Given that the Apollo 12 Part A lunar lander had a mass of 1.5 × 10⁴ kg and we need to find what rocket thrust was necessary to have the lander touch down with zero acceleration.
Formula: The thrust equation is given by;
`T = (m*g) + (m*a)`
where, T = rocket thrust m = mass of the lander g = acceleration due to gravity a = acceleration Since we know the mass of the lander, and the acceleration due to gravity, all we need to do is set the net force equal to zero to find the required rocket thrust.
Then, we can solve for the acceleration (a) as follows:
Mass of the lander,
m = 1.5 × 10⁴ kg Acceleration due to gravity,
g = 9.81 m/s²Acceleration of lander, a = 0 (since it touches down with zero acceleration)
Rocket thrust,
T = ?
Using the thrust equation,
T = (m * g) + (m * a)T = m(g + a)T = m(g + 0) [because the lander touches down with zero acceleration]
T = m * gT = 1.5 × 10⁴ kg × 9.81 m/s² = 1.47135 × 10⁵ N Therefore,
the rocket thrust was 1.47135 × 10⁵ N (Newtons).
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maximum. The slit-screen distance is L=91.2 cm, and a=0.600 mm. What is the wavelength (in nm ) of the incident light? nm
The wavelength of the incident light is 152 nm.
When the intensity pattern is measured by a diffraction pattern created by a double-slit, the maximum intensity is obtained by the center of the pattern. The slit-screen distance L=91.2 cm and a=0.600 mm.
What is the wavelength (in nm ) of the incident light?
The formula to calculate wavelength λ of the incident light is given as
:λ = xL/a
Where, x = 1,
for the maximum So, putting the values in the above formula,
we get: λ = xL/aλ
= (1 × 91.2)/0.600
=152
The wavelength of the incident light is 152 nm.
To calculate the wavelength of incident light, λ using double slit experiment. It is given that the maximum intensity is obtained by the center of the pattern, thus according to the formula derived by Young for the maxima and minima is:
dsinθ = mλ
where, d is the distance between the slits, θ is the angle of diffraction, m is the order of diffraction.
By putting the values in the above formula, we get:
mλ = d sin θ
Where, m = 1λ
= d sin θ
The distance between the slits is not given in the question. Hence, we will use another formula,λ = xL/a
Where, x = 1, for the maximum
λ = xL/aλ
= (1 × 91.2)/0.600
=152
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A 0.210-kg wooden rod is 1.10 m long and pivots at one end. It
is held horizontally and then released. What is the linear
acceleration of a spot on the rod that is 0.704 m from the axis of
rotation?
The linear acceleration of the spot on the rod that is 0.704 m from the axis of rotation is 49.919 m/s².
The given values are Mass of the rod = 0.210 kgLength of the rod = 1.10 m
Distance of the spot from the axis of rotation = 0.704 m
The rod is released horizontally.
This means that the rotation of the rod will be around an axis perpendicular to the rod.
Moment of inertia of a rod about an axis perpendicular to its length is given by the formula,
I=1/12ml²I = Moment of inertia of the rodm = Mass of the rodl = Length of the rod
Substitute the values in the formula and find I.I = 1/12 × 0.210 kg × (1.10 m)²= 0.0205 kg m²
Linear acceleration of a spot on the rod, a is given by the formula:
a = αrwhereα = angular acceleration of the rodr = Distance of the spot from the axis of rotation
Angular acceleration of the rod is given by the formula,τ = Iατ = τorque on the rodr = Distance of the spot from the axis of rotation
Substitute the values in the formula and find α.τ = Iαα = τ/I
The torque on the rod is due to its weight. Weight of the rod, W = mgW = 0.210 kg × 9.8 m/s² = 2.058 N
The torque on the rod is due to the weight of the rod.
It can be found as,τ = W × rτ = 2.058 N × 0.704 mτ = 1.450 Nm
Substitute the values in the formula and find α.α = τ/Iα = 1.450 Nm / 0.0205 kg m²α = 70.732 rad/s²
Substitute the values in the formula and find a.a = αr = 70.732 rad/s² × 0.704 m = 49.919 m/s²
Therefore, the linear acceleration of the spot on the rod that is 0.704 m from the axis of rotation is 49.919 m/s².
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In order for any object to be moving in a circular path at constant speed, the centripetal and centrifugal forces acting on the object must cancel out. there must be a centrifugal force acting on the
For an object to move in a circular path at a constant speed, the centripetal force and the centrifugal force acting on the object must cancel each other out.
To understand this concept, let's break it down step by step:
Circular motion: When an object moves in a circular path, it experiences a force called the centripetal force. This force is always directed towards the center of the circle and acts as a "pull" or inward force.
Centripetal force: The centripetal force is responsible for keeping the object moving in a curved path instead of a straight line. It ensures that the object continuously changes its direction, creating circular motion. Examples of centripetal forces include tension in a string, gravitational force, or friction.
Constant speed: The question mentions that the object is moving at a constant speed. This means that the magnitude of the object's velocity remains the same throughout its circular path. However, the direction of the velocity is constantly changing due to the centripetal force.
Centrifugal force: Now, the concept of centrifugal force comes into play. In reality, there is no actual centrifugal force acting on the object. Instead, centrifugal force is a pseudo-force, which means it is a perceived force due to the object's inertia trying to move in a straight line.
Inertia and centrifugal force: The centrifugal force appears to act outward, away from the center of the circle, in the opposite direction to the centripetal force. This apparent force arises because the object's inertia wants to keep it moving in a straight line tangent to the circle.
Canceling out forces: In order for the object to move in a circular path at a constant speed, the centripetal force must be equal in magnitude and opposite in direction to the centrifugal force. By canceling each other out, these forces maintain the object's motion in a circular path.
To summarize, while the centripetal force is a real force that acts inward, the centrifugal force is a perceived force due to the object's inertia. For circular motion at a constant speed, the centripetal and centrifugal forces appear to cancel each other out, allowing the object to maintain its circular path.
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Please Explain itThe current in an LC circuit with capacitance C0 and inductance L0 obeys the following equation.
Determine the energy in the circuit.
i = I0 sin(at + φ)
Answer: Using maximum current we get
E = L 0 I02 /2
The equation given represents the current in an LC (inductor-capacitor) circuit with capacitance C0 and inductance L0. To determine the energy in the circuit, we use the equation E = (L0 * I0^2) / 2, where I0 represents the maximum current in the circuit.
The equation i = I0 * sin(at + φ) represents the current in an LC circuit, where I0 is the maximum current, a is the angular frequency, t is time, and φ is the phase angle. This equation describes the sinusoidal nature of the current in the circuit.
To calculate the energy in the circuit, we can use the formula E = (L0 * I0^2) / 2, where E represents the energy stored in the circuit, and L0 is the inductance of the circuit.
In this case, since the equation provided gives us the maximum current (I0), we can directly substitute this value into the energy equation. Thus, the energy in the circuit is given by E = (L0 * I0^2) / 2.
The formula represents the energy stored in the magnetic field of the inductor and the electric field of the capacitor in the LC circuit. It is derived from the equations governing the energy stored in inductors and capacitors separately.
By calculating the energy in the circuit using this equation, we can evaluate and quantify the amount of energy present in the LC circuit, which is crucial for understanding and analyzing its behavior and characteristics.
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An Inductor (solenoid) with length 6m, diameter 15cm and 10 number of turns, Nis connected to an AC power source. (3 marks) Calculate the Inductance.
The inductance of the solenoid is approximately 7.35 × 10^-5 H.
To calculate the inductance of the solenoid, we can use the formula:
L = (μ₀ * N² * A) / l
where L is the inductance, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
Number of turns (N) = 10
Length of the solenoid (l) = 6 m
Diameter of the solenoid (d) = 15 cm
First, we need to calculate the cross-sectional area (A) of the solenoid using the diameter:
Radius (r) = d / 2 = 15 cm / 2 = 7.5 cm = 0.075 m
A = π * r² = π * (0.075 m)² ≈ 0.01767 m²
Now, we can calculate the inductance (L) using the formula:
[tex]L = (μ₀ * N² * A) / lμ₀ = 4π × 10^-7 T·m/A\\\\L = (4π × 10^-7 T·m/A) * (10²) * (0.01767 m²) / 6 m\\\\L = (4 * 3.1416 * 10^-7 * 10² * 0.01767) / 6L ≈ 7.35 × 10^-5 H[/tex]
Therefore, the inductance of the solenoid is approximately 7.35 × 10^-5 H.
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Consider two equal point charges separated by a distance d. At what point (other than infinity) would a third test charge experience no net force?
A third test charge placed at the midpoint between two equal point charges separated by a distance d would experience no net force.
When two equal point charges are separated by a distance d, they create an electric field in the space around them. The electric field lines extend radially outward from one charge and radially inward toward the other charge. These electric fields exert forces on any other charges present in their vicinity.
To find the point where a third test charge would experience no net force, we need to locate the point where the electric fields from the two charges cancel each other out. This occurs at the midpoint between the two charges.
At the midpoint, the electric field vectors due to the two charges have equal magnitudes but opposite directions. As a result, the forces exerted by the electric fields on the third test charge cancel each other out, resulting in no net force.
Therefore, the point at the midpoint between the two equal point charges is where a third test charge would experience no net force.
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A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 722 nm and its wavelength in the glass is 543 nm. If the ray in water makes an angle of 45.0 ∘
with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?
The refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
To determine the angle of the refracted ray in the glass, we can use Snell's Law, which relates the angles and indices of refraction of light as it passes through different mediums. Snell's Law states that the ratio of the sines of the angles of incidence (θ₁) and refraction (θ₂) is equal to the ratio of the indices of refraction (n₁ and n₂) of the two mediums.
In this case, the incident angle in water (θ₁) is given as 45.0°, the wavelength of light in water (λ₁) is 722 nm, and the wavelength of light in glass (λ₂) is 543 nm.
We know that the index of refraction (n) of a medium is inversely proportional to the wavelength of light passing through it, so we can use the ratio of the wavelengths to calculate the ratio of the indices of refraction:
n₁ / n₂ = λ₂ / λ₁
Substituting the given values, we have:
n₁ / n₂ = 543 nm / 722 nm
To simplify the calculation, we can convert the wavelengths to meters:
n₁ / n₂ = (543 nm / 1) / (722 nm / 1) = 0.751
Now, we can apply Snell's Law:
sin(θ₁) / sin(θ₂) = n₂ / n₁
sin(θ₂) = (n₁ / n₂) * sin(θ₁)
Plugging in the values, we get:
sin(θ₂) = 0.751 * sin(45.0°)
To find the angle θ₂, we can take the inverse sine (or arcsine) of both sides:
θ₂ = arcsin(0.751 * sin(45.0°))
Evaluating this expression, we find:
θ₂ ≈ 48.4°
Therefore, the refracted ray in the glass makes an angle of approximately 48.4° with respect to the normal.
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beginning with h=4.136x10-15 eV.s and c = 2.998x108 m/s , show that hc =1240 eV-nm.
Beginning with h=4.136x10-15 eV.s and c = 2.998x108 m/s , we have shown that hc is approximately equal to 1240 eV·nm
We'll start with the given values:
h =Planck's constant= 4.136 x 10^(-15) eV·s
c = speed of light= 2.998 x 10^8 m/s
We want to show that hc = 1240 eV·nm.
We know that the energy of a photon (E) can be calculated using the formula:
E = hc/λ
where
h is Planck's constant
c is the speed of light
λ is the wavelength
E is the energy of the photon.
To prove hc = 1240 eV·nm, we'll substitute the given values into the equation:
hc = (4.136 x 10^(-15) eV·s) ×(2.998 x 10^8 m/s)
Let's multiply these values:
hc ≈ 1.241 x 10^(-6) eV·m
Now, we want to convert this value from eV·m to eV·nm. Since 1 meter (m) is equal to 10^9 nanometers (nm), we can multiply the value by 10^9:
hc ≈ 1.241 x 10^(-6) eV·m × (10^9 nm/1 m)
hc ≈ 1.241 x 10^3 eV·nm
Therefore, we have shown that hc is approximately equal to 1240 eV·nm
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A 50 kg student bounces up from a trampoline with a speed of 3.4 m/s. Determine the work done on the student by the force of gravity when she is 5.3 m above the trampoline.
The work done on the student by the force of gravity when she is 5.3 m above the trampoline is approximately 2574 Joules.
To determine the work done on the student by the force of gravity, we need to calculate the change in potential-energy. The gravitational potential energy (PE) of an object near the surface of the Earth is given by the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity, and h is the height above the reference level.
In this case, the student's mass is 50 kg and the height above the trampoline is 5.3 m. We can calculate the initial potential energy (PEi) when the student is on the trampoline and the final potential energy (PEf) when the student is 5.3 m above the trampoline.
PEi = m * g * h_initial
PEf = m * g * h_final
The work done by the force of gravity is the change in potential energy, which can be calculated as:
Work = PEf - PEi
Let's calculate the work done on the student by the force of gravity:
PEi = 50 kg * 9.8 m/s² * 0 m (height on the trampoline)
PEf = 50 kg * 9.8 m/s² * 5.3 m (height 5.3 m above the trampoline)
PEi = 0 J
PEf = 50 kg * 9.8 m/s² * 5.3 m
PEf ≈ 2574 J
Work = PEf - PEi
Work ≈ 2574 J - 0 J
Work ≈ 2574 J
Therefore, the work done on the student by the force of gravity when she is 5.3 m above the trampoline is approximately 2574 Joules.
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You are measuring the bulk air temperature in a closed-loop benchtop wind tunnel. You take five readings of the temperature and determine the average temperature is 77°C with a standard deviation of 4°C. You report the following information: T = 77°C ± 1.8°C
(68% confidence level) You decide that you want to improve the confidence level of your data set to 95%, keeping the same standard deviation of 4°C with an average temperature of 77°C. (a) What are your new temperature limits with a sample size of N = 10. (3) (b) Compare your answer to the 68% confidence level. What is the AT between the two limits? Explain your answer. (6) (c) Compute the mean temperature's precision limits if you increase your confidence level to 99.7% and keep all other parameters the same. (3) (d) If you improve your measurement technique and reduce the standard deviation by 2°C, how will your precision change? Explain. You can use any confidence level to explain/prove your answer. (3)
(a) At the 95% confidence level, the new temperature limits with a sample size of N = 10 are as follows:Lower temperature limit= 77 °C - 2.31 x (4°C / sqrt(10))= 74.08 °C
Upper temperature limit= 77 °C + 2.31 x (4°C / (10))= 79.92 °C
Thus, the new temperature limits are 74.08°C and 79.92°C, respectively.(b) The new temperature limits with a 95% confidence level are wider than the limits with a 68% confidence level.
The AT is the difference between the upper and lower limits. Therefore, the AT is increased as the confidence level increases. The AT at the 68% confidence level is less than the AT at the 95% confidence level because of the wider temperature range at the 95% confidence level. (c) Precision limits are determined using the same formula as temperature limits.
The formula for computing precision limits is as follows:Lower precision limit = Mean temperature - Z x (Standard deviation / sqrt(N))Upper precision limit = Mean temperature + Z x (Standard deviation / (N))
(d) Reducing the standard deviation will increase the precision of the temperature measurement. The precision limits are calculated using the formula
:Lower precision limit = Mean temperature - Z x (Standard deviation / sqrt(N))Upper precision limit = Mean temperature + Z x (Standard deviation / (N))
As a result, reducing the standard deviation of the temperature measurement will decrease the precision limits, making the temperature range smaller and allowing for a more accurate measurement.
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A spinning wheel is suspended from a string and rotates as shown below. As the time goes by, what is the direction in which the angular momentum will change (Hinttime derivative of L) N A w O positi
The direction in which the angular momentum will change is O positive (clockwise).
Angular momentum is a quantity that expresses the rotational momentum of a system. It is proportional to the moment of inertia and angular velocity of a body. L is the symbol for angular momentum, and its formula is:L = Iω, where I is the moment of inertia and ω, is the angular velocity. In this case, a spinning wheel is suspended from a string and rotates as shown below. The direction in which the angular momentum will change is given by the time derivative of L (dL/dt), which is known as the rate of change of angular momentum.dL/dt = I(dω/dt). By applying Newton's second law of motion, we can say that the rate of change of angular momentum is equal to the torque acting on the system: dL/dt = τwhere τ is the torque acting on the system. According to the right-hand rule, the direction of torque acting on the system is perpendicular to the plane of rotation and perpendicular to the force acting on it. Therefore, in this case, the direction of torque acting on the system will be perpendicular to the plane of rotation and directed into the page (towards the observer). Thus, the direction in which the angular momentum will change is O positive (clockwise)
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The pre-exponential and activation energy for the diffusion of chromium in nickel are 1.1 x 10-4 m²/s and 272,000 J/mol, respectively. At what temperature will the diffusion coefficient have a value of 1.2 x 10-14 m²/s? Give your answer in Kelvin.
The temperature at which the diffusion coefficient will have a value of 1.2 x 10^-14 m²/s is 943.16 K given the pre-exponential and activation energy for the diffusion of chromium in nickel are 1.1 x 10^-4 m²/s and 272,000 J/mol, respectively.
The Arrhenius equation relates the rate constant (or diffusion coefficient) to the activation energy and the temperature. The Arrhenius equation is given as k = Ae^(-Ea/RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the temperature. Rearranging this equation, we have log k = log A - (Ea/2.303RT).
This equation suggests that a plot of log k versus (1/T) will give a straight line with slope = -Ea/2.303R and y-intercept = log A. We can use this to find the temperature at which the diffusion coefficient will have a value of 1.2 x 10^-14 m²/s. For this, we need to calculate the value of log k for the given diffusion coefficient and then use it to find the temperature. Log k = log 1.2 x 10^-14 = -32.92
Substituting the values of A and Ea into the equation, we get-32.92 = log 1.1 x 10^-4 - (272,000/2.303RT)
Solving this equation for T gives T = 943.16 K
Therefore, the temperature at which the diffusion coefficient will have a value of 1.2 x 10^-14 m²/s is 943.16 K.
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The principal component of natural gas is methane
(CH4). How many moles of CH4 are present in
131.96 g of methane? (Molar mass of carbon = 12.011 g/mol and molar
mass of hydrogen = 1.0080 g/mol (refer
There are 4.705 moles of CH₄ present in 131.96 g of methane.
The molar mass of CH₄ can be calculated as:
Molar mass of CH₄ = (4 × Molar mass of hydrogen) + Molar mass of carbon
Molar mass of CH₄ = (4 × 1.0080) + 12.011
Molar mass of CH₄ = 16.043 + 12.011
Molar mass of CH₄ = 28.054 g/mol
Number of moles = Mass of substance / Molar mass
Number of moles of CH₄ = 131.96 / 28.054
Number of moles of CH₄ = 4.705 moles
Therefore, there are 4.705 moles of CH₄ present in 131.96 g of methane.
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A solar water heater system is to be installed in a single-family dwelling to provide 300 L (300 kg) of 55°C hot water per day (normal use for a family of four). Estimate the required surface area of the collecting panels, assuming that water is supplied to the house at 15°C, that the average power per unit area from the sun is 130 W/m2 (typical of much of the central United States), and that the collecting panels are 60% efficient.
To provide 300 L (300 kg) of hot water at 55°C per day for a family of four, the solar water heater system requires an estimated surface area of collecting panels. [tex]A = (300 kg × 4186 J/kg·°C × (55°C - 15°C)) / (130 W/m² × 0.60)[/tex]
Assuming an average power per unit area from the sun of 130 W/m² and a panel efficiency of 60%, the required surface area can be calculated based on the energy needed to heat the water.
By considering the temperature difference between the initial water temperature (15°C) and the desired hot water temperature (55°C), along with the specific heat capacity of water, the required surface area can be determined.
The energy needed to heat the water can be calculated using the equation:
Energy = mass × specific heat capacity × temperature difference
For heating 300 kg of water from 15°C to 55°C, and considering the specific heat capacity of water (approximately 4186 J/kg·°C), the energy needed is:
Energy = [tex]300 kg × 4186 J/kg·°C × (55°C - 15°C)[/tex]
To estimate the energy provided by the solar panels, we multiply the average power per unit area from the sun (130 W/m²) by the collecting panel efficiency (60%), and then by the surface area of the panels (A):
Energy provided = [tex]130 W/m² × 0.60 × A[/tex]
Setting the energy needed equal to the energy provided, we can solve for the required surface area:
[tex]300 kg × 4186 J/kg·°C × (55°C - 15°C) = 130 W/m² × 0.60 × A[/tex]
Simplifying the equation, we can calculate the required surface area:
[tex]A = (300 kg × 4186 J/kg·°C × (55°C - 15°C)) / (130 W/m² × 0.60)[/tex]
Therefore, the required surface area of the collecting panels can be estimated by evaluating the right side of the equation.
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The total energy of a moving proton is 1150 MeV. What is the speed of the electron in terms of c, to two significant figures? (Recall that the mass of a proton can be written as 938 MeV/c2.)
The speed of the electron in terms of c, to two significant figures is 0.71 c.
The total energy of a moving proton is 1150 MeV.
Mass of a proton = 938 MeV/c²
Formula:
The relativistic kinetic energy of a proton in terms of its speed is given by K = (γ – 1)mc², where K is kinetic energy, γ is the Lorentz factor, m is mass, and c is the speed of light.
The Lorentz factor is given by γ = (1 – v²/c²)^(–1/2), where v is the speed of the proton.
Solution:
From the question,
Total energy of a moving proton = 1150 MeV
Therefore,
Total energy of a moving proton = Kinetic energy + Rest energy of the proton
1150 MeV = K + (938 MeV/c²)c²
K = (1150 – 938) MeV/c² = 212 MeV/c²
The relativistic kinetic energy of a proton is given by,
K = (γ – 1)mc²
Hence,
γ = (K/mc²) + 1
γ = (212 MeV/c²) / (938 MeV/c²) + 1
γ = 1.2264 (approx)
Using this Lorentz factor, we can find the speed of the proton,
v = c√[1 – (1/γ²)]
v = c√[1 – (1/1.2264²)]
v = c x 0.7131
v = 0.7131c
Therefore, the speed of the proton is 0.7131c (to two significant figures).
Hence, the required answer is 0.71 c (to two significant figures).
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: A 480 nm argon-ion laser passes through a narrow slit and the diffraction pattern is observed on a screen 5.048 m away. On the viewing screen, the distance between the centers of the second minima on either side of the central bright fringe is 36 mm. Consider the angle is small. a) Which formula can be used to calculate the location of a minima on the viewing screen? b) Find the width of the slit.
a) The formula used to calculate the location of a minima on the viewing screen in the case of diffraction through a single slit is given by the equation: y = (mλL) / w. b) Width of the slit is approximately 0.1336 mm.
The formula is:
y = (mλL) / w
where:
y is the distance from the central maximum to the minima on the screen,
m is the order of the minima (m = 1 for the first minima, m = 2 for the second minima, and so on),
λ is the wavelength of light,
L is the distance between the slit and the screen (5.048 m in this case),
w is the width of the slit.
b) To find the width of the slit, we can rearrange the above equation:
w = (mλL) / y
Given:
λ = 480 nm = 480 x 10^-9 m,
L = 5.048 m,
y = 36 mm = 36 x 10^-3 m,
m = 2 (since we are considering the second minima on either side of the central bright fringe),
Substituting these values into the equation, we can calculate the width of the slit (w): w = (mλL) / y
= (2)(480 x 10^-9 m)(5.048 m) / (36 x 10^-3 m)
w ≈ 0.1336 mm
Therefore, the width of the slit is approximately 0.1336 mm.
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HAIS Please Consider a inner & outer radil Ry 3 R₂, respectively. R₂ A HR I J= R1 hollow longmetalic Acylinder of I current of current density I 15 flowing in the hollow cylinder, Please find the magnetic field energy within the men per unit length
To find the magnetic field energy within a hollow long metallic cylinder with inner radius R₁ and outer radius R₂, through which a current density of J = 15 is flowing, we can use the formula for magnetic field energy per unit length. The calculation involves integrating the energy density over the volume of the cylinder and then dividing by the length.
The magnetic field energy within the hollow long metallic cylinder per unit length can be calculated using the formula:
Energy per unit length = (1/2μ₀) ∫ B² dV
where μ₀ is the permeability of free space, B is the magnetic field, and the integration is performed over the volume of the cylinder.
For a long metallic cylinder with a hollow region, the magnetic field inside the cylinder is given by Ampere's law as B = μ₀J, where J is the current density.
To evaluate the integral, we can assume the current flows uniformly across the cross-section of the cylinder, and the magnetic field is uniform within the cylinder. Thus, we can express the volume element as dV = Adx, where A is the cross-sectional area of the cylinder and dx is the infinitesimal length.
Substituting the values and simplifying the integral, we have:
Energy per unit length = (1/2μ₀) ∫ (μ₀J)² Adx
= (1/2) J² A ∫ dx
= (1/2) J² A L
where L is the length of the cylinder.
Therefore, the magnetic field energy within the hollow long metallic cylinder per unit length is given by (1/2) J² A L, where J is the current density, A is the cross-sectional area, and L is the length of the cylinder.
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A coaxial cable used in a transmission line has an inner radius of 0.20 mm and an outer radius of 0.60 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with polystyrene. (Also assume that the outer conductor is infinitesimally thin.) 104 X pF/m Two parallel plates of area 55 cm² are given charges of equal magnitudes 9.8 x 10-7 C but opposite signs. The electric field within the dielectric material filling the space between the plates is 4.6 x 10 V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.
The capacitance per meter of the coaxial cable is 104 pF/m. The magnitude of the charge induced on each dielectric surface is 9.9 x 10⁻⁷C.
Given:
Inner radius of a coaxial cable (r1) = 0.20 mm,
Outer radius of a coaxial cable (r2) = 0.60 mm,
Polystyrene Dielectric medium. (ε = 2.6),
Electric Field (E) = 4.6 x 10³ V/m,
Charge given (q) = 9.8 x 10⁻⁷C,
Area (A) = 55 cm² = 5.5 x 10⁻² m²
(a) Capacitance of Coaxial Cable:
The Capacitance of a coaxial cable is given by:
C = 2πε / ln (r₂ / r₁)
C = (2π x 2.6) / ln (0.6 / 0.2)C = 104 pF/m
Therefore, capacitance per meter of the coaxial cable is 104 pF/m
(b) Dielectric Surface:
The surface charge density induced on each dielectric surface is given by
σ = q / Aσ
= 9.8 x 10⁻⁷C / 5.5 x 10⁻² m²σ
= 1.8 x 10⁻⁵ C/m²
Now, the magnitude of the charge induced on each dielectric surface is given byq' = σ x Aq' = (1.8 x 10⁻⁵ C/m²) x (5.5 x 10⁻² m²)q' = 9.9 x 10⁻⁷C
Therefore, the magnitude of the charge induced on each dielectric surface is 9.9 x 10⁻⁷C.
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a A 10-kg block is attached to a very light horizontal spring on a smooth horizontal table. A force of 40 Nis required to compress the spring 20 cm. Suddenly, the block is struck by a 4-kg stone traveling to the right at a speed v, - 3.90 m/s. The stone rebounds at 20 m/s horizontally to the left, while the block starts to oscillate. Find the Amplitude of the oscillation. (10 points)
Considering the conservation of linear momentum before and after the collision between the stone and the block, we find that the amplitude of the oscillation is approximately 2.14 meters.
Mass of the block (m1) = 10 kg
Mass of the stone (m2) = 4 kg
Initial velocity of the stone (v1) = -3.90 m/s (to the right)
Final velocity of the stone (v2) = 20 m/s (to the left)
Compression of the spring (x) = 20 cm = 0.20 m
Force required to compress the spring (F) = 40 N
Before the collision, the block is at rest, so its initial velocity (v1') is zero. The stone's momentum before the collision is given by:
m2 * v1 = -4 kg * (-3.90 m/s) = 15.6 kg·m/s (to the left)
After the collision, the stone rebounds and moves to the left with a velocity of 20 m/s. The block starts to oscillate, and we want to find its amplitude (A).
The conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision:
(m1 * v1') + (m2 * v1) = (m1 * v2') + (m2 * v2)
Substituting the known values:
(10 kg * 0 m/s) + (4 kg * (-3.90 m/s)) = (10 kg * v2') + (4 kg * 20 m/s)
0 + (-15.6 kg·m/s) = 10 kg * v2' + 80 kg·m/s
-15.6 kg·m/s = 10 kg * v2' + 80 kg·m/s
-95.6 kg·m/s = 10 kg * v2'
Now, we calculate the velocity of the block (v2'):
v2' = -95.6 kg·m/s / 10 kg
v2' = -9.56 m/s (to the left)
The velocity of the block at the extreme points of the oscillation is given by:
v_max = ω * A
where ω is the angular frequency, which is calculated using Hooke's law:
F = k * x
where F is the force applied, k is the spring constant, and x is the compression of the spring. Rearranging the equation, we get:
k = F / x
Substituting the known values:
k = 40 N / 0.20 m
k = 200 N/m
The angular frequency (ω) is calculated using:
ω = sqrt(k / m1)
Substituting the known values:
ω = sqrt(200 N/m / 10 kg)
ω = sqrt(20 rad/s)
Now, we is calculate the maximum velocity (v_max):
v_max = ω * A
A = v_max / ω
A = (-9.56 m/s) / sqrt(20 rad/s)
A ≈ -2.14 m
The amplitude of the oscillation is approximately 2.14 meters. The negative sign indicates the direction of the oscillation.
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The force of attraction between the Earth (m = 5.98 x
1024 kg) and Halley’s Comet (m = 2.2 x 1014
kg) when it is closest to the sun is 1.14 x 107 N.
Calculate the distance of separation.
The distance of separation between the Earth and Halley's Comet when it is closest to the sun is approximately 4.87 x 10^11 meters.
The distance of separation between the Earth and Halley's Comet can be calculated using the formula for gravitational force:
F = G * (m1 * m2) / r^2
Rearranging the formula, we have:
r = sqrt((G * (m1 * m2)) / F)
Plugging in the given values:
r = sqrt((6.67 x 10^-11 N(m/kg)^2 * (5.98 x 10^24 kg * 2.2 x 10^14 kg)) / (1.14 x 10^7 N)
Calculating the result:
r ≈ 4.87 x 10^11 meters
Therefore, the distance of separation between the Earth and Halley's Comet when it is closest to the sun is approximately 4.87 x 10^11 meters.
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2. Describe what happens when a transverse wave travels from a low density medium to a high density medium. (3 marks)
When a transverse wave travels from a low density medium to a high density medium, it undergoes reflection and inversion of the wave.
When a wave travels from one medium to another medium, the wave undergoes a change in its speed and direction of propagation. It also undergoes reflection and inversion, if there is a boundary present between the two media. The direction of propagation changes at the boundary surface of two media due to the variation of refractive indices of two media. The wave inversion occurs at the boundary surface of two media. So, when a transverse wave travels from a low density medium to a high density medium, it undergoes reflection and inversion of the wave.The inversion of the wave is when the wave goes from an upside-down position to a right-side-up position.
This is what happens when the wave goes from a lower density medium to a higher density medium. When the wave hits the boundary between the two media, it is reflected back in the opposite direction, with the same frequency and wavelength. The speed of the wave is determined by the medium through which it is traveling, so when the wave hits the boundary, it slows down as it enters the higher density medium.
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a
wire carrting a 4A current is placed at an angle of 40 degrees with
the respect to a magnetic field of strength 0.7T. if the length of
the wire is 1.6m what is the magnatude of the magnetic force ac
The magnitude of the magnetic force acting on the wire is 2.22 N
The given parameters are:
Current (I) = 4A,
Angle (θ) = 40°,
Magnetic Field (B) = 0.7T,
Length of wire (L) = 1.6m.
The formula for calculating the magnitude of the magnetic force acting on the wire is given by:
F = BILsinθ
Where,
F is the magnitude of the magnetic force acting on the wire,
B is the magnetic field strength,
I is the current passing through the wire,
L is the length of the wire,
θ is the angle between the wire and the magnetic field.
So, substituting the given values in the above formula:
F = BILsinθ
F = (0.7T) (4A) (1.6m) sin 40°
F = 2.22 N (approx)
Therefore, the magnitude of the magnetic force acting on the wire is 2.22 N (approx).
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2. Gases are very useful for converting heat into work, since they easily expand or contract with temperature.
2.1 Suppose you have 2 liters of air at 300 K, in a cylinder closed on all sides, except that the lid is a piston that can slide along the x-axis. Draw a picture showing the gas and how it can be expanded or compressed.
2.2 The gas is maintained at a relative pressure of 2.0 x 105 N/m². Using PV = n R T, calculate the number of moles.
2.3 The gas is now immersed in boiling water and now reaches a temperature of 373 K. The same pressure is maintained. Calculate the volume reached by the gas at the new temperature.
A cylinder containing air at a temperature of 300 K has a lid that slides along the x-axis. The air in the cylinder can expand or compress. The diagram below illustrates this situation. Figure Picture of the gas and how it can be expanded or compressed
The number of moles in the air inside the cylinder can be calculated using PV = nRT. Where R = 8.31 J/(mol K).T = 300 Kn = number of moles. PV = n RT n = PV/RT Substitute the given values into the formula Therefore, there are 161.1 moles of air inside the cylinder.
The volume reached by the gas at a new temperature of 373 K can be determined using the following formula V2 = volume reached by the gas at a new temperature. Substitute the given values into the formula (2 L/300 K) = (V2/373 K) V2 = (2 L/300 K) x 373 K V2 = 2.49 liters Therefore, the volume reached by the gas at a temperature of 373 K is 2.49 liters.
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Problem 1 (30 points) Consider two objects of masses m = 7.133 kg and m2 = 0.751 kg. The first mass (m) is traveling along the negative y-axis at 45.5 km/hr and strikes the second stationary mass m2, locking the two masses together. a) (5 Points) What is the velocity of the first mass before the collision? Vm= > m/s b) (3 Points) What is the velocity of the second mass before the collision? V m2=C m/s c) (1 Point) The final velocity of the two masses can be calculated using the formula number: (Note: use the formula-sheet given in the introduction section) d) (5 Points) What is the final velocity of the two masses? > m/s e) (4 Points) Choose the correct answer: f) (4 Points) What is the total initial kinetic energy of the two masses? Ki J g) (5 Points) What is the total final kinetic energy of the two masses? KE J h) (3 Points) How much of the mechanical energy is lost due to this collision? AEint-
The velocity of the first mass before the collision is [tex]$V_{m_1} = < -12.64 > \, \text{m/s}$[/tex].
the second mass is stationary (not moving) before the collision, its velocity before the collision is zero:
[tex]$V_{m_2} = < 0, 0, 0 > \, \text{m/s}$[/tex]. the final velocity of the two masses is [tex]$V_{m_f} = < -91.19 > \, \text{m/s}$[/tex]. the total initial kinetic energy of the two masses is [tex]$K_i = 570.305 \, \text{J}$[/tex].
Given:
Mass of the first object, m1 = 7.133 kg
Mass of the second object, m2 = 0.751 kg
Velocity of the first object before the collision, V1 = -45.5 km/hr
To solve the problem, we need to convert the given velocity to meters per second (m/s) and use the principles of conservation of momentum and kinetic energy.
a) To find the velocity of the first mass before the collision:
Given velocity, V1 = -45.5 km/hr
Converting km/hr to m/s:
V1 = (-45.5 km/hr) * (1000 m/km) * (1 hr/3600 s)
V1 = -12.64 m/s (rounded to two decimal places)
Therefore, the velocity of the first mass before the collision is [tex]$V_{m_1} = < -12.64 > \, \text{m/s}$[/tex].
b) Since the second mass is stationary (not moving) before the collision, its velocity before the collision is zero:
[tex]$V_{m_2} = < 0, 0, 0 > \, \text{m/s}$[/tex].
c) The final velocity of the two masses can be calculated using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
[tex]$m_1 \cdot V_{m_1} + m_2 \cdot V_{m_2} = (m_1 + m_2) \cdot V_{m_f}$[/tex]
d) To find the final velocity of the two masses:
[tex]$m_1 \cdot V_{m_1} + m_2 \cdot V_{m_2} = (m_1 + m_2) \cdot V_{m_f}$[/tex]
Substituting the known values:
[tex]$(7.133 \, \text{kg}) \cdot (-12.64 \, \text{m/s}) + (0.751 \, \text{kg}) \cdot (0 \, \text{m/s}) = (7.133 \, \text{kg} + 0.751 \, \text{kg}) \cdot V_{m_f}$[/tex]
Solving for [tex]$V_{m_f}$[/tex]:
[tex]$V_{m_f} = -91.19 \, \text{m/s}$[/tex] (rounded to two decimal places)
Therefore, the final velocity of the two masses is [tex]$V_{m_f} = < -91.19 > \, \text{m/s}$[/tex].
f) To calculate the total initial kinetic energy of the two masses:
Initial kinetic energy of the first mass, [tex]$K_1 = \frac{1}{2} \cdot m_1 \cdot \left| V_{m_1} \right|^2$[/tex]
[tex]$K_1 = \frac{1}{2} \cdot 7.133 \, \text{kg} \cdot \left| -12.64 \, \text{m/s} \right|^2$[/tex]
Initial kinetic energy of the second mass, [tex]$K_2 = \frac{1}{2} \cdot m_2 \cdot \left| V_{m_2} \right|^2$[/tex]
[tex]$K_2 = \frac{1}{2} \cdot 0.751 \, \text{kg} \cdot \left| 0 \, \text{m/s} \right|^2$[/tex]
Total initial kinetic energy, [tex]$K_i = K_1 + K_2$[/tex]
Calculating the values:
[tex]$K_1 = 570.305 \, \text{J}$[/tex] (rounded to three decimal places)
[tex]$K_2 = 0 \, \text{J}$[/tex] (since the second mass is stationary)
[tex]$K_i = 570.305 \, \text{J}$[/tex]
Therefore, the total initial kinetic energy of the two masses is [tex]$K_i = 570.305 \, \text{J}$[/tex].
g) To calculate the total final kinetic energy of the two masses:
Final kinetic energy of the combined masses, [tex]$K_f = \frac{1}{2} \cdot (m_1 + m_2) \cdot \left| V_{m_f} \right|^2$[/tex]
[tex]$K_f = \frac{1}{2} \cdot (7.133 \, \text{kg} + 0.751 \, \text{kg}) \cdot \left| -91.19 \, \text{m/s} \right|^2$[/tex]
Calculating the value:
[tex]$K_f = 30263.929 \, \text{J}$[/tex] (rounded to three decimal places)
Therefore, the total final kinetic energy of the two masses is [tex]$K_f = 30263.929 \, \text{J}$[/tex].
h) The change in mechanical energy can be calculated as:
[tex]$\Delta E_{\text{int}} = K_f - K_i$[/tex]
Calculating the value:
[tex]$\Delta E_{\text{int}} = 30263.929 \, \text{J} - 570.305 \, \text{J}$[/tex]
[tex]$\Delta E_{\text{int}} = 29693.624 \, \text{J}$[/tex] (rounded to three decimal places)
Therefore, the change in mechanical energy due to this collision is [tex]$\Delta E_{\text{int}} = 29693.624 \, \text{J}$[/tex].
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A polar bear walks toward Churchill, Manitoba. The pola bear's displacement is 25.0 km [S 30.0°E]. Determine th components of the displacement. a)dx= 25 cos30° [E], dy= 25 sin 30°[S] b)dx= 25 cos 30° [W], d = 25 sin 30°[N] c) dx= 25 sin 30° [E], dy= 25 cos30°[S] d)dx= 25 cos 30º[E], d = 25 sin30°[N]
The components of the polar bear's displacement are (A) dx = 25 cos 30° [E], dy = 25 sin 30° [S].
In this case, option (a) is the correct answer. The displacement of the polar bear is given as 25.0 km [S 30.0°E]. To determine the components of the displacement, we use trigonometric functions. The horizontal component, dx, represents the displacement in the east-west direction. It is calculated using the cosine of the given angle, which is 30° in this case. Multiplying the magnitude of the displacement (25.0 km) by the cosine of 30° gives us the horizontal component, dx = 25 cos 30° [E].
Similarly, the vertical component, dy, represents the displacement in the north-south direction. It is calculated using the sine of the given angle, which is 30°. Multiplying the magnitude of the displacement (25.0 km) by the sine of 30° gives us the vertical component, dy = 25 sin 30° [S].
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7. The steady state and pseudosteady state flow equations in a circular drainage area, and productivity Index are given as q=7.081*10¯^3 kh/Bμ (rhoe-rhowf)/({In(re/rw)+s}) q=7.081*10¯^3 kh/Bμ (p-rhowf)/({In(re-rw)-0,75+s}) J=q/Δp Similary the dimensionless pressure, dimensionless rate and dimensionless productivity index are defined as: pn=Δp/pch qn=q/qch Jn=J/Jch
a. You are asked to find out the what are the characteristic variables that make those varaibles dimensionless and write the dimensionless pressure, rate and productivity index variables? b. Also find out how do these three dimensionless variables relate to each other? Or aren't they related at all?
a. The characteristic variables that make those variables dimensionless and write the dimensionless pressure, rate, and productivity index variables are as follows:Dimensionless Pressure (pn):
(Δp/pch)Dimensionless Rate (qn): (q/qch)Dimensionless Productivity Index (Jn): (J/Jch)The characteristic variables (pch, qch, and Jch) are obtained by choosing appropriate reference values for pressure (pch), rate (qch), and productivity index (Jch).b. These three dimensionless variables are related by the equationJn = pn/qnProductivity index (J) is related to pressure (p) and rate (q) through the following equation:
J = q/ΔpFor dimensionless variables, we divide both sides of the above equation by qch/Jch, which gives usJn = pn/qnThus, the dimensionless productivity index is equal to the dimensionless pressure divided by the dimensionless rate.About Characteristic variablesCharacteristic variables come from experimental observations or obtained from experimental intuition on the process.
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Provide two examples of experiments or phenomena that Planck's /
Einstein's principle of EMR quantization cannot explain
Planck's and Einstein's principle of EMR quantization, which states that energy is quantized in discrete packets, successfully explains many phenomena such as the photoelectric effect and the resolution of the ultraviolet catastrophe. However, there may still be experiments or phenomena that require further advancements in our understanding of electromagnetic radiation beyond quantization principles.
The Photoelectric Effect: The photoelectric effect is the phenomenon where electrons are ejected from a metal surface when it is illuminated with light.
According to the classical wave theory of light, the energy transferred to the electrons should increase with the intensity of the light. However, in the photoelectric effect, it is observed that the energy of the ejected electrons depends on the frequency of the incident light, not its intensity. This behavior is better explained by considering light as composed of discrete energy packets or photons, as proposed by the quantization principle.
The Ultraviolet Catastrophe: The ultraviolet catastrophe refers to a problem in classical physics where the Rayleigh-Jeans law predicted that the intensity of blackbody radiation should increase infinitely as the frequency of the radiation approached the ultraviolet region.
However, experimental observations showed that the intensity levels off and decreases at higher frequencies. Planck's quantization hypothesis successfully resolved this problem by assuming that the energy of the radiation is quantized in discrete packets, explaining the observed behavior of blackbody radiation.
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Three people are pulling on a 50N rope. The first person is pulling to the right with a force of 445N. The second person weighs 65kg and is pulling to the right with a force of 235N. The rope is moving to the right at an acceleration of 1.4m/s^2. With how much force is the third person pulling if they are pulling to the left?
Please show steps clearly with equations if possible
force the third person pulling if they are pulling to the left:
680 N - Force to the left = (m1 + 65 kg + m3) * 1.4 m/s^2
To solve this problem, we can apply Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's calculate the total force exerted to the right:
Total force to the right = Force by the first person + Force by the second person
= 445 N + 235 N
= 680 N
Next, let's determine the force exerted to the left by the third person. Since the rope is moving to the right with an acceleration of 1.4 m/s^2, we can calculate the net force acting on the system:
Net force = Total force to the right - Force to the left
= 680 N - Force to the left
Since the system is accelerating to the right, the net force must be equal to the mass of the system multiplied by its acceleration:
Net force = Mass of the system * Acceleration
= (Mass of the first person + Mass of the second person + Mass of the third person) * Acceleration
We know the mass of the second person (65 kg), so let's assume the masses of the first and third persons are m1 and m3, respectively. Therefore, the equation becomes:
680 N - Force to the left = (m1 + 65 kg + m3) * 1.4 m/s^2
Finally, rearranging the equation to solve for the force to the left (Force to the left = 680 N - (m1 + 65 kg + m3) * 1.4 m/s^2), we need additional information about the masses of the first and third persons to determine the force exerted by the third person.
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5. A laser travels through two slits onto a screen behind the slits. Thecentral maximum of the diffraction contains nine, smaller
individual interference bright spots – four on each side of the
middle.
a. The diffraction pattern is due to the
A. width of the slits B. distance between the slits
b. The interference pattern is due to the
A. width of the slits B. distance between the slits
c. The first diffraction minimum (p=1) aligns with one of the interference minimums. What is
the order for the interference minimum (i.e. the value for m) that aligns with the diffraction
minimum? Explain your answer.
d. What is the ratio between the slit spacing to the slit's width (d/a)?
The diffraction pattern is due to the width of the slits.b. The interference pattern is due to the distance between the slits.
The order for the interference minimum (i.e. the value for m) that aligns with the diffraction minimum is m = 5. A diffraction pattern is produced when a wave is forced to pass through a small opening or around a sharp corner. Diffraction is the bending of light around a barrier or through an aperture in the barrier. It occurs as a result of interference between waves that must compete for the same space.
Diffraction pattern is produced when light is made to pass through a narrow slit or opening. This light ray diffracts from the slit and produces a pattern of interference fringes on a screen behind it. The spacing between the fringes and the size of the pattern depend on the wavelength of the light and the size of the opening. Therefore, the diffraction pattern is due to the width of the slits.
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An electron moves at velocity 0.9c. How fast must it move to double its momentum?
To double its momentum, the electron must move at a velocity v2 given by (2 * p_rel1) / (γ * m).
The momentum of an object is given by the equation:
p = m * v
where p is the momentum, m is the mass of the object, and v is its velocity.
To double the momentum, we need to find the velocity at which the momentum becomes twice its initial value.
Let's assume the initial momentum is p1 and the initial velocity is v1. We want to find the velocity v2 at which the momentum doubles, so the new momentum becomes 2 * p1.
Since momentum is directly proportional to velocity, we can set up the following equation:
2 * p1 = m * v2
Since we want to find the velocity v2, we can rearrange the equation:
v2 = (2 * p1) / m
However, we need to take into account relativistic effects when dealing with velocities close to the speed of light. The relativistic momentum is given by:
p_rel = γ * m * v
where γ is the Lorentz factor, given by:
γ = 1 / sqrt(1 -[tex](v/c)^2)[/tex]
In this case, the initial velocity v1 = 0.9c.
Now, let's substitute the initial velocity and momentum into the relativistic momentum equation:
p_rel1 = γ * m * v1
To find the velocity v2 that doubles the momentum, we can set up the equation:
2 * p_rel1 = γ * m * v2
Rearranging the equation, we have:
v2 = (2 * p_rel1) / (γ * m)
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