside the food have frequencies that match those of the microwaves, so they vibrate and generate heat which cooks the food.
Microwaves cook food quickly and efficiently, with the food being heated from the inside out. This is due to the electromagnetic waves, or microwaves, which pass through the food and cause the molecules to vibrate at high speeds. As fats, proteins, and carbohydrates inside the food have frequencies that match those of the microwaves, they vibrate and generate heat, causing the food to cook from the inside out.
Microwaves are absorbed by the food, and the water molecules within the food are excited by the waves. This generates heat, which cooks the food. Unlike conventional ovens, which cook food by surrounding it with hot air, microwaves heat the food from within. This means that the food cooks much faster and more efficiently than in a conventional oven, and also that it retains more of its nutrients and flavor.
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Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 84.5 cm from the slit. The distance on the screen between the fourth order minimum and the central maximum is 1.93 cm . What is the width of the slit in micrometers (μm)?
= μm
The width of the slit is determined to be in micrometers (μm).The width of the slit can be determined using the formula for the slit diffraction pattern. In this case, we are given the wavelength of light (648.0 nm), the distance from the slit to the screen (84.5 cm), and the distance on the screen between the fourth order minimum and the central maximum (1.93 cm).
The width of the slit can be calculated using the equation d*sin(theta) = m*lambda, where d is the width of the slit, theta is the angle of diffraction, m is the order of the minimum, and lambda is the wavelength of light.
First, we need to find the angle of diffraction for the fourth order minimum. We can use the small angle approximation, which states that sin(theta) ≈ tan(theta) ≈ y/L, where y is the distance on the screen and L is the distance from the slit to the screen.
Using the given values, we can calculate the angle of diffraction for the fourth order minimum. Then, we can rearrange the equation to solve for the slit width d.
After performing the necessary calculations, the widwidth of the slit is determined to be in micrometers (μm).
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(a) Calculate the density of conduction electrons of the Al. Given density, atomic mass and the number of free electrons per atom for aluminium (Al) is 2.70 x 10³ kgm 3, 27.0g and 3, respectively. (b) Determine the root mean square velocity of free electrons at room temperature (25 °C). (c) Calculate the relaxation time for the electron in the Al, if the electrical conductivity of Al at room temperature is 3.65 x 107-¹m-1
(a) The density of conduction electrons in aluminum is 3.00 x 10²² electrons/m³,(b) The root mean square velocity of free electrons at room temperature is approximately 1.57 x 10⁶ m/s and (c) 9.26 x 10⁻¹⁵ s.
(a) The density of conduction electrons can be calculated using the formula:
Density of conduction electrons = (Number of free electrons per atom) * (Density of aluminum) / (Atomic mass of aluminum).
Plugging in the given values:
Density of conduction electrons = (3) * (2.70 x 10³ kg/m³) / (27.0 g/mol) = 3.00 x 10²² electrons/m³.
(b) The root mean square velocity of free electrons at room temperature can be calculated using the formula:
Root mean square velocity = √((3 * Boltzmann constant * Temperature) / (Mass of the electron)).
Substituting the values:
Root mean square velocity = √((3 * 1.38 x 10⁻²³ J/K * 298 K) / (9.11 x 10⁻³¹ kg)) ≈ 1.57 x 10⁶ m/s.
(c) The relaxation time for the electron can be calculated using the formula:
Relaxation time = (1 / (Electrical conductivity * Density of conduction electrons)).
Substituting the given values:
Relaxation time = (1 / (3.65 x 10⁷ Ω⁻¹m⁻¹ * 3.00 x 10²² electrons/m³)) ≈ 9.26 x 10⁻¹⁵ s.
Therefore, the density of conduction electrons in aluminum is 3.00 x 10²² electrons/m³, the root mean square velocity of free electrons at room temperature is approximately 1.57 x 10⁶ m/s, and the relaxation time for the electron in aluminum is approximately 9.26 x 10⁻¹⁵ s.
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A fishermen is standing nearly above a fish. The apparent depth
is 1.5m. What is the actual depth?
( Use snell's law, and law of refraction. )
The question asks for the actual depth of a fish when the apparent depth is given, and it suggests using Snell's law and the law of refraction to solve the problem.
Snell's law relates the angles of incidence and refraction of a light ray at the interface between two media with different refractive indices. In this scenario, the fisherman is observing the fish through the interface between air and water. The apparent depth is the perceived depth of the fish, and it is different from the actual depth due to the refraction of light at the air-water interface.
To find the actual depth, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the two media. By knowing the angle of incidence and the refractive indices of air and water, we can determine the angle of refraction and calculate the actual depth.
The law of refraction, also known as the law of Snellius, states that the ratio of the sines of the angles of incidence and refraction is equal to the reciprocal of the ratio of the refractive indices of the two media. By applying this law along with Snell's law, we can determine the actual depth of the fish based on the given apparent depth and the refractive indices of air and water.
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A 0.812-nm photon collides with a stationary electron. After the collision, the electron moves forward and the photon recoils backwards. (a) Find the momentum of the electron.
A 0.812-nm photon collides with a stationary electron. After the collision, the electron moves forward and the photon recoils backwards. (a)The momentum of the electron after the collision is approximately -8.193 × 10^-28 kg·m/s (taking into account the negative sign to indicate the opposite direction of motion compared to the photon)
To find the momentum of the electron after the collision, we can use the principle of conservation of momentum. In this case, we assume the system is isolated, and there are no external forces acting on it.
The momentum of a particle is given by the product of its mass and velocity:
Momentum = mass × velocity
However, for objects moving at speeds close to the speed of light, we need to consider relativistic effects. The relativistic momentum of an object is given by:
Momentum = (mass × velocity) / √(1 - (velocity^2 / c^2))
where c is the speed of light in a vacuum.
In this case, we're dealing with a photon and an electron. Photons have no rest mass, so their momentum is given by:
Photon Momentum = photon energy / c
Given that the photon has a wavelength of 0.812 nm, we can use the equation:
Photon Energy = (Planck's constant × speed of light) / wavelength
Let's calculate the momentum of the photon:
Photon Energy = (6.626 × 10^-34 J·s × 3 × 10^8 m/s) / (0.812 × 10^-9 m)
≈ 2.458 × 10^-19 J
Photon Momentum = (2.458 × 10^-19 J) / (3 × 10^8 m/s)
≈ 8.193 × 10^-28 kg·m/s
Now, let's consider the recoil of the electron. Since the photon recoils backwards, we assume the electron moves forward.
To find the momentum of the electron, we'll use the law of conservation of momentum:
Initial Momentum (before collision) = Final Momentum (after collision)
Since the electron is initially at rest, its initial momentum is zero. Therefore:
Final Momentum (electron) + Final Momentum (photon) = 0
Final Momentum (electron) = -Final Momentum (photon)
Final Momentum (electron) ≈ -8.193 × 10^-28 kg·m/s
The momentum of the electron after the collision is approximately -8.193 × 10^-28 kg·m/s (taking into account the negative sign to indicate the opposite direction of motion compared to the photon).
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A hollow cylinder with an inner radius of 4.0 mm and an outer radius of 24 mm conducts a 5.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 16 mm from its center ?
The magnitude of the magnetic field at a point 16 mm from the center of the hollow cylinder is 0.0625 T.
To calculate the magnitude of the magnetic field at a point 16 mm from the center of the hollow cylinder, we can use Ampere's law.
Ampere's law states that the magnetic field around a closed loop is directly proportional to the current passing through the loop.
The formula for the magnetic field produced by a current-carrying wire is:
B = (μ₀ * I) / (2π * r)
where B is the magnetic field, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), I is the current, and r is the distance from the center of the wire.
In this case, the current I is 5.0 A, and the distance r is 16 mm, which is equivalent to 0.016 m.
Plugging the values into the formula, we have:
B = (4π × 10^-7 T·m/A * 5.0 A) / (2π * 0.016 m)
B = (2 × 10^-6 T·m) / (0.032 m)
B = 0.0625 T
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what is solution ?
with steps
1- A ball is thrown vertically upward with a speed 18 m/s, Find: a. Find the time taken to reach 10m ? b. Find the speed at position 10m? c. Find the position of the ball after 2s?
The problem involves a ball being thrown vertically upward with an initial speed of 18 m/s. The task is to determine: a) the time taken to reach a height of 10m, b) the speed of the ball at a height of 10m, and c) the position of the ball after 2 seconds.
To solve this problem, we can use the equations of motion for vertical motion under constant acceleration. The key parameters involved are time, speed, and position.
a) To find the time taken to reach a height of 10m, we can use the equation: h = u*t + (1/2)*g*t^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. By substituting the given values, we can solve for t.
b) To find the speed of the ball at a height of 10m, we can use the equation: v = u + g*t, where v is the final velocity. We can substitute the known values of u, g, and the previously calculated value of t to find the speed.
c) To find the position of the ball after 2 seconds, we can again use the equation: h = u*t + (1/2)*g*t^2. By substituting the known values of u, g, and t = 2s, we can calculate the position of the ball after 2 seconds.
In summary, we can determine the time taken to reach 10m by solving an equation of motion, find the speed at 10m using another equation of motion, and calculate the position after 2 seconds using the same equation.
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When performing Young's double slit experiment, at what angle
(in degrees) is the first-order maximum for 638 nm wavelength light
falling on double slits if the separation distance is 0.0560
mm?
When performing Young's double slit experiment, at 6132.64 angle
(in degrees) is the first-order maximum for 638 nm wavelength light
falling on double slits if the separation distance is 0.0560
mm.
In Young's double-slit experiment, the angle for the first-order maximum can be determined using the formula:
θ = λ / (d * sin(θ))
Where:
θ is the angle for the first-order maximum,
λ is the wavelength of light,
d is the separation distance between the slits.
Given:
λ = 638 nm = 638 × 10^(-9) meters
d = 0.0560 mm = 0.0560 × 10^(-3) meters
Let's calculate the angle θ:
θ = (638 × 10^(-9)) / (0.0560 × 10^(-3) * sin(θ))
To solve this equation, we can make an initial guess for θ and then iteratively refine it using numerical methods. For a rough estimate, we can assume that the angle is small, which allows us to approximate sin(θ) ≈ θ (in radians). Therefore:
θ ≈ (638 × 10^(-9)) / (0.0560 × 10^(-3) * θ)
Simplifying the equation:
θ^2 ≈ (638 × 10^(-9)) / (0.0560 × 10^(-3))
θ^2 ≈ (638 / 0.0560) × (10^(-9) / 10^(-3))
θ^2 ≈ 11428.6
Taking the square root of both sides:
θ ≈ √11428.6
θ ≈ 106.97 radians (approximately)
To convert this angle from radians to degrees, we multiply by the conversion factor:
θ ≈ 106.97 * (180 / π)
θ ≈ 6132.64 degrees
Therefore, the approximate angle for the first-order maximum in Young's double-slit experiment with 638 nm wavelength light falling on double slits with a separation distance of 0.0560 mm is approximately 6132.64 degrees.
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A charge of +12 nC is placed on the x-axis at x = 4.4 m, and a charge of -25 nC is placed at x = -5.6 m. What is the magnitude of the electric field at the origin? Give your answer to a decimal place.
The magnitude of the electric field at the origin, rounded to one decimal place, is approximately 7.2 × 10⁶ N/C.
We need to calculate the electric field contribution from each charge and then add them together to find the net electric field at the origin.
Given:
Charge Q1 = +12 nC
Position x1 = 4.4 m
Charge Q2 = -25 nC
Position x2 = -5.6 m
Electrostatic constant k ≈ 8.99 × 10⁹ Nm²/C²
First, let's calculate the electric field contribution from Q1:
E1 = (8.99 × 10⁹ Nm²/C²) * (12 × 10⁻⁹ C) / (4.4 m)²
Substituting the values and performing the calculation:
E1 = 2.40 × 10⁶ N/C
Next, we calculate the electric field contribution from Q2:
E2 = (8.99 × 10⁹ Nm²/C²) * (-25 × 10⁻⁹) C) / (5.6 m)²
Substituting the values and performing the calculation:
E2 = -9.59 × 10⁶ N/C
Now, let's find the net electric field at the origin by summing the contributions:
E_net = E1 + E2
Substituting the values and performing the calculation:
E_net = (2.40 × 10⁶ N/C) + (-9.59 × 10⁶ N/C)
E_net = -7.19 × 10⁶ N/C
Finally, we take the magnitude of E_net to find the absolute value of the electric field at the origin:
|E_net| = |-7.19 × 10⁶ N/C|
|E_net| = 7.19 × 10⁶ N/C
After calculating the net electric field at the origin as -7.19 × 10⁶ N/C, rounding to one decimal place gives us:
|E_net| ≈ 7.2 × 10⁶ N/C
Therefore, the magnitude of the electric field at the origin, rounded to one decimal place, is approximately 7.2 × 10⁶ N/C.
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A standing wave on a 2-m stretched string is described by: y(x,t) = 0.1 sin(3x) cos(50rt), where x and y are in meters and t is in seconds. Determine the shortest distance between a node and an antinode
The shortest distance between a node and an antinode is π/3 meters.
In a standing wave, a node is a point where the amplitude of the wave is always zero, while an antinode is a point where the amplitude is maximum.
In the given equation, y(x,t) = 0.1 sin(3x) cos(50t), the node occurs when sin(3x) = 0, which happens when 3x = nπ, where n is an integer. This implies x = nπ/3.
The antinode occurs when cos(50t) = 1, which happens when 50t = 2nπ, where n is an integer. This implies t = nπ/25.
To find the shortest distance between a node and an antinode, we need to consider the difference in their positions. In this case, the difference in x-values is Δx = (n+1)π/3 - nπ/3 = π/3
Therefore, the shortest distance between a node and an antinode is π/3 meters.
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Consider a 0.100 g pin dropped from a height of 1.75 m onto a hard surface, where 0.050 % of its energy is converted into a pulse of sound that has a duration of 0.100 s. If you are in an environment where the intensity of the quietest audible sound is 5 x 10-6 W/m², how close do you need to be to the pin to hear it drop?
Summary:
To hear the sound of a 0.100 g pin dropped from a height of 1.75 m, we need to determine how close we need to the pin. Given that 0.050% of the pin's energy is converted into a sound pulse with a duration of 0.100 s, and the intensity of the quietest audible sound is 5 x 10^-6 W/m², we can calculate the required distance.
Explanation:
To find the distance at which we can hear the sound of the pin dropping, we can start by calculating the energy of the sound pulse. Since 0.050% of the pin's energy is converted into sound, we can determine the sound energy by multiplying 0.050% (0.0005) by the gravitational potential energy of the pin. The potential energy is given by mgh, where m is the mass of the pin (0.100 g) and h is the height (1.75 m). Converting the mass to kilograms and performing the calculation, we find that the sound energy is 1.715 x 10^-4 J.
Next, we can determine the power of the sound pulse by dividing the sound energy by the duration of the pulse. The power is given by P = E / t, where P is the power, E is the energy, and t is the duration of the sound pulse. Substituting the values, we get P = 1.715 x 10^-4 J / 0.100 s, which equals 1.715 x 10^-3 W.
Now, we can use the equation for sound intensity to calculate the required distance. The equation is I = P / A, where I is the sound intensity, P is the power, and A is the area through which the sound is spreading. Since we are given the sound intensity (5 x 10^-6 W/m²) and the power (1.715 x 10^-3 W), we can rearrange the equation to solve for A. Rearranging, we get A = P / I = 1.715 x 10^-3 W / 5 x 10^-6 W/m², which equals 3.43 x 10^2 m².
Since the area of a sphere is given by A = 4πr², where r is the radius, we can solve for r by rearranging the equation as r = √(A / (4π)). Substituting the value of A, we find that r is approximately 2.09 meters. Therefore, one needs to be about 2.09 meters away from the pin to hear the sound of it dropping, assuming no other factors affect the sound propagation.
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a
3.0 kg block is attached to spring. I supply 15J or energy to
stretch the spring. the block is then released and oscillating with
period or 0.40 s. what is the amplitude?
The amplitude of the oscillation is 0.35 meters.
When a block is attached to a spring and released, it undergoes oscillatory motion with a period of 0.40 seconds. To find the amplitude of this oscillation, we need to use the energy conservation principle and the formula for the period of oscillation.
Calculate the spring constant (k)
To find the amplitude, we first need to determine the spring constant. The energy supplied to stretch the spring can be written as:
E = (1/2)kx^2
where E is the energy, k is the spring constant, and x is the displacement from the equilibrium position. We know that the energy supplied is 15 J, and the block's mass is 3.0 kg. Rearranging the equation, we have:
k = (2E) / (m * x^2)
Substituting the given values, we get:
k = (2 * 15 J) / (3.0 kg * x^2)
k = 10 / x^2
Calculate the angular frequency (ω)
The period of oscillation (T) is given as 0.40 seconds. The period is related to the angular frequency (ω) by the equation:
T = 2π / ω
Rearranging the equation, we find:
ω = 2π / T
ω = 2π / 0.40 s
ω ≈ 15.7 rad/s
Calculate the amplitude (A)
The angular frequency is related to the spring constant (k) and the mass (m) by the equation:
ω = √(k / m)
Rearranging the equation to solve for the amplitude (A), we get:
A = √(E / k)
Substituting the given values, we have:
A = √(15 J / (10 / x^2))
A = √(15x^2 / 10)
A = √(3/2)x
Since we want the amplitude in meters, we can calculate it by substituting the given values:
A = √(3/2) * x
A ≈ √(3/2) * 0.35 m
A ≈ 0.35 m
Therefore, the amplitude of the oscillation is approximately 0.35 meters.
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Two firecrackers explode at the same place in a rest frame with a time separation of 11 s in that frame. Find the time between explosions according to classical physics, as measured in a frame moving with a speed 0.8 c with respect to the rest frame. Answer in units of s.
According to classical physics, the time between explosions measured in the frame moving with a speed of 0.8c is approximately 18.33 seconds.
To find the time between explosions according to classical physics, we can use the concept of time dilation. In special relativity, time dilation occurs when an observer measures a different time interval between two events due to relative motion.
The time dilation formula is given by:
Δt' = Δt / √[tex](1 - (v^2 / c^2))[/tex]
Where
Δt' is the time interval measured in the moving frame,
Δt is the time interval measured in the rest frame,
v is the relative velocity between the frames, and
c is the speed of light.
In this case, the time interval measured in the rest frame is 11 seconds (Δt = 11 s), and the relative velocity between the frames is 0.8c (v = 0.8c).
Plugging these values into the time dilation formula, we have:
Δt' = 11 / √[tex](1 - (0.8c)^2 / c^2)[/tex]
Δt' = 11 / √(1 - 0.64)
Δt' = 11 / √(0.36)
Δt' = 11 / 0.6
Δt' = 18.33 s
Therefore, according to classical physics, the time between explosions measured in the frame moving with a speed of 0.8c is approximately 18.33 seconds.
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Question 1) 2 charges are placed on the x-axis: a charge of +12.6nC at the origin and a charge of -31.3nC placed at x=24cm. What is the electric field vector on the y-axis at y=31cm?
To find the electric field vector on the y-axis at y = 31 cm due to the two charges, we can use the principle of superposition. The electric field at a point is the vector sum of the electric fields produced by each charge individually.
Given:
Charge q1 = +12.6 nC at the origin (x = 0)
Charge q2 = -31.3 nC at x = 24 cm = 0.24 m
Point of interest: y = 31 cm = 0.31 m
We can use Coulomb's law to calculate the electric field produced by each charge at the point of interest.
Electric field due to q1 (E1):
Using Coulomb's law, the electric field at point P due to charge q1 is given by:
[tex]E1 = k * (q1 / r1^2) * u[/tex], where k is the Coulomb's constant, r1 is the distance from q1 to P, and u is the unit vector pointing from q1 to P.
Since q1 is located at the origin, the distance r1 is the distance from the origin to P, which is equal to the y-coordinate of P.
r1 = y = 0.31 m
Plugging in the values:
E1 = [tex]k * (q1 / r1^2) * u1[/tex]
Electric field due to q2 (E2):
Similarly, the electric field at point P due to charge q2 is given by:
E2 = k * (q2 / r2^2) * u, where r2 is the distance from q2 to P, and u is the unit vector pointing from q2 to P.
The distance r2 is the horizontal distance from q2 to P, which is given by:
r2 = x2 - xP
= 0.24 m - 0
= 0.24 m
Plugging in the values:
E2 =[tex]k * (q2 / r2^2) * u2[/tex]
Total Electric Field (E):
The total electric field at point P is the vector sum of E1 and E2:
E = E1 + E2
Calculating the magnitudes and directions:
1. Calculate E1:
E1 = k * [tex](q1 / r1^2) * u1[/tex]
2. Calculate E2:
E2 = k [tex]* (q2 / r2^2) * u2[/tex]
3. Calculate E:
E = E1 + E2
Remember to include the appropriate signs and directions for the electric field vectors based on the signs and electric of the .
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Suppose a 58.0-kg gymnast climbs a rope. What is the tension in
the rope if he
accelerates upward at a rate of 2.37 m/s^2?
The numerical value of the tension in newtons (N).58.0 kg * 2.37 m/s²) + (58.0 kg * 9.8 m/s²)
To determine the tension in the rope, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
The gymnast's mass is given as 58.0 kg, and the acceleration upward is 2.37 m/s². We need to find the tension in the rope.
Considering the forces acting on the gymnast, we have two forces: the tension force in the rope pulling upward and the force of gravity pulling downward. These two forces will be equal in magnitude but opposite in direction to maintain equilibrium.
The net force can be expressed as:
Net force = Tension - Weight
where Weight = mass * gravity, and gravity is approximately 9.8 m/s².
Using the given values, the weight can be calculated as:
Weight = 58.0 kg * 9.8 m/s²
Next, we can set up the equation:
Net force = Tension - Weight = mass * acceleration
Substituting the values, we have:
Tension - (58.0 kg * 9.8 m/s²) = 58.0 kg * 2.37 m/s²
Now, we can solve for the tension:
Tension = (58.0 kg * 2.37 m/s²) + (58.0 kg * 9.8 m/s²)
Calculate the numerical value of the tension in newtons (N).
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A physical pendulum consists of a uniform rod that can swing freely from one end, with a small, heavy bob attached to the other end. If the length of the rod is 2.55 m, and the mass of the bob and the rod are both 1.4 kg, what is the period of this pendulum?
The period of the physical pendulum with a uniform rod of length 2.55 m and a mass of 1.4 kg for both the bob and the rod is approximately 3.35 seconds.
The period of a physical pendulum depends on the length of the pendulum and the acceleration due to gravity. The formula to calculate the period of a physical pendulum is:
T = 2π√(I / (mgh))
Where T is the period, I is the moment of inertia of the pendulum, m is the mass of the pendulum, g is the acceleration due to gravity, and h is the distance between the center of mass of the pendulum and the pivot point.
For a uniform rod rotating about one end, the moment of inertia is given by:
I = (1/3) * m * L²
Where L is the length of the rod.
Plugging in the given values, we have:
I = (1/3) * 1.4 kg * (2.55 m)² = 2.45 kg·m²
Substituting this value and the known values of m = 1.4 kg, g = 9.8 m/s², and h = L/2 = 1.275 m into the period formula, we get:
T = 2π√(2.45 kg·m²/ (1.4 kg * 9.8 m/s² * 1.275 m)) ≈ 3.35 s
Therefore, the period of this physical pendulum is approximately 3.35 seconds.
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Q|C A Carnot heat engine operates between temperatures Th and Tc . (d) Does the answer to part (c) depend on Tc ? Explain.
Yes, the answer to part (c) does depend on Tc. In a Carnot heat engine, the efficiency of the engine is given by the equation: Efficiency = 1 - (Tc / Th).
Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. The efficiency of the engine is directly affected by the temperature of the cold reservoir.
As Tc increases, the efficiency of the engine decreases. Therefore, the answer to part (c) does depend on Tc. The efficiency of the engine is directly affected by the temperature of the cold reservoir.
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A standing wave on a 2-m stretched string is described by: y(x,t) = 0.1 sin(3πx) cos(50πt), where x and y are in meters and t is in seconds. Determine the shortest distance between a node and an antinode.
D = 25 cm
D = 12.5 cm
D = 16.67 cm
D = 50 cm
D = 33.34 cm
A standing wave on a 2-m stretched string is described by y(x,t) = 0.1 sin(3πx) cos(50πt), where x and y are in meters and t is in seconds.The shortest distance between a node and an anti node is 100 cm, or 1 m.So option 2 is correct.
The distance between a node and an anti node in a standing wave is equal to half of the wavelength of the wave.
The wavelength of a wave can be calculated using the following formula:wavelength = v / f
where:
v ,is the speed of the wave.
f, is the frequency of the wave.
In this case, the speed of the wave is equal to the speed of sound in a stretched string, which is about 200 m/s. The frequency of the wave is equal to the reciprocal of the period of the wave, which is equal to 1/50 s.
wavelength = v / f
= 200 m/s / (1/50 s)
= 1000 m / 50
= 20 m
The shortest distance between a node and an antinode is therefore equal to half of the wavelength, which is equal to:
distance = wavelength / 2
= 20 m / 2
= 10 m
= 1000 cm / 10
= 100 cm
Since the string is 2 m long, there are 2 nodes and 2 antipodes on the string. The shortest distance between a node and an antinode is therefore 100 cm, or 1 m.
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In a physics laboratory experiment, a coil with 150 turns enclosing an area of 12 cm2 is rotated in a time interval of 0.050 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is 6.0×10−5 T.
A. What is the magnetic flux through each turn of the coil before it is rotated?
B. What is the magnetic flux through each turn of the coil after it is rotated?
C.What is the average emf induced in the coil?
The magnetic flux through each turn of the coil before it is rotated is 7.2 × 10⁻⁹ Wb. The magnetic flux through each turn of the coil after it is rotated is 7.2 × 10⁻⁹ Wb. The average emf induced in the coil is zero.
Area of the coil, A = 12 cm²Number of turns, N = 150Magnetic field, B = 6.0×10−5 T Time interval, t = 0.050 sThe induced emf can be calculated using Faraday’s law. According to Faraday’s law,The induced emf is given as,ε = -NdΦ/dtWhere N is the number of turns in the coil, dΦ/dt is the time rate of change of the magnetic flux through a single turn of the coil.
A. Before rotation, the plane of the coil is perpendicular to the magnetic field.The magnetic flux through each turn of the coil before it is rotated is,Φ = BA = (6.0 × 10⁻⁵ T) × (12 × 10⁻⁴ m²) = 7.2 × 10⁻⁹ WbThe magnetic flux through each turn of the coil before it is rotated is 7.2 × 10⁻⁹ Wb.
B. After rotation, the plane of the coil is parallel to the magnetic field.The magnetic flux through each turn of the coil after it is rotated is,Φ = BA = (6.0 × 10⁻⁵ T) × (12 × 10⁻⁴ m²) = 7.2 × 10⁻⁹ Wb.The magnetic flux through each turn of the coil after it is rotated is 7.2 × 10⁻⁹ Wb.
C. The change in flux is,ΔΦ = Φf - ΦiΔΦ = (7.2 × 10⁻⁹) - (7.2 × 10⁻⁹) = 0Since the time interval of rotation is very small, the average emf induced in the coil is equal to the instantaneous emf at the midpoint of the time interval.The average emf induced in the coil is,ε = -NdΦ/dtε = -150 × (0)/0.050ε = 0. The average emf induced in the coil is zero.
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A 21 N Tension force is applied to a 120 N crate at a 20 degree angle relative to the horizon causing it to move with a constant speed across the horizontal surface. What is the coefficient of
friction between the crate and the surface?
The coefficient of friction between the crate and the surface is 0.17.
Since the crate is moving with a constant speed, the net force acting on it must be zero.
In other words, the force of friction must be equal and opposite to the tension force applied.
The force of friction can be calculated using the following formula:
frictional force = coefficient of friction * normal force
where the normal force is the force perpendicular to the surface and is equal to the weight of the crate, which is given as 120 N.
In the vertical direction, the tension force is balanced by the weight of the crate, so there is no net force.
In the horizontal direction, the tension force is resolved into two components:
21 N * cos(20°) = 19.8 N acting parallel to the surface and
21 N * sin(20°) = 7.2 N acting perpendicular to the surface.
The frictional force must be equal and opposite to the parallel component of the tension force, so we have:
frictional force = 19.8 N
The coefficient of friction can now be calculated
:coefficient of friction = frictional force / normal force
= 19.8 N / 120 N
= 0.165 or 0.17 (rounded to two significant figures)
Therefore, the coefficient of friction between the crate and the surface is 0.17.
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QUESTION 6 Find REQ of the following: with R₁ = R2 = R3 = 8 ohms, R4 = 2 ohms, R5 = 10 ohms and Rg = 12 ohms. Find REQ. R₁ R4 1 wwwww R₂ w R3 00 PAGE R6 un ERG
Answer:
The equivalent resistance (REQ) of the given circuit is 14 ohms.
Explanation:
To find the equivalent resistance (REQ) in the given circuit, we can start by simplifying the circuit step by step.
First, let's simplify the series combination of R₁ and R₄:
R₁ and R₄ are in series, so we can add their resistances:
R₁ + R₄ = 8 ohms + 2 ohms = 10 ohms
The simplified circuit becomes:
R₁ R₄
1 w
10Ω
Next, let's simplify the parallel combination of R₂ and R₃:
R₂ and R₃ are in parallel, so we can use the formula for calculating the equivalent resistance of two resistors in parallel:
1/REQ = 1/R₂ + 1/R₃
Substituting the values:
1/REQ = 1/8 ohms + 1/8 ohms = 1/8 + 1/8 = 2/8 = 1/4
Taking the reciprocal on both sides:
REQ = 4 ohms
The simplified circuit becomes:
R₁ R₄
1 w
10Ω
REQ
4Ω
Now, let's simplify the series combination of R₅ and REQ:
R₅ and REQ are in series, so we can add their resistances:
R₅ + REQ = 10 ohms + 4 ohms = 14 ohms
The final simplified circuit becomes:
R₁ R₄
1 w
10Ω
REQ
4Ω
R₅
10Ω
14Ω
Therefore, the equivalent resistance (REQ) of the given circuit is 14 ohms.
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. The hottest place on the Earth is Al-'Aziziyah, Libya, where the temperature has soared to 136.4 ∘ F. The coldest place is Vostok, Antarctica, where the temperature has plunged to −126.9 ∘ F. Express these temperatures in degrees Celsius and in Kelvins.
Here are the temperatures in degrees Celsius and Kelvins
Temperature | Degrees Fahrenheit | Degrees Celsius | Kelvins
Al-'Aziziyah, Libya | 136.4 | 58.0 | 331.15
Vostok, Antarctica | −126.9 | −88.28 | 184.87
To convert from degrees Fahrenheit to degrees Celsius, you can use the following formula:
°C = (°F − 32) × 5/9
To convert from degrees Celsius to Kelvins, you can use the following formula:
K = °C + 273.15
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quick answer
please
QUESTION 10 4 In a mass spectrometer, a charged particle enters a uniform magnetic field that is perpendicular to the velocity of the particle itself. The subsequent motion of the particle is a circul
In a mass spectrometer, the radius of the circular path decreases if the magnetic field strength decreases.
The correct answer is d. The value of the magnetic field strength decreases.
In a mass spectrometer, the radius of the circular path followed by a charged particle is directly proportional to the momentum of the particle and inversely proportional to the product of the charge and the magnetic field strength. Mathematically, the radius (r) is given by:
r = (p) / (qB),
where p is the momentum of the particle, q is the charge of the particle, and B is the magnetic field strength.
If the magnetic field strength decreases (option d), while the other factors remain constant, the radius of the circular path will decrease. This is because a weaker magnetic field will exert less force on the charged particle, resulting in a tighter and smaller circular path.
The other options (a, b, c, e) do not directly affect the radius of the circular path in a mass spectrometer.
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The complete question is:
In a mass spectrometer, a charge particle enters a uniform magnetic field that is perpendicular to theparticle itself. The subsequent motion of the particle is a circular path. The radius of the circular path will decrease if __________ .
a. the value of the speed increases.
b. the sign of the charge is flipped.
c. the value of the charge increases.
d. the value of the magnetic field strength decreases.
e. the value of the mass increases.
A photon of wavelength 1.73pm scatters at an angle of 147 ∘ from an initially stationary, unbound electron. What is the de Broglie wavelength of the electron after the photon has been scattered?
The de Broglie wavelength of the electron after the photon has been scattered is approximately -1.12 picometers (-1.12 pm).
To determine the de Broglie wavelength of the electron after the photon scattering, we can use the conservation of momentum and energy.
Given:
Wavelength of the photon before scattering (λ_initial) = 1.73 pm
Scattering angle (θ) = 147°
The de Broglie wavelength of a particle is given by the formula:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant, and p is the momentum of the particle.
Before scattering, both the photon and the electron have momentum. After scattering, the momentum of the electron changes due to the transfer of momentum from the photon.
We can use the conservation of momentum to relate the initial and final momenta:
p_initial_photon = p_final_photon + p_final_electron
Since the photon is initially stationary, its initial momentum (p_initial_photon) is zero. Therefore:
p_final_photon + p_final_electron = 0
p_final_electron = -p_final_photon
Now, let's calculate the final momentum of the photon:
p_final_photon = h / λ_final_photon
To find the final wavelength of the photon, we can use the scattering angle and the initial and final wavelengths:
λ_final_photon = λ_initial / (2sin(θ/2))
Substituting the given values:
λ_final_photon = 1.73 pm / (2sin(147°/2))
Using the sine function on a calculator:
sin(147°/2) ≈ 0.773
λ_final_photon = 1.73 pm / (2 * 0.773)
Calculating the value:
λ_final_photon ≈ 1.73 pm / 1.546 ≈ 1.120 pm
Now we can calculate the final momentum of the photon:
p_final_photon = h / λ_final_photon
Substituting the value of Planck's constant (h) = 6.626 x 10^-34 J·s and converting the wavelength to meters:
λ_final_photon = 1.120 pm = 1.120 x 10^-12 m
p_final_photon = (6.626 x 10^-34 J·s) / (1.120 x 10^-12 m)
Calculating the value:
p_final_photon ≈ 5.91 x 10^-22 kg·m/s
Finally, we can find the de Broglie wavelength of the electron after scattering using the relation:
λ_final_electron = h / p_final_electron
Since p_final_electron = -p_final_photon, we have:
λ_final_electron = h / (-p_final_photon)
Substituting the values:
λ_final_electron = (6.626 x 10^-34 J·s) / (-5.91 x 10^-22 kg·m/s)
Calculating the value:
λ_final_electron ≈ -1.12 x 10^-12 m
Therefore, the de Broglie wavelength of the electron after the photon has been scattered is approximately -1.12 picometers (-1.12 pm).
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