The correct answer is (d). The value of ΔS (change in entropy) when one mole of N₂(g) is expanded from 20.0 L at 273 K to 300 L at 400 K is approximately -21.7 J/K.
To calculate the change in entropy, we can use the formula ΔS = nCmp ln[tex](\frac{T_{2} }{T_{1} } )[/tex] - nR ln[tex](\frac{V_{2} }{V_{1} })[/tex], where ΔS is the change in entropy, n is the number of moles (in this case, 1 mole), Cmp is the molar heat capacity at constant pressure, T₁ and T₂ are the initial and final temperatures respectively, and V₁ and V₂ are the initial and final volumes respectively. R is the ideal gas constant.
Substituting the given values, we have ΔS = (1 mol) × (29.4 J K⁻¹ mol⁻¹) ln(400 K/273 K) - (1 mol) × (8.314 J K⁻¹ mol⁻¹) × ln(300 L/20.0 L).
Simplifying the calculation, we get ΔS ≈ -21.7 J/K.
Therefore, the value of ΔS when one mole of N₂(g) is expanded from 20.0 L at 273 K to 300 L at 400 K is approximately -21.7 J/K. The negative sign indicates a decrease in entropy during the process. Hence, the correct answer is AS = -21.7 J/K.
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From the first principles, calculate the technical requirements to build the electromagnetic rail gun on the Moon to defend the Earth from asteroids. Technical requirements to be calculated include the following parameters:
1. Energy and power required to operate the rail gun at the required firing frequency.
2. Physical dimensions and weight required for the rail gun.
3. Probability of hitting an asteroid at various locations in space.
4. Realistic dimensions of heat dissipating radiators around the railgun operating in vacuum with the required frequency of projectile firing with 100% target hitting probability.
Rail gun uses metallic Moon’s rocks as projectiles and electricity as a propellant. Assume frictionless motion for electromagnetically propelled metallic projectiles inside the rail gun. Use at least three values for kinetic energy and physical dimensions of asteroids. Assume asteroid bundling with various distributions in dimensions and distance between them (proximity).
To build an electromagnetic rail gun on the Moon for defending the Earth from asteroids, the technical requirements include determining the energy and power required for operation, the physical dimensions and weight of the rail gun, the probability of hitting asteroids in different locations, and the realistic dimensions of heat dissipating radiators.
Energy and power: Calculating the energy and power required depends on the firing frequency of the rail gun. It involves considering the mass of the metallic Moon rocks used as projectiles, the desired projectile velocity, and the desired firing rate. By determining the kinetic energy per projectile and the desired firing frequency, the total energy and power requirements can be calculated.Physical dimensions and weight: The physical dimensions and weight of the rail gun are influenced by factors such as the desired projectile velocity, the mass of the projectiles, and the required structural integrity. The dimensions need to accommodate the projectiles and provide sufficient space for the electromagnetic acceleration mechanism. The weight should be manageable for deployment on the Moon.Probability of hitting asteroids: The probability of hitting an asteroid depends on factors such as the accuracy of the rail gun, the distance to the target, and the size and speed of the asteroid. Detailed calculations and simulations can be performed to assess the probability of hitting asteroids at various locations in space.Heat dissipating radiators: Operating a rail gun at a high firing frequency generates significant heat, which needs to be dissipated efficiently to prevent overheating. Realistic dimensions of heat dissipating radiators can be determined by considering factors such as the power requirements, cooling efficiency, and operating conditions in a vacuum.Learn more about electromagnetic
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what is it called when two mirrors facing each other
When two mirrors are placed facing each other, it creates a phenomenon known as "mirror reflection" or "infinite reflection." This occurs as the light reflects back and forth between the mirrors, creating multiple reflections that appear to stretch infinitely into the distance.
The reflection continues on and on until it becomes too small to see. In this way, a person sees many reflections of themselves, and each reflection is smaller than the previous one. This is called an infinity mirror or a mirror tunnel.An infinity mirror is a visual illusion that looks like the mirror has no end. It is accomplished by placing a mirror in front of another and allowing a small amount of space between the two. Then, light is reflected back and forth in the space between the mirrors, generating an infinite loop of images.
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Aluminium of mass 0.55 kg, with an initial temperature of 22° C, is heated for 90 minutes using a 71.474W power source. During this time the Aluminium reaches its melting temperature of 660.3° C and is partially melted. Assume no energy is lost to the surroundings. Calculate how much energy is supplied to the aluminium during this time. Round your answer to 3 significant figures.
The amount of energy supplied to the aluminum during this time is approximately 385,257 J.
To calculate the energy supplied to the aluminum, we can use the formula: Energy = Power × Time. Given that the power source has a power of 71.474 W and the heating time is 90 minutes (which needs to be converted to seconds), we can compute the energy supplied as Energy = 71.474 W × 90 minutes × 60 seconds/minute = 385,257 J.
The energy supplied to the aluminum is obtained by multiplying the power (in watts) by the time (in seconds). In this case, the power source provides a constant power of 71.474 W throughout the 90 minutes of heating. To ensure consistent units, we convert the time from minutes to seconds by multiplying by 60. By performing the calculation, we find that the energy supplied to the aluminum is approximately 385,257 J.
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which is the process used in fire investigation to determine if and where electrical circuits were energized at the time of the fire?
The process used in fire investigation to determine if and where electrical circuits were energized at the time of the fire is live circuit analysis.
This is a technique in which investigators use specialized equipment to measure the voltage and amperage levels of electrical circuits in the building. Live circuit analysis is conducted once the electrical power supply is re-established on-site for investigating the fire's origin and cause. It involves checking electrical outlets, appliances, and other devices that might have been connected to electrical circuits in the building.
This process is vital for determining whether an electrical fault or malfunction caused the fire and identifying the responsible parties for negligence. In summary, the live circuit analysis is a standard fire investigation procedure that can determine the presence and location of electrical faults that led to a fire. The technique provides insights for experts to reconstruct the origin and causes of the fire to prevent future tragedies.
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Two charged concentric spherical theirs have radi of 11.0 cm and 14.5 cm. The charge on the inoer sheil is 4.20×10
−3
C and thut on,the outer sheil is 1.90×10
−1
C. Find the magntude of the electric feld at the following points. (a) at f=12.5 cm N/C (b) ate +205 cm MC
The magnitude of the electric field at point P, located 12.5 cm from the center of the concentric spherical shells, is 8.75 × 10^3 N/C.
To calculate the electric field at point P, we can use the formula for the electric field due to a charged spherical shell. The electric field at a point outside a charged spherical shell is given by:
E = (k * Q) / r^2
Where E is the electric field, k is the electrostatic constant (8.99 × 10^9 N m^2/C^2), Q is the charge on the shell, and r is the distance from the center of the shell.
For the inner shell (radius = 11.0 cm, charge = 4.20 × 10^-3 C):
E_inner = (k * Q_inner) / r_inner^2
E_inner = (8.99 × 10^9 N m^2/C^2 * 4.20 × 10^-3 C) / (0.11 m)^2
E_inner = 3.19 × 10^5 N/C
For the outer shell (radius = 14.5 cm, charge = 1.90 × 10^-1 C):
E_outer = (k * Q_outer) / r_outer^2
E_outer = (8.99 × 10^9 N m^2/C^2 * 1.90 × 10^-1 C) / (0.145 m)^2
E_outer = 2.09 × 10^4 N/C
The net electric field at point P is the vector sum of the electric fields due to the inner and outer shells:
E_net = E_inner + E_outer
E_net = 3.19 × 10^5 N/C + 2.09 × 10^4 N/C
E_net = 3.40 × 10^5 N/C
Therefore, the magnitude of the electric field at point P is 3.40 × 10^5 N/C.
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For an intrinsic semiconductor, as the temperature increases, the number of electrons excited to conduction band, CB increases. Sketch a diagram of the probability function, f(E) for electrons at T>0 K and show that in the tail region, the value of f(E) increases with T.
In an intrinsic semiconductor, as temperature increases, more electrons are excited to the conduction band due to thermal energy, leading to an increase in the probability of finding electrons at higher energy levels, especially in the tail region beyond the band gap.
In an intrinsic semiconductor, as the temperature increases, more electrons are excited to the conduction band. This is due to thermal energy provided to the electrons, allowing them to overcome the band gap energy and move from the valence band to the conduction band.
To sketch a diagram of the probability function, f(E), we can use an energy axis (E) and a vertical axis representing the probability of finding an electron at a given energy level.
At absolute zero temperature (T=0 K), the probability function, f(E), is represented by a step function with a sharp cutoff at the energy corresponding to the band gap. This is because at T=0 K, there is no thermal energy available for the electrons to overcome the band gap and move to higher energy levels.
As the temperature increases (T > 0 K), the probability function, f(E), starts to show a gradual increase in the tail region of the diagram. The tail region represents energy levels closer to the conduction band edge. This increase in f(E) with temperature is due to the higher thermal energy available, allowing more electrons to be excited to higher energy levels.
The diagram would show a smooth, gradual increase in the value of f(E) as we move from lower energies (valence band) to higher energies (conduction band) along the energy axis. The slope of the probability function in the tail region would become steeper as the temperature increases, indicating a higher probability of finding electrons at higher energy levels.
It's important to note that the diagram would still exhibit a sharp cutoff at the band gap energy, as there is still an energy barrier that needs to be overcome for electrons to move from the valence band to the conduction band. However, with increasing temperature, the probability of electrons being present in the tail region beyond the band gap energy would significantly increase.
Overall, the sketch of the probability function, f(E), for electrons at T > 0 K would show a gradual increase in the tail region with increasing temperature, indicating a higher probability of finding electrons at higher energy levels.
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What is the weight of a 2.50-kg sandbag on the surface of the earth?
d. 49.0N
e. 98.0N
c. 24.5N
a. 2.50N
b. 9.80N
A rock is suspended from a rope and is accelerating upward. Which of the following statements is true regarding the tension in the string?
c. The tension is the same as the weight of the rock.
b. The tension is less than the weight of the rock.
a. The tension points down.
d. The tension is greater than the weight of the rock.
e. The stress is independent of the magnitude of the rock's acceleration.
The tension in a rope supporting a rock that is accelerating upward is greater than the weight of the rock. The weight of a 2.50 kg sandbag on the surface of the Earth is 24.5 N.
The weight of an object is the force exerted on it due to gravity. On the surface of the Earth, the weight of an object can be calculated by multiplying its mass by the acceleration due to gravity (9.8 m/s^2):
Weight = mass * acceleration due to gravity
Weight = 2.50 kg * 9.8 m/s^2
Weight = 24.5 N
Therefore, the weight of a 2.50 kg sandbag on the surface of the Earth is 24.5 N, so the correct answer is option c. 24.5N.
When a rock is suspended from a rope and accelerating upward, the tension in the string is greater than the weight of the rock. This is because the tension in the rope must provide an additional force to overcome the gravitational force acting on the rock and accelerate it upward. The tension in the rope is equal to the sum of the weight of the rock and the additional force required to produce the acceleration. Therefore, the correct answer is option d. The tension is greater than the weight of the rock.
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3 objects, A B, and C, all carry electric charge, The amount of charge and the sign of the charge on these objects is not necessarily the same, but we know that A is positively charged Initially the object are held in place on the x-axis by an external force. A is located at x =0,B is located at x=5 units and C at x=10 units. At t=0 object B is released and begins to move in the positive x-direction. Which of the following statements must be true. You can choose more than 1. The given information does not allow us to determine the sign of the charge on B and C. B is positively charged an C is negatively charged. The net force on B points in the positive x-direction. C is positively charged, but carries less charge than A. B and C are both negatively charged.
The given information allows us to determine that B is positively charged and that the net force on B points in the positive x-direction. So, the following statements must be true:a) B is positively charged. b) The net force on B points in the positive x-direction.
There is no information available that indicates that C is positively charged and carries less charge than A. So, the statement "C is positively charged, but carries less charge than A" is not true. Moreover, the sign of the charge on C is not given.
Therefore, the statement "The given information does not allow us to determine the sign of the charge on B and C" is true.B and C cannot be both negatively charged since the given information indicates that A is positively charged. Therefore, the statement "B and C are both negatively charged" is not true.
Answer: The given information does not allow us to determine the sign of the charge on B and C and The net force on B points in the positive x-direction.
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what evidence can you cite that galactic cannibalism really happens
Galactic cannibalism, also known as galactic mergers or galactic interactions, occurs when one galaxy combines with or absorbs material from another galaxy. There is abundant observational evidence supporting the existence of galactic cannibalism.
Here are some key pieces of evidence:
Galaxy collisions have been observed, revealing various stages of merging or interaction between galaxies. These observations include distorted shapes, tidal tails, bridges of stars and gas connecting interacting galaxies, and clear signs of galactic collisions. Computer simulations based on our understanding of gravitational interactions and galaxy dynamics can replicate the features observed in interacting galaxies. These simulations provide additional evidence that galactic cannibalism is a natural outcome of gravitational interactions between galaxies. Stellar and gas streams are formed when galaxies merge, leading to the gravitational forces stripping stars and gas from the involved galaxies. These stripped materials create elongated streams or tidal tails that can be observed, providing strong evidence of past or ongoing galactic interactions. Galaxy mergers can trigger intense bursts of star formation and activate supermassive black holes at the centers of galaxies, known as active galactic nuclei (AGN). The presence of AGN and starburst activity in interacting galaxies serves as evidence for the energetic effects of galactic cannibalism. The distribution and characteristics of dwarf galaxies, which are smaller companion galaxies often found near larger galaxies, offer insights into galactic cannibalism. The presence of dwarf galaxies around larger galaxies aligns with the idea that they were captured or absorbed during galactic interactions.These lines of evidence, supported by numerous observational studies and theoretical models, strongly indicate the occurrence of galactic cannibalism. They contribute to our understanding of the dynamics involved in the evolution of galaxies.
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How much energy is required to accelerate a spaceship with a rest mass of 133 metric tons to a speed of 0.537c ? Tries 0/20 Every day our Earth receives 1.55×10^22J energy from the Sun. If we were able to use 0.85 percent of this enerav to accelerate spaceships, then how many missions would be possible in one year? Tries 0/20
To accelerate a spaceship with a mass of 133 metric tons to a speed of 0.537c, the energy required can be calculated using Einstein's mass-energy equivalence principle. By converting the mass to kilograms and applying the equation E = [tex]mc^2[/tex], the energy can be determined. If 0.85 percent of the daily solar energy received (1.55×[tex]10^22[/tex] J) is available for spaceship acceleration, the number of missions possible in one year can be calculated by dividing the available energy by the energy required per mission.
To calculate the energy required to accelerate a spaceship with a rest mass of 133 metric tons to a speed of 0.537c, we can use Einstein's mass-energy equivalence principle, E = [tex]mc^2[/tex]. First, convert the mass of the spaceship to kilograms by multiplying it by 1000. Then, calculate the energy using the formula:
E = (mass) *[tex](speed of light)^2[/tex] * sqrt(1 -[tex](velocity/speed of light)^2[/tex])
For the second question, if we can use 0.85 percent of the daily energy received from the Sun (1.55×[tex]10^22[/tex] J), multiply this value by the number of days in a year (365) to find the total available energy. Divide this energy by the energy required for each mission to determine the number of missions possible in one year.
Number of missions = (Available energy) / (Energy required per mission)
Calculating these expressions will provide the complete answers.
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Suppose an electic field exerts a 4.8 × 10-17 N westward force on an electron.
Find the horizontal component of the force that this field exerts on a proton, taking east to be positive.
The horizontal component of the force that the electric field exerts on the proton is 1.6 × 10^(-19) C times the electric field strength.
The horizontal component of the force exerted by the electric field on a proton can be determined using Newton's second law and the principle of superposition. Since both the electron and proton experience the same electric field, we can assume that the electric field strength is the same for both particles.
The force experienced by a charged particle in an electric field can be expressed as F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.
Given that the force exerted on the electron is 4.8 × 10^(-17) N, we can use this information to find the charge of the electron. The charge of an electron is -1.6 × 10^(-19) C.
F = qE
4.8 × 10^(-17) N = (-1.6 × 10^(-19) C)E
Now, let's determine the charge of a proton. The charge of a proton is +1.6 × 10^(-19) C.
Using the charge of the proton, we can find the horizontal component of the force by rearranging the equation:
F = qE
F = (1.6 × 10^(-19) C)E
Therefore, the horizontal component of the force that the electric field exerts on the proton is 1.6 × 10^(-19) C times the electric field strength.
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An LC circuit consists of a 3.25 mH inductor and a 3.5 μF
capacitor.
a) Find its impedance Z at 65 Hz in Ω.
b) Find its impedance Z at 7 kHz in Ω.
a) At a frequency of 65 Hz, the impedance (Z) of the LC circuit can be calculated using the formula Z = √(R² + (XL - XC)²). Given that the resistance (R) is 0 ohms, the inductive reactance (XL) is 1.0648 ohms, and the capacitive reactance (XC) is 400.18 ohms, we can substitute these values into the formula. Thus, Z = √(0² + (1.0648 - 400.18)²) = 400.17 ohms, approximately.
b) At a frequency of 7 kHz, using the same formula, the resistance (R) being 0 ohms, the inductive reactance (XL) is 66.617 ohms, and the capacitive reactance (XC) is 2.2144 ohms. Plugging in these values, we get Z = √(0² + (66.617 - 2.2144)²) = 66.63 ohms, approximately.
Therefore, the impedance of the LC circuit at a frequency of 65 Hz is approximately 400.17 ohms, and at a frequency of 7 kHz is approximately 66.63 ohms.
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which part of a centrifugal pump transmits energy in the form of velocity to the water? (299)
The impeller of a centrifugal pump is the component that transmits energy in the form of velocity to the water. It consists of curved blades or vanes that rotate, creating a centrifugal force that accelerates the fluid, increasing its velocity.
In a centrifugal pump, the impeller is responsible for transferring energy to the water in the form of velocity. The impeller is typically a wheel-like structure with curved blades or vanes.
When the pump is operational, the impeller rotates rapidly, drawing in water through the inlet. As the water enters the impeller, the curved blades exert a force on it, imparting angular momentum and causing it to move in a tangential direction.
Due to the centrifugal force generated by the rotating impeller, the water is propelled outward and accelerates as it moves away from the impeller's center.
This acceleration increases the water's velocity, transforming the potential energy into kinetic energy. The high-velocity water is then discharged from the impeller into the pump's volute or diffuser section, where the kinetic energy is gradually converted back into pressure energy.
The impeller is the crucial component of a centrifugal pump that transmits energy in the form of velocity to the water. Through its rotation and curved blades, it imparts angular momentum to the water, resulting in increased velocity and kinetic energy, which drives the flow of water through the pump system.
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A force of 60 N has a x-component of 28 N. What is the y-component? OA. 2800 N OB. 53 N OC.57N OD. 66 N OE. 94 N
To determine the y-component of a force given its x-component, we need to use vector addition. The force vector can be represented as the sum of its x-component and y-component, forming a right triangle. The y-component can be calculated using trigonometry.
Given that the x-component of the force is 28 N, and the total force is 60 N, we can use the Pythagorean theorem to find the magnitude of the y-component. Let's denote the y-component as [tex]F_{y}[/tex]. The equation is:
[tex]F^{2}=F_{x}^{2} + F_{y}^{2}[/tex]
Substituting the given values:
[tex]60^{2}=28^{2} + F_{y}^{2}[/tex]
[tex]3600=784 + F_{y}^{2}[/tex]
[tex]F_{y}^{2} = 2816[/tex]
Taking the square root of both sides:
[tex]F_{y}[/tex] ≈ 53 N
Therefore, the y-component of the force is approximately 53 N. The correct option is B.
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An orifice plate with diameter 10 cm inserted in a pipe of 20 cm diameter. Pressure difference is measured by Hg differential manometer on two sides of the orifice plate gives reading 50 cm of Hg. Find the fluid flow rate. Coefficient of discharge Ca=0.64 and specific gravity of fluid is 0.90. (density of mercury is 13.6 g/cm³)
Given data:Diameter of orifice plate = 10 cm = 0.1 of pipe = 20 cm = 0.2 mPressure difference = 50 cm of Hg
Coefficient of discharge, C_a = 0.64Specific of fluid
SG = 0.9Density of mercury
ρ_m = 13.6 g/cm³ = 13600 kg/m³
We need to find the fluid flow rate.
From Bernoulli's principle of fluid flow, the difference, ∆P between the two points in a flow is related to the flow rate, Q by the formula:
∆P = KQ²where K is a constant for a given flow system known as the coefficient of discharge.
Now, the area of the orifice plate is given by:
[tex]A = π/4 × d² = π/4 × (0.1)² = 0.00785 m²[/tex]
The area of the pipe is given by:
[tex]A' = π/4 × d'² = π/4 × (0.2)² = 0.0314 m²[/tex]
Now, the flow rate is given by:
[tex]Q = A√(2g∆h/ρ)(C_a/C_c)[/tex]
Where g is the acceleration due to gravity and ∆h is the difference in the levels of the mercury in the two legs of the differential manometer.g = 9.8 m/s²∆h = 50 cm of Hg =50/100 m of Hg = 0.5 m of Hg
Now, to convert the pressure of mercury to the equivalent fluid pressure, we use the formula:
P = ρghwhere P is the pressure,
ρ is the density, g is the acceleration due to gravity and h is the height of the fluid column.
[tex]P_m = ρ_mgh_m = 13600 × 9.8 × 0.5 = 66640 N/m²[/tex]
The fluid pressure is half the mercury pressure, therefore:
[tex]P = P_m/2 = 66640/2 = 33320 N/m²[/tex]
Substituting the given values in the formula for Q, we get:
[tex]Q = 0.00785√(2 × 9.8 × 0.5/1000 × 33320)(0.64/C_c)C_[/tex]
c is the coefficient of contraction of the orifice plate which is assumed to be 0.6 for a standard orifice plate.
The value of Q can be calculated as follows:
Q = 0.0269 m³/
The fluid flow rate is 0.0269 m³/s.33518187
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A race car starts from rest and accelerates at 20.0 m/s2 for 5.0 s. It then moves with a uniform velocity for 60 s before it decelerates at 10.0 m/s2 until it comes to a stop. *Note: Show the complete solution by showing all of your work! (
a)Determine the velocity of the race car after 5.0 s.
(b)Determine the total distance traveled by the car between from the instant it started to move until it came to a stop.
(c)What was the total time that the car was in motion?
(a)The velocity of the race car after 5.0 s is 100.0 m/s. (b) The total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m. (c) The total time that the car was in motion is 75.0 s.
Let's calculate the values step by step:
Given:
Acceleration during the first phase (a1) = 20.0 m/s²
Time during the first phase (t1) = 5.0 s
Uniform velocity phase (t2) = 60 s
Deceleration during the third phase (a3) = -10.0 m/s²
(a) To determine the velocity of the race car after 5.0 s, we can use the formula:
v = u + a × t
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
t is the time.
Since the race car starts from rest, the initial velocity (u) is 0 m/s.
v = 0 + (20.0 m/s²) × (5.0 s)
v = 100.0 m/s
Therefore, the velocity of the race car after 5.0 s is 100.0 m/s.
(b) To determine the total distance traveled by the car, we need to calculate the distance covered during each phase and sum them up.
Distance during the first phase:
Using the equation of motion:
s1 = u × t1 + (1/2) × a1 × t1²
Since the initial velocity (u) is 0 m/s:
s1 = (1/2) × (20.0 m/s²) × (5.0 s)²
s1 = 250.0 m
Distance during the second phase:
Since the car moves with a uniform velocity, the distance covered is:
s2 = v × t2
s2 = 100.0 m/s × 60 s
s2 = 6000.0 m
Distance during the third phase:
Using the equation of motion:
s3 = v × t3 + (1/2) ×a3 × t3
Since the final velocity (v) is 0 m/s:
s3 = (1/2) × (-10.0 m/s²) × t3²
The time during the third phase (t3) can be found by equating the final velocity to 0:
v = u + a × t
0 = 100.0 m/s + (-10.0 m/s²) × t3
t3 = 10.0 s
Substituting the value of t3:
s3 = (1/2) × (-10.0 m/s²) × (10.0 s)²
s3 = -500.0 m
Total distance traveled by the car:
Total distance = s1 + s2 + s3
= 250.0 m + 6000.0 m + (-500.0 m)
= 6250.0 m
Therefore, the total distance traveled by the car from the instant it started to move until it came to a stop is 6250.0 m.
(c) To determine the total time that the car was in motion, we add the
time for each phase:
Total time = t1 + t2 + t3
= 5.0 s + 60 s + 10.0 s
= 75.0 s
Therefore, the total time that the car was in motion is 75.0 s.
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A cylinder made out of Steel has a radius of 4.0 mm±0.1 mm and a length of 150 mm±4 mm at a temperature of 15
∘
C±0.3
∘
C Use table 19.1 in Katz to find the coefficients of linear exansion, a, for a given material. If a range is given for α, use the lowest value. Assume that the thermal expan What is the change in length of the the cylinder after it has been cooled to a temperature of −36
∘
C±0.3
∘
C? ΔL=−0.099 mm (2.s.f) (3.33 points) What is the absolute uncertainty in this change in length?
The absolute uncertainty in the change in length of the cylinder is 0.001 mm.
To calculate the change in length of the cylinder, we need to consider the coefficient of linear expansion (α) of the steel material. The coefficient of linear expansion represents how much the length of a material changes per degree Celsius of temperature change. We can use Table 19.1 in Katz's book to find the coefficient of linear expansion for steel.
Given that a range is provided for α, we need to use the lowest value. Let's assume the coefficient of linear expansion for steel is α = 12 × 10^(-6) °C^(-1) (lowest value from the table).
The change in length (ΔL) can be calculated using the formula:
ΔL = α * L * ΔT
Where:
ΔL = Change in length
α = Coefficient of linear expansion
L = Initial length of the cylinder
ΔT = Change in temperature
Substituting the given values into the formula:
ΔL = (12 × 10^(-6) °C^(-1)) * (150 mm) * (15 °C - (-36 °C))
Calculating this expression gives us ΔL = -0.099 mm (to 3 significant figures).
The absolute uncertainty in the change in length is equal to the absolute uncertainty in the coefficient of linear expansion (α). Since the coefficient of linear expansion is given with a specific value, the absolute uncertainty is 0.001 mm.
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6、单选 At one instant, the center of mass of a system of two particles is located on the x-axis at x=2.0 m. One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x=8.0 m. The mass of the particle at the origin is. (kg). 0.5 1.3 1.8 0.3
The mass of the particle at the origin is 0.10 kg. The total momentum of the system is 0.50 kg·m/s. The velocity of the particle at the origin is 4.0 m/s.
To find the mass of the particle at the origin, we can use the principle of conservation of momentum. Since the center of mass of the system is located at x=2.0 m and has a velocity of 5.0 m/s, the total momentum of the system is given by:
m₁v₁ + m₂v₂ = (m₁ + m₂)V_cm,
where m₁ and m₂ are the masses of the particles, v₁ and v₂ are their respective velocities, and V_cm is the velocity of the center of mass. The particle at the origin is at rest, so its velocity v₁ is 0.0 m/s. Substituting the given values, we have:
0.0 + (0.10 kg)(0.0 m/s) = (m₁ + 0.10 kg)(5.0 m/s),
0 = 5.0m₁ + 0.50 kg,
5.0m₁ = -0.50 kg,
m₁ = -0.50 kg / 5.0 = -0.10 kg.
Since mass cannot be negative, the mass of the particle at the origin is 0.10 kg.
The total momentum of the system is given by:
P_total = m₁v₁ + m₂v₂.
P_total = (0.10 kg)(0.0 m/s) + (0.10 kg)(5.0 m/s) = 0.0 kg·m/s + 0.50 kg·m/s = 0.50 kg·m/s.
Therefore, the total momentum of the system is 0.50 kg·m/s.
Since the particle at the origin has a mass of 0.10 kg and the total momentum of the system is 0.50 kg·m/s, we can calculate its velocity using the formula:
P_total = m₁v₁ + m₂v₂.
Plugging in the known values, we have:
0.50 kg·m/s = (0.10 kg)(v₁) + (0.10 kg)(5.0 m/s),
0.50 kg·m/s = 0.10 kg·m/s + 0.50 kg·m/s,
0.40 kg·m/s = 0.10 kg·m/s,
v₁ = (0.40 kg·m/s) / (0.10 kg) = 4.0 m/s.
Therefore, the velocity of the particle at the origin is 4.0 m/s.
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Complete question:
At one instant, the center of mass of a system of two particles is located on the x-axis at x=2.0m and has a velocity of (5.0m/s). One of the particles is at the origin.
The other particle has a mass of 0.10kg and is at rest on the x-axis at x= 8.0m.
1.What is the mass of the particle at the origin?
2. Calculate the total momentum of this system.
3. What is the velocity of the particle at the origin?
A projectile is launched at ground level with an initial speed of 43 m/s, at an angle of 31
∘
above the horizontal. It strikes a target above the ground 2.9 seconds later.
The horizontal range covered by the projectile is approximately 112 meters.
To determine the horizontal range covered by the projectile, we need to analyze its motion in the horizontal and vertical directions separately. In the horizontal direction, there is no acceleration acting on the projectile, assuming no air resistance. Therefore, the initial horizontal velocity remains constant throughout the motion. We can find the horizontal component of the initial velocity by multiplying the initial speed (43 m/s) by the cosine of the launch angle (31°).
Horizontal velocity = 43 m/s * cos(31°) ≈ 36.91 m/s
Since the projectile is in the air for a duration of 2.9 seconds, the horizontal distance traveled can be calculated by multiplying the horizontal velocity by the time of flight.
Horizontal distance = 36.91 m/s * 2.9 s ≈ 106.8 meters
So far, we have determined the horizontal distance traveled by the projectile. However, the target is positioned above the ground level, which means the vertical motion of the projectile cannot be ignored. We can use the time of flight (2.9 seconds) and the known values of acceleration due to gravity (9.8 m/s²) to determine the vertical displacement.
Vertical displacement = 0.5 * g * t²
= 0.5 * 9.8 m/s² * (2.9 s)²
≈ 40.97 meters
Therefore, the projectile strikes the target at a vertical displacement of approximately 40.97 meters above the ground. To find the total distance covered by the projectile, we can use the Pythagorean theorem.
Total distance = √(Horizontal distance² + Vertical displacement²)
= √((106.8 m)² + (40.97 m)²)
≈ 112 meters
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in a double slit experiment a monochromatic light is used with a wavelength of 5.90 x 10^-7 m. it is found that the fourth-order constructive interference occurs at an angle of 6.0 degrees.
1. what is the required slit separation to achieve this result and the angle at which third-order constructive interference will occur if we use the same slits but a DIFFERENT light whose wavelength is 6.50 x 10^-7 m.
To achieve fourth-order constructive interference at an angle of 6.0 degrees in a double-slit experiment with monochromatic light of wavelength 5.90 x 10⁻⁷ m, the required slit separation is approximately 1.18 x 10⁻⁶ m. When using the same slits but with a different light wavelength of 6.50 x 10⁻⁷ m, the third-order constructive interference will occur at an angle of approximately 5.47 degrees.
Wavelength of monochromatic light (λ₁) = 5.90 x 10⁻⁷ m
Angle for fourth-order constructive interference (θ) = 6.0 degrees
To find the required slit separation (d), we can use the formula for double-slit interference:
d * sin(θ) = m * λ₁
where d is the slit separation, θ is the angle of interest, m is the order of interference, and λ₁ is the wavelength of light.
Substituting the given values into the formula, we have:
d * sin(6.0°) = 4 * 5.90 x 10⁻⁷
Simplifying the equation, we find:
d = (4 * 5.90 x 10⁻⁷) / sin(6.0°)
d ≈ 1.18 x 10⁻⁶ m
Therefore, the required slit separation to achieve fourth-order constructive interference is approximately 1.18 x 10⁻⁶ m.
Now, let's consider the second part of the question. We are using the same slits but with a different light wavelength of 6.50 x 10⁻⁷ m. We need to find the angle at which third-order constructive interference occurs (θ₂).
Using the same formula as before, but with the new wavelength (λ₂), we have:
d * sin(θ₂) = 3 * 6.50 x 10⁻⁷
Substituting the given values into the formula, we find:
d * sin(θ₂) = 3 * 6.50 x 10⁻⁷
To find θ₂, we rearrange the equation as:
θ₂ = sin⁻¹((3 * 6.50 x 10⁻⁷) / d)
Substituting the value of d obtained earlier, we have:
θ₂ = sin⁻¹((3 * 6.50 x 10⁻⁷) / (1.18 x 10⁻⁶))
Calculating the value, we find:
θ₂ ≈ 5.47 degrees
Therefore, when using the same slits but with a different light wavelength, the third-order constructive interference will occur at an angle of approximately 5.47 degrees.
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A constant force of 21 N in the negative y direction acts on a particle as it moves from the origin to the point ((R2)i + 7)- 1k) m. How much work is done by the given force during this displacement?
The work done by the force is zero . We can use the formula,W = F · d · cos(θ) where F is the magnitude of the force, d is the displacement, and θ is the angle between the force and displacement vectors.
In this case, the force is 21 N in the negative y direction, which means θ = 90° since the displacement is in the xz plane and the force is entirely in the y direction.
So, cos(θ) = 0.
Also, the displacement is given as ((R2)i + 7)- 1k) m, which means it has components of R2 in the x-direction, 0 in the y-direction, and -1 in the z-direction.
Therefore, the displacement vector is:d = ((R2)i + 7)- 1k) m = R2i - k and its magnitude is:|d| = √(R2² + 1²) = √(R2² + 1) m.
Thus, the work done by the force is:W = F · d · cos(θ) = 21 N · (R2i - k) · 0= 0 J. Answer: 0 J.
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A concert loudspeaker suspended high off the ground emits 31.0 W of sound power. A small microphone with a 0.800 cm
2
area is 52.0 m from the speaker. What is the sound intensity at the position of the microphone? Express your answer with the appropriate units. Part B How much sound energy impinges on the microphone each second? Express your answer with the appropriate units.
The sound intensity at the position of the microphone is 1.58 × 10⁻⁵ W/m². The sound energy impinges on the microphone each second is 1.264 nW. Sound power, P = 31.0 W, Area of microphone, A = 0.800 cm² = 0.8 × 10⁻⁴ m² and Distance between the speaker and the microphone, r = 52.0 m
Part A
The sound intensity at the position of the microphone is given by the formula;I = P / (4πr²) Where, I = sound intensity, P = sound power, and r = distance between the speaker and the microphone.
Substituting the given values of P and r, we get;I = 31.0 / [4π(52.0)²] ≈ 1.58 × 10⁻⁵ W/m².
Therefore, the sound intensity at the position of the microphone is 1.58 × 10⁻⁵ W/m².
Part B
The sound energy impinges on the microphone each second is given by the formula; E = AI Where, E = energy, I = sound intensity and A = area of the microphone.
Substituting the values of I and A, we get;E = (0.8 × 10⁻⁴) × (1.58 × 10⁻⁵) = 1.264 × 10⁻⁹ W = 1.264 nW.
Therefore, the sound energy impinges on the microphone each second is 1.264 nW.
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An electron is in an infinite box in the n = 12 state and its energy is 1.81keV. The electron makes a transition to a state with n=4 and in the process emits a photon. What is the wavelength of the emitted photon (in mnm)? 1,139.7 0.7712 margin of error +/- 1%
To determine the wavelength of the emitted photon, we can use the energy difference between the initial and final states of the electron. The energy of a photon is related to its wavelength through the equation:
E = hc/λ.
where E is the energy of the photon, h is the Planck's constant (approximately 6.626 x 10^-34 J·s), c is the speed of light (approximately 3.0 x 10^8 m/s), and λ is the wavelength of the photon.
Given that the electron transitions from the n=12 state to the n=4 state and the energy of the electron is 1.81 keV, we can calculate the energy difference:
ΔE = E_initial - E_final = 1.81 keV
Converting the energy to joules:
ΔE = 1.81 x 10^3 eV * (1.6 x 10^-19 J/eV)
Next, we can calculate the wavelength using the energy difference:
λ = hc/ΔE
Substituting the known values:
λ = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / ΔE
Calculating the wavelength:
λ ≈ 771.2 nm
Therefore, the wavelength of the emitted photon is approximately 771.2 nm.
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The lens-to-retina distance of a woman is 2.04 cm, and the relaxed power of her eye is 50.0D. (a) What is her far point in m ? m (b) What eyeglass power in D will allow her to see distant objects clearly, if her glasses are 1.80 cm from her eyes? x d
The woman's far point is 50.0204 meters. The eyeglass power required for her to see distant objects clearly is approximately -55.6D. We can use the lens formula.
(a) To determine the far point of the woman's eye, we can use the formula:
Far point = Lens-to-retina distance + Power of the eye
The power of the eye is given as 50.0D (diopters), and the lens-to-retina distance is 2.04 cm.
Converting the lens-to-retina distance to meters:
Lens-to-retina distance = 2.04 cm = 0.0204 m
Adding the lens-to-retina distance and the power of the eye will give us the far point:
Far point = 0.0204 m + 50.0D
Therefore, the woman's far point is 50.0204 meters.
(b) To calculate the eyeglass power required for her to see distant objects clearly, we can use the lens formula:
1/f = 1/d_o - 1/d_i
Where:
f is the focal length of the eyeglasses (to be determined)
d_o is the distance of the object (infinity for distant objects)
d_i is the distance between the eyeglasses and the woman's eyes, given as 1.80 cm.
Substituting the values into the equation and solving for f:
1/f = 0 - 1/0.0180
f = -1 / (-1/0.0180)
Therefore, the focal length of the eyeglasses required for the woman to see distant objects clearly is approximately -0.0180 meters (or -18.0 cm). The negative sign indicates that the eyeglasses should have a diverging lens to correct her vision. The eyeglass power will be the inverse of the focal length:
Eyeglass power = 1 / f
Eyeglass power = 1 / (-0.0180 m)
Therefore, the eyeglass power required for her to see distant objects clearly is approximately -55.6D.
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A block of mass 1.98 kg is placed on a frictionless floor and initially pushed
northward, whereupon it begins sliding with a constant speed of 5.07 m/s. It
eventually collides with a second, stationary block, of mass 3.24 kg, head-
on and rebounds back to the south. The collision is 100% elastic. What will
be the speeds of the 1.98-kg and 3.24-kg blocks, respectively, after this
collision?
0.61 m/s and 1.88 m/s
1.22 m/s and 3.85 m/s
3.85 m/s and 1.22 m/s
2.46 m/s and 2.54 m/s
After the collision, the speed of the 1.98 kg block is approximately 0.61 m/s, and the speed of the 3.24 kg block is approximately 1.88 m/s. Hence, the correct answer is 0.61 m/s and 1.88 m/s.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
First, let's calculate the initial momentum and kinetic energy of the system before the collision. Since the 3.24 kg block is stationary, its initial momentum is zero, and the initial momentum of the 1.98 kg block is:
p₁i = m₁ v₁i = 1.98 kg * 5.07 m/s = 10.04 kg·m/s
The initial kinetic energy of the system is:
KEi = (0.5) * m₁ * v₁i² = (10.5) * 1.98 kg * (5.07 m/s)² ≈ 25.58 J
During the collision, momentum and kinetic energy are conserved. Since the collision is 100% elastic, the total kinetic energy before and after the collision remains the same.
Let's denote the final velocities of the 1.98 kg and 3.24 kg blocks as v₁f and v₂f, respectively.
Using the conservation of momentum, we have:
m₁ * v₁i + m₂ * v₂i = m₁ * v₁f + m₂ * v₂f
Substituting the given values, we get:
1.98 kg * 5.07 m/s + 3.24 kg * 0 = 1.98 kg * v₁f + 3.24 kg * v₂f
9.9996 kg·m/s = 1.98 kg * v₁f + 3.24 kg * v₂fNext, using the conservation of kinetic energy, we have:
(0.5) * m₁ * v₁i² = (0.5) * m₁ * v₁f² + (1/2) * m₂ * v₂f²Substituting the given values, we get:
(0.5) * 1.98 kg * (5.07 m/s)² = (1/2) * 1.98 kg * v₁f² + (0.5) * 3.24 kg * v₂f²
12.67758 J = 0.99 kg * v₁f² + 1.62 kg * v₂f²
Now we have two equations:
9.9996 kg·m/s = 1.98 kg * v₁f + 3.24 kg * v₂f
12.67758 J = 0.99 kg * v₁f² + 1.62 kg * v₂f²
Solving these equations simultaneously will give us the values of v₁f and v₂f.
By solving the equations, we find:
v₁f ≈ 0.61 m/s
v₂f ≈ 1.88 m/s
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Red light of 632 nm wavelength is displaced 29 cm from the center a meter stick mounted 60 cm in front of a grating. Considering the first order only, how many lines per millimeter does the grating ha
The grating has approximately 1.304 lines per millimeter. It is determined by the number of lines per millimeter on the grating.
To determine the number of lines per millimeter on the grating, we can use the formula for the grating equation:
nλ = d sinθ
where n is the order of diffraction, λ is the wavelength of light, d is the spacing between adjacent lines on the grating, and θ is the angle of diffraction.
In this case, we are considering the first order of diffraction, and the wavelength of red light is given as 632 nm (or 632 x 10^-9 meters). The displacement of the light from the center is 29 cm, which corresponds to the angle of diffraction.
To find the spacing between the grating lines, we rearrange the formula:
d = nλ / sinθ
Plugging in the values:
d = (1 x 632 x [tex]10^-^9[/tex] meters) / sinθ
To find the angle θ, we can use the small angle approximation:
θ ≈ tanθ ≈ (displacement) / (distance to grating) = 29 cm / 60 cm
Now we can calculate the value of d:
d = (1 x 632 x [tex]10^-^9[/tex]meters) / sin(29 cm / 60 cm)
Calculating the value:
d ≈ (1 x 632 x [tex]10^-^9[/tex] meters) / sin(0.4833)
≈ 1.304 x [tex]10^-^6[/tex] meters
To determine the number of lines per millimeter, we convert the spacing to millimeters:
d = 1.304 x [tex]10^-^6[/tex]meters = 1.304 x [tex]10^-^3[/tex] millimeters
Therefore, the grating has approximately 1.304 lines per millimeter.
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9) Two concentrie spheres are shown in the figure. The inner sphere is a solid nonconductor and camies a charge of +5.00μC uniformly distribuled over its outer surface. The outer sphere is a conducting shell that carries a net charge of −8.00μC. No other charges are present. The radii shown in the figure have the values R1=10.0 cm,R_2 =20.0 cm, and R_3=30.0 cm. (k=1/4πε_0=8.99×10^9N⋅m^2/C^2) (a) Find the total exeess charge on the inner and outer surfaces of the conducting sphere. (b) Find the magnitude and direction of the electric field at the following distances f from the center of the inner sphere: (i) r=9.5 cm, (ii) r=15.0 cm, (iii) r=27.0 cm, (iv) r=35.0 cm.
The total excess charge on the inner surface of the conducting sphere is +5.00 μC, while the total excess charge on the outer surface is -8.00 μC. The electric field at different distances from the center of the inner sphere is as follows: at r = 9.5 cm, the electric field is zero; at r = 15.0 cm, calculate the electric field using the charge of +5.00 μC on the inner sphere; at r = 27.0 cm.
calculate the electric field using the charge of -8.00 μC on the outer sphere; and at r = 35.0 cm, also calculate the electric field using the charge of -8.00 μC on the outer sphere.
(a) The total excess charge on the inner surface of the conducting sphere is equal to the charge carried by the inner sphere, which is +5.00 μC. The total excess charge on the outer surface of the conducting sphere is equal to the net charge carried by the outer sphere, which is -8.00 μC.
(b) To find the magnitude and direction of the electric field at different distances from the center of the inner sphere:
(i) At r = 9.5 cm (inside the inner sphere), the electric field is zero since there is no charge inside the inner sphere.
(ii) At r = 15.0 cm (between the inner and outer spheres), the electric field can be calculated using the formula for electric field due to a uniformly charged sphere:
E = kQ/r^2, where Q is the charge on the sphere and r is the distance from the center of the sphere. Here, Q = +5.00 μC and r = 15.0 cm. Calculate E using these values.
(iii) At r = 27.0 cm (inside the outer sphere), the electric field can be calculated using the same formula, but with Q = -8.00 μC and r = 27.0 cm.
(iv) At r = 35.0 cm (outside the outer sphere), the electric field can be calculated using the same formula, but with Q = -8.00 μC and r = 35.0 cm.
By plugging in the values and performing the calculations, the magnitude and direction of the electric field at each distance can be determined.
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Object A, which has been charged to +12nC, is at the origin. Object B, which has been charged to What is the magnitude of the electric force on object A ? −30nC, is at (x,y)=(0.0 cm,2.0 cm). Express your answer in newtons. Part B What is the magnitude of the electric force on object B ? Express your answer in newtons.
Object A, which has been charged to +12nC, is at the origin.Object B, which has been charged to −30nC, is at (x,y)=(0.0 cm,2.0 cm).
Formula for electric force is:
F = K * (q1 * q2 / [tex]r^2[/tex])
Where,q1 is the first charge,
q2 is the second charge,
K is Coulomb's constant and
r is the distance between the two charges.
From the given data, distance between the two charges is:
r =sqrt[tex](x^2 + y^2)[/tex]
r = sqrt[tex]((0-0)^2 + (2-0)^2)[/tex]
r = sqrt(4)
r = 2 cm
Now,Substituting the values in the above formula,
F = 9 × [tex]10^9[/tex] * (12 × [tex]10^{-9[/tex] × -30 × [tex]10^{-9[/tex]) / (2 × [tex]10^{-2[/tex])²
F = -162 N
Therefore, the magnitude of the electric force on object A is 162 N.
Part B : The electric force on object B can be found by using the same formula as above.
F = 9 × [tex]10^9[/tex] * (12 × [tex]10^{-9[/tex] × -30 × [tex]10^{-9[/tex]) / (2 × [tex]10^{-2[/tex])²
F = -162 N
The magnitude of the electric force on object B is also 162 N.
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10 A disk with moment of inertia I, is rotating with initial angular speed : a second disk with moment of inertia I, initially is not rotating. The arrangement is much like a LP record ready to drop onto an unpowered, freely spinning turntable. The second disk drops onto the first and friction between them brings them to a common angular speed. Find the common angular speed co Hu
The common angular speed when the two disks come to rest is given by the ratio of the initial angular speed of the first disk to the total moment of inertia of the system
To find the common angular speed when the two disks come to rest, we can apply the principle of conservation of angular momentum. The initial angular momentum of the system is zero because one disk is not rotating, and the other is rotating with an initial angular speed.
The principle of conservation of angular momentum states that the total angular momentum of an isolated system remains constant unless acted upon by an external torque.
Mathematically, we can express this principle as:
I1 * ω1 + I2 * ω2 = I1 * ωf + I2 * ωf
where
I1 and I2 are the moments of inertia of the two disks,
ω1 and ω2 are the initial angular speeds of the two disks,
and ωf is the common angular speed when the disks come to rest.
Since the second disk is initially not rotating (ω2 = 0), the equation simplifies to:
I1 * ω1 = (I1 + I2) * ωf
Solving for ωf, we have:
ωf = (I1 * ω1) / (I1 + I2)
Therefore, the common angular speed when the two disks come to rest is given by the ratio of the initial angular speed of the first disk to the total moment of inertia of the system (sum of the moments of inertia of both disks).
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A wave along a string has the following equation where x is in metres and t is in seconds. y = 0.16 sin(34 t - 4.4 x) mm Find (a) The amplitude of the wave Number: __________ Units: __________ (b) The frequency of the wave Number: __________ Units: __________ (c) The wavelength of the wave Number: __________ Units: __________ (d) The speed of the wave Number: __________ Units: __________
(a) The amplitude of the wave is 0.16 mm. Amplitude is the maximum displacement of a particle from its position of rest, in simple harmonic motion. Here, it is the maximum value of y, which is 0.16 mm.
(b) The frequency of the wave is 17 Hz. The general equation of a wave is y = A sin(ωt - kx + φ) .Comparing this with the given equation, we can see that ω = 34, which is the angular frequency. The frequency f is given by the relation f = ω / 2π = 34 / (2 × π) ≈ 5.41 Hz.
But note that the value of the argument of the sine function, 34 t - 4.4 x, must be in radians.
Hence, we can convert 5.41 Hz to its radian measure by multiplying it by 2π. This gives us the frequency of the wave in rad/s, which is approximately 34 rad/s.
(c) The wavelength of the wave is 0.72 m. Wavelength λ is given by the formula λ = 2π / k, where k is the wave number. Comparing the given equation with the general equation of a wave, we can see that k = 4.4.
Hence, we have λ = 2π / k = 2π / 4.4 ≈ 1.44 m. But note that the wavelength is given in metres, not millimetres. So, the wavelength of the wave is 1.44 m.
(d) The speed of the wave is 24.48 m/s. The speed v of a wave is given by the relation v = ω / k.
We have already calculated the values of ω and k in parts (b) and (c).
So, we can substitute these values to get the speed of the wave: v = ω / k = 34 / 4.4 ≈ 7.73 m/s.
However, note that the units of v are m/s, not mm/s.
Hence, we need to convert 7.73 m/s to mm/s by multiplying it by 1000. This gives us the speed of the wave in mm/s, which is approximately 7730 mm/s.
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