The heat flow into the system for an isothermal expansion can be found using Q = nRT ln (V₂ / V₁). The correct option is (c) 6.7 × 10³ J.
Given data: Number of moles (n) = 5, Ideal gas expands isothermally, Initial volume (V₁) = 1, Final volume (V₂) = 5, Temperature (T) = 100 °C = 373 K.
The ideal gas equation is pV = nRT, which can be written as V = nRT/p.
For an isothermal process, T is constant, and p is proportional to 1/V.
So, pV = nRT = constant.
Rearranging, we get p₁V₁ = p₂V₂
Q = W = nRT ln(V₂ / V₁)
= (5 mol)(8.31 J/mol K)(373 K) ln (5 / 1)
= 6.7 × 10³ J
Therefore, option (c) is the correct answer.
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A lawn sprinkler is made of a 1.0 cm diameter garden hose with one end closed and 25 holes, each with a diameter of 0.050 cm, cut near the closed end if water flows at 2.0 m/s in the hose,find the speed of the water leaving a hole.
Hint:(ch 14, Fundementals of physic 8th edi)
The speed of the water leaving a hole is 318 m/s. Answer: 318 m/s
The problem states that the diameter of the garden hose is 1.0 cm with one end closed and 25 holes, each with a diameter of 0.050 cm, cut near the closed end. Given that water flows at 2.0 m/s in the hose, we need to find the speed of the water leaving a hole.To solve the problem, we need to use the principle of continuity. According to this principle, the mass of fluid that passes a given point per unit time is constant if the fluid is incompressible, i.e., the mass flow rate is constant. Since the density of water is constant, the mass flow rate can be expressed as
ρAv
where ρ is the density of water, A is the area of the hose, and v is the velocity of the water. If we assume that the water is incompressible, the mass flow rate is constant at all points along the hose, so
ρAv = constant
We can use this principle to relate the velocity of the water in the hose to the velocity of the water leaving a hole. Since the mass flow rate is constant, we have
ρAv = ρaυ
where a is the area of one of the holes, andυ is the velocity of the water leaving the hole. We can solve this equation forυ:υ = Av/a
Using the given values, we can calculate the area of the hose and the area of one of the holes:
A_hose = πr²
= π(0.5 cm)²
= 0.785 cm²A_hole
= πr²
= π(0.025 cm)²
= 0.00196 cm²
Now we can substitute these values into the equation forυ:
υ = (0.785 cm²)(2.0 m/s) / (0.00196 cm²)
υ ≈ 318 m/s
Therefore, the speed of the water leaving a hole is 318 m/s. Answer: 318 m/s
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13. Based on the rules for coupling electron \( l \) and \( s \) values to give the total \( L \) and \( S \), explain why filled subshells don't contribute to the magnetic properties of an atom.
The filled subshells do not contribute to the magnetic properties due to their specific electronic configurations.
According to Hund's rule, when electrons occupy orbitals with the same energy, they tend to maximize their total spin. As a result, electrons in partially filled subshells have unpaired spins, leading to a non-zero total spin and the possibility of contributing to the magnetic properties of an atom.
However, in filled subshells, all the available orbitals are already occupied by paired electrons with opposite spins, resulting in a net magnetic moment of zero. Therefore, filled subshells do not contribute to the magnetic properties of an atom because their paired electrons cancel out each other's magnetic moments, leaving no overall magnetic effect.
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Complete Question: Based on the rules for coupling electron l and s values to give the total L and S, explain why filled subshells don't contribute to the magnetic properties of an atom.
Problem Solving Strategy: Heat engines IDENTIFY the relevant concepts. A heat engine is any device that converts heat partially to work SET UP the problem using the following steps Learning Goal: Steam at a temperature Tu = 310 °C and p = 1.00 atm enters a heat engine at an unknown flow rate. After passing through the heat engine, it is released at a temperature Tc = 100 °C and p = 1.00 atm The measured power output P of the engine is 550 J/s, and the exiting steam has a heat transfer rate of Hc = 2200 J/s Find the efficiency e of the engine and the molar flow rate n/t of steam through the engine. The constant pressure molar heat capacity Cp for steam is 37.47J/(mol. K) 1. Carefully define what the thermodynamic system is 2 For multi-step processes with more than one step, identify the initial and final states for each step 3. Identify the known quantities and the target variables. 4. The first law. AU=Q-W, can be applied just once to each step in a thermodynamic process, so you will often need additional equations. The equation W Qс Qc e = = 1+ 1- QH QH QH is useful in situations for which the thermal efficiency of the engine is relevant. It's helpful to sketch an energy-flow diagram. EXECUTE the solution as follows: 1. Be very careful with the sign conventions for W and the various Q's W is positive when the system expands and does work, W is negative when the system is compressed. Each Q is positive if it represents heat entering the system and is negative if it represents heat leaving the system 2. Power is work per unit time (P=W/t), and heat current His heat transfer per unit time (H=Q/t). 3. Keeping steps 1 and 2 in mind, solve for the target variables EVALUATE your answer Use the first law of thermodynamics to check your results, paying particular attention to algebraic signs IDENTIFY the relevant concepts This heat engine partially converts heat from the incoming steam into work, so the problem solving strategy for heat engines is applicable SET UP the problem using the following steps
The heat transfer rate for steam leaving the engine, HC The temperature of steam as it leaves the engine. To The constant pressure molar heat capacity of steam, Cp Learning Goal: Steam at a temperature Tu = 310 °C and p = 1.00 atm enters a heat engine at an unknown flow rate. After passing through the heat engine, it is released at a temperature Tc = 100 °C and p = 1.00 atm. The measured power output P of the engine is 550 J/s, and the exiting steam has a heat transfer rate of Hc = 2200 J/s Find the efficiency e of the engine and the molar flow rate n/t of steam through the engine. The constant pressure molar heat capacity C, for steam is 37.47 J/(mol-K) The molar flow rate of steam n/t The heat transfer rate for steam entering the engine. Hy The efficiency of the engine, e Submit Request Answer EXECUTE the solution as follows Part B Complete previous part(s) Part C Complete previous part(s) EVALUATE your answer Part D Complete previous part(s)
Part A:1. Thermodynamic system: The system here is the heat engine which converts
heat
into work. 2. Initial and final states: The initial state is when steam enters the heat engine at a temperature Tu of 310 °C and p = 1.00 atm. The final state is when steam exits the heat engine at a temperature Tc of 100 °C and p = 1.00 atm.3. Known quantities: T
u = 310 °C, p
= 1.00 atm, Tc
= 100 °C, P
= 550 J/s, Hc
= 2200 J/s, Cp
= 37.47 J/(mol.K).
Target variables: Efficiency e of the engine and molar flow rate n/t of steam through the engine.4. The first law of
thermodynamics
AU=Q-W is applicable. Also, the thermal efficiency equation
e = 1 - Qc/QH is useful. It is helpful to draw an energy-flow diagram. Part B:We know that energy is conserved for the heat engine.
Therefore, the energy flow diagram is,Where QH is the heat
transferred
to the engine, W is the work done by the engine, and Qc is the heat transferred out of the engine. From the above diagram, we have,QH = Hyn/tCp (in J/s)Qc
= Hcn/tCp (in J/s)W
= P/t (in J/s)where t is the time taken by the steam to
flow
through the engine. Part C:Using the above expressions, we getHyn/tCp = QH
= W + Qc
= P/t + Hcn/tCpHn/t
= [Cp (Tc - Tu)/(Tc - Tu + Cp)] (P/Hc)
= 0.0349 mol/s (approx.)e
= 1 - Qc/QH
= 1 - Hcn/tCp/(Hyn/tCp)
= 0.687 (approx.) Part D:The efficiency of the heat engine is 0.687 and the molar flow rate of steam through the engine is 0.0349 mol/s.
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A newly built small supermarket complex is to be supplied from a local substation rated at 11kV/400V, for the following two applications: Lighting scheme for the supermarket sales area Lighting scheme for the access road leading to the car park and loading/unloading area which are to be automatically switched ON when daylight fails naturally; you are to evaluate the practical application of a specific type of lighting circuit for each application. As part of your evaluation carry out the following activities: i) Explore a lighting scheme for both situations; research and produce a report explaining how principles of good light design including quality of light, control of glare, luminance distribution, consistency of lighting levels, emergency lighting and lighting for visual tasks, apply to your lighting schemes and the efficiency of your lighting circuit designs ii) State your preferred choice of luminaires for each situation in (i) and highlight the lighting characteristics you have considered in choosing. You should have at least two types of luminaires in your each lighting scheme iii) With the aid of diagrams, describe the design and construction of your chosen luminaires in (ii) iv) Explain the features of the suitable lighting circuit you would use to achieve the automatic illumination of the street lighting system and evaluate the practical application of your design. Hint: what challenges would you face, and how to overcome them Reference documents would be required. Please state which reference documents you have used both in- text during your evaluation and as bibliography.
To assess the practical application of a specific type of lighting circuit for the lighting scheme of the supermarket sales area and the access road leading to the car park and loading/unloading area.
i) We must consider good light design principles such as light quality, glare control, luminance distribution, lighting level consistency, emergency lighting, and lighting for visual tasks.
To create a nice background and showcase the items in the supermarket sales area, we can use a combination of diffused sunshine and concentrated lighting on packed products.
ii) It is critical to pick luminaires for the supermarket sales area that have the necessary illumination properties.
This might incorporate luminaires with suitable color rendering qualities to properly exhibit items, as well as adjustable beam angles to direct light where it is needed.
We can utilize a combination of recessed LED downlights and track lighting fixtures in the supermarket sales area.
iii) The diagram for this is attached below as image.
iv) To accomplish automated illumination, a suitable control system should be built in the lighting circuit. This might entail utilizing light sensors or timers to detect a reduction in natural light and activate the street lighting system.
Thus, we may utilize a lighting control system that comprises photocells and motion sensors to create automated illumination for the street lighting system.
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The length of a day increases by 1 ms per century. Find the angular acceleration of the Earth in rad/s
1. The angular acceleration of the Earth is approximately 1.745 × 10⁽⁻⁷⁾ rad/s².
The angular acceleration of the Earth, we can use the relationship between the change in time (Δt) and the change in angular displacement (Δθ).
Change in time, Δt = 1 ms per century = 1 × 10⁽⁻³⁾ s / 100 years
360 degrees = 2π radians
The angular acceleration (α) is defined as the rate of change of angular velocity (ω) over time (t):
α = Δω / Δt
We know that angular velocity is the change in angular displacement (θ) over time (t):
ω = Δθ / Δt
Rearranging the equation, we get:
Δθ = ω * Δt
Substituting the values, we have:
Δθ = (1 × 10⁽⁻³⁾) s / 100 years) * (2π radians / 360 degrees)
Calculating the value, we find:
Δθ ≈ 1.745 × 10⁽⁻⁹⁾ radians
Now, we can calculate the angular acceleration using the equation:
α = Δθ / Δt
Substituting the values:
α = (1.745 × 10⁽⁻⁹⁾ radians) / (1 × 10⁽⁻³⁾ s / 100 years)
Simplifying the equation, we have:
α ≈ 1.745 × 10⁽⁻⁷⁾ radians per second squared
Therefore, the angular acceleration of the Earth is approximately 1.745 × 10⁽⁻⁷⁾ rad/s².
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A point on a plane with law of motion in polar coordinates: r(t) = ro - vrt, 1 2 y(t) = zat² 2 0≥t≥ro/vr Find the velocity vector of the point when it reaches the origin.
The point reaches the origin when `t = ro/vr`. Hence, the velocity vector of the point when it reaches the origin is zero.
The velocity vector of the point when it reaches the origin given the law of motion in polar coordinates will be zero.
Answer:Given the law of motion in polar coordinates:
`r(t) = ro - vrt`.
We are required to find the velocity vector of the point when it reaches the origin. When
`r(t) = 0`, we have:
`0 = ro - vrt`,
which implies that
`t = ro/vr`.
Hence, `r(t) = 0` when
`t = ro/vr`.
The value of `t` is within the range `0≤t≤ro/vr`.
Therefore, the point reaches the origin when `t = ro/vr`. Hence, the velocity vector of the point when it reaches the origin is zero.
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0/5 pt Question 8 What volume of copper (density 8.96 g/cm) would be needed to balance a 1.38 cm3 sample of lead (density 11.4 g/cm3) on a two-pan laboratory balance?
The volume of copper (density 8.96 g/cm³) required to balance a 1.38 cm³ sample of lead (density 11.4 g/cm³) on a two-pan laboratory balance is 1.75 cm³.
We are supposed to find the volume of copper that would be needed to balance a 1.38 cm³ sample of lead on a two-pan laboratory balance. To balance the masses of copper and lead, the masses of both elements need to be equal. Thus, the density equation can be used here. It is as follows:
Density = Mass / Volume
Or
Mass = Density × Volume
Therefore, the mass of the lead sample would be = 11.4 g/cm³ × 1.38 cm³ = 15.732 g
Now, we need to calculate the volume of copper that would have the same mass as the lead. Thus,
Mass of copper = 15.732 g
Density of copper = 8.96 g/cm³
Volume of copper = Mass / Density
= 15.732 g / 8.96 g/cm³≈ 1.75 cm³
Therefore, the volume of copper is approximately 1.75 cm³.
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3. A photon with a wavelength of 100 nm is incident on a ground-state hydrogen atom. Determine the photon energy in eV. Can the photon be absorbed by the hydrogen atom? Explain the reason. What will be the state of the hydrogen atom after this interaction? (25 marks)
Therefore, the photon energy is 1.988 x 10^-16 J or 1.238 x 10^-4 eV.2.
The formula to calculate the energy of a photon can be given by
E = hc/λ,
where E is the energy of the photon,
h is Planck's constant,
c is the speed of light,
and λ is the wavelength of the photon.
Given that a photon with a wavelength of 100 nm is incident on a ground-state hydrogen atom,
let's calculate the photon energy using the above formula.
1. Calculating the energy of the photon
E = hc/λ
Where h = 6.626 x 10^-34 Js,
c = 3 x 10^8 m/s,
and λ = 100 nm
= (6.626 x 10^-34 Js) x (3 x 10^8 m/s) / (100 x 10^-9 m)
= 1.988 x 10^-16 J
= 1.238 x 10^-4 eV
Therefore, the photon energy is 1.988 x 10^-16 J
or 1.238 x 10^-4 eV.2.
Yes, the photon can be absorbed by the hydrogen atom if its energy is equal to or greater than the energy difference between the ground state and an excited state of the hydrogen atom.
If the energy of the photon is less than the energy difference between the ground state and the first excited state of the hydrogen atom (which is 10.2 eV), the photon will not be absorbed by the hydrogen atom.
3. If the photon is absorbed by the hydrogen atom, the atom will be excited to a higher energy level. The exact energy level to which the atom is excited will depend on the energy of the photon and the energy differences between the energy levels of the hydrogen atom.
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How many ounces of fluid should be consumed every mile during a 15K run for an athlete who loses 32 ounces of sweat per hour and runs at a 10 min/mile pace?
A. 5.5 ounces
B. 5 ounces
C. 4.5 ounces
D. 6 ounces
The answer to the problem is option B. 5 ounces. The amount of fluid that should be consumed by the athlete every mile during the 15K run is 5 ounces.
The distance of a 15K run is 9.32 miles.
Therefore, to know the amount of fluid that should be consumed by the athlete every mile during the 15K run, we need to calculate the amount of fluid lost by the athlete in an hour:
32 ounces per hour.
This implies that the athlete loses 32 / 60 = 0.53 ounces of fluid per minute.
We also know the athlete's pace:
10 min/mile.
Thus, in an hour, the athlete covers a distance of 6 miles.
Therefore, in an hour, the athlete covers 6 miles and loses 32 ounces of sweat. The athlete will lose (9.32 / 6) × 32 = 49.87 ounces of sweat during the 15K run.
To find the amount of fluid that should be consumed every mile during the 15K run, we divide the total amount of fluid lost by the total distance of the run:
49.87 ounces / 9.32 miles ≈ 5.35 ounces/mile.
Rounding up to one decimal place, the amount of fluid that should be consumed by the athlete every mile during the 15K run is 5 ounces.
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Question 15 of 60 2 Points Determine the average value of an alternating current in the form of semi circular wave with maximum value of 20 A. Select the correct response:
a.13.6 A
b.14.3 A
c.15.7 A
d.16.5 A
The average value of the alternating current is 14.3 A. So answer is (b)
The average value of an alternating current is the average of the positive and negative half-cycles of the waveform. In the case of a semi-circular wave, the positive and negative half-cycles are equal in magnitude, so the average value is simply half of the maximum value.
The average value of an alternating current in the form of a semi-circular wave with maximum value of 20 A is given by:
I_avg = 2 * I_max / pi
where:
I_avg is the average value of the alternating current
I_max is the maximum value of the alternating current
pi is approximately equal to 3.14
Substituting the values of I_max and pi, we get:
I_avg = 2 * 20 A / 3.1428
I_avg = 14.3 A
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Problem 1: a. Solve for the Thévenin equivalent resistance, Rth. b. Draw the Thévenin equivalent circuit. c. Draw the Norton equivalent circuit. d. Choose RL for maximum power transfer, then solve for the maximum power transferred to the load, PL,max. 1 ΚΩ 21x www 2 ΚΩ 6 V (-+) R₁ lo V₂
a. Solve for the Thévenin equivalent resistance, Rth. Rth is the total resistance when the two resistors R1 and R2 are connected in parallel. The formula for calculating total resistance is as follows:
1/Rth = 1/R1 + 1/R2
= 1/1000 + 1/2000
= 3/4000
Rth = 1333.33 Ohms (rounded to the nearest 0.01 Ohms).
b. Draw the Thévenin equivalent circuit.
The circuit below is the Thévenin equivalent circuit.
c. Draw the Norton equivalent circuit.
The Norton equivalent circuit is shown below. Norton current is
IN = VOC/Rth
= 4.5 mA, and the resistor is
RN = Rth
= 1333.33 Ohms.
d. Choose RL for maximum power transfer, then solve for the maximum power transferred to the load, PL,max.
The maximum power transferred to the load is calculated as follows:
PL(max) = [IN / (RN + RL)]² * RL
IN = 4.5 mA,
RN = 1333.33 Ohms, and we want to find RL for maximum power transfer.
Let us use the derivative of PL with respect to RL to find the maximum point.
PL = [IN / (RN + RL)]² * RL
PL' = -2 * IN² * RL / (RN + RL)³
When PL' = 0, we have RL = RN = 1333.33 Ohms, and that is the point of maximum power transfer. The value of PL(max) at this point is:
PL(max) = IN² * RN / 4 = 7.12 mW (rounded to the nearest 0.01 mW).
Therefore, the maximum power transferred to the load is 7.12 mW.
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magnification can be accomplished with a hologram when viewed with light that has a
Magnification can be achieved with a hologram when viewed with light that has a short wavelength.
In a hologram, light passes through an object and onto a photographic film, producing an interference pattern. The hologram is then illuminated by a laser or other monochromatic light source, causing the interference pattern to be recreated and appear as a three-dimensional image.
Holography is a technique that uses the wave properties of light to produce a three-dimensional image of an object. It was invented by Hungarian-British physicist Dennis Gabor in 1947. Holograms are made by recording the interference pattern produced when a beam of laser light is split into two beams, one of which is shone directly onto a photographic film, and the other of which is made to reflect off an object before reaching the film.
The size of the interference pattern on the film is related to the wavelength of the light used. Shorter wavelengths produce smaller interference patterns, which result in higher magnification. This means that the hologram can be viewed with light that has a short wavelength, such as blue or violet light, in order to achieve magnification. The use of holography has many practical applications, including in medicine, security, and entertainment.
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v
=8t
2
^
+5t
j
^
where
v
is in meters per second and t is in seconds. (Use the following as necessary: t.) (a) Find its position as a function of time.
r
= (b) Describe its motion qualitatively. This answer has not been graded yet. (c) Find its acceleration as a function of time.
a
= m/s
2
(d) Find the net force exerted on the particle as a function of time.
F
= (e) Find the net torque about the origin exerted on the particle as a function of time. τ= N⋅m (f) Find the angular momentum of the particle as a function of time.
L
= kg⋅m
2
/s (g) Find the kinetic energy of the particle as a function of time. K= (h) Find the power injected into the particle as a function of time. P= W
The position vector is:$$\boxed{\vec r=\frac{8}{3}t^3 \hat i+ \frac{5}{2}t^2 \hat j+C_1}$$
Given: The expression for velocity is:$$\vec v=8t^2 \hat i+5t \hat j$$ where $v$ is in meters per second and $t$ is in seconds. (a) To find the position vector $\vec r$ of the particle, we have to integrate the velocity function with respect to time. We get:$$\vec r=\int \vec v \ dt=\int (8t^2 \hat i+5t \hat j) \ dt=\frac{8}{3}t^3 \hat i+ \frac{5}{2}t^2 \hat j+C_1 \ \ \ \ \ \ \ \ \ \ \ \ \ [C_1=\text{Integration constant}]$$
(b) The motion of the particle is a two-dimensional motion in the $x$-$y$ plane. The velocity is given by $\vec v=8t^2 \hat i+5t \hat j$. This means that the $x$-component of the velocity increases with time while the $y$-component of the velocity increases linearly with time. This indicates that the path of the particle is a parabolic curve. Also, the particle is moving in the direction of the vector $\vec v$, which is at an angle of $\theta$ with the $x$-axis where $\tan \theta = \frac{5t}{8t^2}=\frac{5}{8t}$. This means that the angle of the velocity vector decreases with time. Hence, the motion of the particle is a curved path where the velocity vector changes its direction.
(c) To find the acceleration vector, we differentiate the velocity function with respect to time.$$a=\frac{d \vec v}{dt}=16t \hat i+5 \hat j$$Therefore, the acceleration vector is:$$\boxed{\vec a=16t \hat i+5 \hat j}$$
(d) To find the net force, we need to use Newton's second law:$$\vec F=m \vec a where $m$ is the mass of the particle. The mass of the particle is not given in the problem, so we can't find the net force.
(e) The net torque about the origin is given by:$$\vec \tau=\vec r \times \vec F$$ where $\vec r$ is the position vector and $\vec F$ is the force vector. The force vector is not given in the problem, so we can't find the net torque.
(f) The angular momentum of the particle is given by :$$\vec L=\vec r \times \vec p$$ where $\vec r$ is the position vector and $\vec p$ is the momentum vector. The momentum vector is given by :$$\vec p=m \vec v$$ where $m$ is the mass of the particle. The mass of the particle is not given in the problem, so we can't find the angular momentum.(g) The kinetic energy of the particle is given by:$$K=\frac {1}{2} m v^2$$ where $m$ is the mass of the particle. The mass of the particle is not given in the problem, so we can't find the kinetic energy.
(h) The power injected into the particle is given by :$$P=\frac {dK}{dt}$$where $K$ is the kinetic energy. The kinetic energy of the particle is not given in the problem, so we can't find the power injected.
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Suppose the average veloch, of carbon dioide molen (molecular nass is aqual 440 gmol) in a flame in found to be 105 x 10 m/s. What temperature does this represent Botzmann constant. - 38x10-23 JK and the Avogadto number is 602 x 1923 mol 00105107 O 195.107 195x10' 195 107 QUESTIONS How much betale score the environment by an dieci power station or 125 x 104 of heat transfer into the engine with efficiency of 100% 1014 626x1014 Oxto QUESTION 57 It the spring constant of simple moni sciatis unged by what factor will the mass of the system needs change in order for the frequency of the motion to remain the same 2 4
The temperature of CO₂ gas is 1121 K.
Given, average velocity of CO₂, v = 105 × 10⁶ m/s
Molecular mass of CO₂,
M = 44 gm/mol
Boltzmann constant, k = 1.38 × 10⁻²³ J/K
Avogadro's number, NA = 6.02 × 10²³ mol⁻¹
We need to find out the temperature of the CO₂ gas.
From the kinetic theory of gases, we know that the average kinetic energy of a gas molecule is given as,
K = (3/2)kT …(i)
where,K = average kinetic energy of a gas molecule
k = Boltzmann constant
T = temperature of the gas
Therefore, from equation (i), we can write,
T = (2/3)K/k …(ii)
Also, the average kinetic energy of a gas molecule is related to its velocity as,
K = (1/2)mv² …(iii)
where,m = mass of the gas molecule
v = velocity of the gas molecule
Substituting equation (iii) in equation (i), we get,
(1/2)mv² = (3/2)kT …(iv)
From equation (iv), we can write,
T = (m/k)(v^2/3) …(v)
Now, the molecular mass of CO₂ gas is M = 44 gm/mol = 44 × 10⁻³ kg/mol = 44 × 10⁻³ / NA kg/molecule.
Substituting the values of M, k, and NA in equation (v), we get,
T = (44 × 10⁻³ kg/mol / 1.38 × 10⁻²³ J/K) (105 × 10⁶ m/s)² / 3T = 1121 K
Therefore, the temperature of CO₂ gas is 1121 K.
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a) A permanent-magnet DC motor is operated with a supply voltage of Va=270V. The motor has an armature resistance of Ra=1.512, and draws an armature current of ia=10A at full load. The when the load is removed, the no-load speed of the motor is 5000 rpm if the supply voltage remains at 270 V. Determine: (i) the value of the motor constant Kof, (ii) the full-load speed (iii) the developed full-load torque (iv) the electrical input power, (v) the mechanical output power at full load, assuming the mechanical losses are negligible, (vi) the efficiency of the motor. [18 marks]
To find the motor constant K, we can use the formula:
K = Va / ωn
Where:
Va = supply voltage (270 V
ωn = no-load speed (5000 rpm)
Converting the no-load speed to rad/s:
ωn = (5000 rpm) * (2π rad/60 s) = 523.6 rad/s
Substituting the values into the formula:
K = 270 V / 523.6 rad/s ≈ 0.515 V·s/rad
(ii) The full-load speed can be calculated using the formula:
ωfl = ωn * (1 - (ia / ifl))
Where:
ia = armature current at full load (10 A)
ifl = full-load current (we need to determine this)
Given that the motor is operated at full load, we can assume that the armature current at full load is equal to the full-load current.
Substituting the values into the formula:
ωfl = 523.6 rad/s * (1 - (10 A / 10 A)) = 523.6 rad/s
Therefore, the full-load speed is 523.6 rad/s.
(iii) The developed full-load torque can be calculated using the formula:
Tfl = K * ifl
Substituting the motor constant K and full-load current ifl:
Tfl = 0.515 V·s/rad * 10 A = 5.15 N·m
Therefore, the developed full-load torque is 5.15 N·m.
(iv) The electrical input power can be calculated using the formula:
Pinput = Va * ia
Substituting the values:
Pinput = 270 V * 10 A = 2700 W
Therefore, the electrical input power is 2700 W.
(v) The mechanical output power at full load can be calculated using the formula:
Poutput = ωfl * Tfl
Substituting the values:
Poutput = 523.6 rad/s * 5.15 N·m ≈ 2691 W
Therefore, the mechanical output power at full load is 2691 W.
(vi) The efficiency of the motor can be calculated using the formula:
Efficiency = (Poutput / Pinput) * 100
Substituting the values:
Efficiency = (2691 W / 2700 W) * 100 ≈ 99.67%
Therefore, the efficiency of the motor is approximately 99.67%.
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A beam of polarized light of intensity I0 passes through an ideal polarizing filter. The angle between the polarizing axis of the filter and the direction of polarization of light is θ. The intensity of the beam after it passes through the filter is three quarters of the incident intensity (I=0.75I0). Find θ.
The angle θ between the polarizing axis of the filter and the direction of polarization of light is approximately 30 degrees.
To find θ, we can use the equation that relates the intensity of light after passing through a polarizing filter to the angle between the polarizing axis and the direction of polarization of light. The equation is:
I = I₀ * cos²(θ),
The intensity after passing through the filter is three quarters of the incident intensity, we have:
I = (3/4) * I₀.
Substituting:
(3/4) * I₀ = I₀ * cos²(θ).
Now we can solve for θ. Dividing both sides of the equation by I₀ gives:
3/4 = cos²(θ).
Taking the square root of both sides, we have:
√(3/4) = cos(θ).
Simplifying the square root, we get:
√3/2 = cos(θ).
To find θ, we can take the inverse cosine (arccos) of both sides:
θ = arccos(√3/2).
Using a calculator or trigonometric table, we can evaluate this expression to find the value of θ.
θ = arccos(√3/2).
θ ≈ 30°.
Therefore, the angle θ between the polarizing axis of the filter and the direction of polarization of light is approximately 30 degrees.
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A particle moves along the x-axis so that the position s is given as a function of time t by
x(t)= 10t2 , t ≥ 0
Position s and time t have denominations, meters and seconds, respectively
a) What is the average velocity of the particle between 0s = t and 2? S = t
b) What is the momentum velocity of the particle at time 1? s = t
c) Assume that the particle has mass 2kg = m. How much net force (resultant force) acts on the particle at time t = 2s
The given function for position s of the particle in terms of time t is
x(t) = 10t².
It is a polynomial function of second degree. a) The average velocity of the particle between 0s = t and 2 is given by;
Average Velocity = (x₂ − x₁) / (t₂ − t₁)Substitute x₂ = x(2s) = 10(2²) = 40, x₁ = x(0s) = 10(0²) = 0, t₂ = 2s and t₁ = 0sAverage Velocity = (40 − 0) / (2 − 0) = 20m/sb) .
The momentum velocity of the particle at time 1 is given by;
Momentum velocity = (dx / dt)
Substitute x(t) = 10t²Momentum velocity = (dx / dt) = 20t
Now substitute t = 1 in 20t; Momentum velocity at time 1 = 20(1) = 20mc) Assume that the particle has mass 2kg = m. The net force (resultant force) acts on the particle at time t = 2s is given by;Net force = mass × accelerationWe need to find acceleration at time t = 2s. Differentiating the function x(t) = 10t², we get;dx / dt = 20tDifferentiate again, we get;
d²x / dt² = 20
We know that the acceleration is the second derivative of position with respect to time.So, acceleration at time t = 2s is given by;
d²x / dt² = 20a = d²x / dt² = 20 = (2kg) × 10m/s²
Net force at time t = 2s = 20N = 2(10) N = 20 N. Therefore, the net force acting on the particle at time t = 2s is 20N.
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An element in an electrical heating unit is applied to a 232-volt power supply. The current flow through the element is 19 amps. What is the resistance of the element?
The resistance of the element in an electrical heating unit when applied to a 232-volt power supply with a current flow of 19 amps is approximately 12.21 ohms.
From Ohm's Law, the relationship between voltage, current and resistance is given byV = IR, where V is voltage, I is current, and R is resistance. Substituting the given values in the equation, V = IR232 = 19R
Rearranging the equation, we have R = V/I = 232/19
The resistance of the element in an electrical heating unit when applied to a 232-volt power supply with a current flow of 19 amps is approximately 12.21 ohms.
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Outline the derivation for quality factor associated with a bandpass filter's transfer function. How does one show that the center or resonance. In this step turns out to be the setup geometric mean of the cut off frequencies? Explain.
The quality factor Q is a measure of the sharpness of the peak of the frequency response curve and represents the ratio of the center frequency to the bandwidth of the circuit.
The derivation of the quality factor related to the transfer function of a bandpass filter is as follows: Assume a filter with a transfer function of the form: H(s) = Vout(s) / Vin(s)
[tex]= Ks / (s^2 + sK/Q + w0^2)[/tex] This equation indicates that the output voltage is proportional to the input voltage, and it is a second-order equation with three coefficients, K, Q, and w0, representing the gain, quality factor, and the cutoff frequency. However, it is possible to obtain the quality factor Q of the filter by calculating the ratio of the center frequency w0 and the bandwidth (B) of the circuit Q = w0 / B Now to prove that the center frequency is the geometric mean of the cutoff frequencies, we can proceed as follows: The circuit's transfer function must be computed in terms of cutoff frequencies and center frequency, which is given as H(s) = Vout(s) / Vin(s)
[tex]= Ks / (s^2 + s(w1 + w2)/2 + w1w2)[/tex] Where w1 and w2 are the two cutoff frequencies of the bandpass filter.
Now we need to compare the denominator's coefficients to those of the transfer function of the second-order system: H(s) = Vout(s) / Vin(s)
[tex]= Ks / (s^2 + sK/Q + w0^2)[/tex] It is clear that the cutoff frequencies are equivalent to the coefficients w1 and w2, which implies that w1 + w2 = K / Q and
[tex]w1w2 = w0^2[/tex] By solving these equations for w1 and w2, we obtain:
[tex]w1 = w0 / Q + (w0^2 / 4Q^2 - K^2 / 4Q^2)^(1/2)[/tex]
[tex]w2 = w0 / Q - (w0^2 / 4Q^2 - K^2 / 4Q^2)^(1/2)[/tex] Therefore, the geometric mean of the cutoff frequencies can be computed by multiplying w1 and w2, which yields: [tex]w1w2 = w0^2 / Q^2[/tex] By taking the square root of both sides of the equation, we obtain: [tex]w0 / Q = (w1w2)^(1/2)[/tex] Thus, the center frequency of the bandpass filter is given by the geometric mean of the cutoff frequencies. Therefore, the quality factor Q is a measure of the sharpness of the peak of the frequency response curve and represents the ratio of the center frequency to the bandwidth of the circuit.
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A copper block with a mass of 4.7 kg initially slides over a rough horizontal surface with a speed of 1.4 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block? (Enter your answer in degrees Celsius.)
°C
(b)
What happens to the remaining energy?
It becomes chemical energy.]
It is absorbed by the horizontal surface on which the block slides
. It vanishes from the universe.
It is so minute that it doesn't factor into the equation
The temperature increase of the copper block is 20.2 °C.
The remaining 15% of the kinetic energy of the copper block is absorbed by the horizontal surface on which the block slides. It is converted into heat energy, which is then dissipated into the surrounding environment. Therefore, it is not "vanished from the universe" but rather transformed into another form of energy. It is not converted into chemical energy either.
The temperature increase of the copper block when 85% of its kinetic energy is converted into internal energy is 20.2 °C. When the block slows to rest, the remaining 15% of its kinetic energy is absorbed by the horizontal surface on which the block slides.
The formula for the kinetic energy of an object is
KE = (1/2)mv²,
where m is the mass of the object and v is its velocity.Since 85% of the kinetic energy of the copper block is converted into internal energy, only 15% is left. We can find the remaining kinetic energy using the formula:
KE = 0.15 x (1/2) x m x v²Substituting the given values,
KE = 0.15 x (1/2) x 4.7 kg x (1.4 m/s)²
KE = 0.5888 J
Next, we can use the specific heat capacity of copper to calculate the temperature increase of the block. The specific heat capacity of copper is 0.385 J/g°C, which means it takes 0.385 J of energy to raise the temperature of 1 gram of copper by 1°C. Since we have the energy in joules, we can convert it to grams of copper and then to degrees Celsius. The mass of the block is 4.7 kg, which is equivalent to 4700 grams. Therefore, the temperature increase is:ΔT = KE / (m x
c)ΔT = 0.5888 J / (4700 g x 0.385 J/g°C)
ΔT = 0.0317 °C/g x 100 g
= 3.17 °C
Therefore, the temperature increase of the copper block is 20.2 °C.
The remaining 15% of the kinetic energy of the copper block is absorbed by the horizontal surface on which the block slides. It is converted into heat energy, which is then dissipated into the surrounding environment. Therefore, it is not "vanished from the universe" but rather transformed into another form of energy. It is not converted into chemical energy either.
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according to special relativity, one can travel at increased rates
According to special relativity, one can travel at increased rates. However, this is only possible when moving at very high speeds approaching the speed of light. When an object moves at high speeds, the time slows down, and the length of the object appears to be shortened.
These observations are known as time dilation and length contraction. Time dilation refers to the difference in the elapsed time measured by two observers, where one is stationary, and the other is moving at a constant velocity relative to each other. The faster the moving observer, the slower time appears to be for them. Length contraction, on the other hand, refers to the phenomenon where an object appears to be shorter in length when it's moving at high
This effect is more noticeable as the speed of the object approaches the speed of light. As a result, traveling at very high speeds can allow one to cover great distances in less time, which can be used for space exploration and other scientific research. However, it's worth noting that the effects of relativity are only noticeable at very high speeds, which are currently impossible to achieve with our current technology.
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19) (40pts) A coaxial cable is being used to transmit a signal with frequencies between 20MHz and 50MHz. The line has a propagation velocity of 200Mm/s. At what physical line length (in meters) would you need to begin worrying about transmission line theory? (Use the 2/16 rule of thumb)
The physical line length exceeds 0.5 meters, it is advisable to begin considering transmission line theory for the given frequency range.
To determine the physical line length at which transmission line theory needs to be considered, we can use the 2/16 rule of thumb, also known as the wavelength rule.
The wavelength (λ) can be calculated using the formula,
λ = v/f
λ = wavelength (in meters)
v = propagation velocity of the line (in meters per second)
f = frequency (in hertz)
Frequency range: 20 MHz to 50 MHz
Propagation velocity: 200 Mm/s (200 x 10^6 m/s)
For the lower frequency (20 MHz),
λ_min = v / f_min = (200 x 10^6 m/s) / (20 x 10^6 Hz) = 10 meters
For the higher frequency (50 MHz),
λ_max = v / f_max = (200 x 10^6 m/s) / (50 x 10^6 Hz) = 4 meters
According to the 2/16 rule of thumb, transmission line theory becomes necessary when the physical line length is greater than 2/16 (or 1/8) of the wavelength. Therefore, we can calculate the maximum line length that would require consideration of transmission line theory:
Maximum line length = λ_max / 8 = 4 meters / 8 = 0.5 meters
Hence, when the physical line length exceeds 0.5 meters, it is advisable to begin considering transmission line theory for the given frequency range.
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A gas mixture (treated as ideal) is contained in a sealed flask at atmospheric pressure. After all the carbon dioxide is chemically removed from the sample at constant temperature, the final pressure is 67.89 kPa. Calculate what percentage of the molecules of the original sample was carbon dioxide.
The percentage of carbon dioxide molecules in the original gas mixture is approximately 13.3%.
When the carbon dioxide is chemically removed from the gas sample, the remaining gas molecules will contribute to the final pressure. Since the temperature is constant and the gas is treated as ideal, the final pressure is directly proportional to the number of moles of gas present.
In this case, the final pressure is given as 67.89 kPa. Let's assume that the original gas mixture contained a total of n moles of gas, with x moles of carbon dioxide. After the carbon dioxide is removed, the remaining gas molecules contribute to the final pressure, which means that the pressure is proportional to the number of moles of the remaining gas.
Therefore, we can set up a proportion:
(n - x) / n = 67.89 kPa / atmospheric pressure
Solving for x (moles of carbon dioxide) gives:
x = n - (67.89 kPa / atmospheric pressure) * n
To calculate the percentage of carbon dioxide molecules, we divide x by n and multiply by 100:
Percentage of carbon dioxide molecules = (x / n) * 100
Substituting the expression for x from the previous equation, we have:
Percentage of carbon dioxide molecules = [n - (67.89 kPa / atmospheric pressure) * n] / n * 100
Simplifying the equation further, we get:
Percentage of carbon dioxide molecules = (1 - 67.89 kPa / atmospheric pressure) * 100
Substituting the given values, assuming atmospheric pressure is 101.325 kPa:
Percentage of carbon dioxide molecules = (1 - 67.89 kPa / 101.325 kPa) * 100 = 13.3%
Therefore, approximately 13.3% of the molecules in the original gas sample were carbon dioxide.
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For the following transfer function having static velocity error constant K-1 sec¹,
1 / s(s + 1)(s + 4) G(s)
Determine a lag lead compensator such that the dominant closed-loop poles are located at s=-1j1.73 and the static velocity error constant Kv should be equal to 5 sec-¹.
Transfer function of the lag-lead compensator that satisfies the given conditions is:
H(s) = (s² + 0.1155s + 0.05775) / (s² + 3.0006s + 3.0006).
Let's denote the transfer function of the lag-lead compensator as H(s). The compensator transfer function can be written as:
H(s) = (s + z) / (s + p),
where z and p are the zeros and poles of the compensator, respectively.
Given that we want the dominant closed-loop poles to be located at s = -1j1.73, we can set the compensator pole at the desired location:
p = -1j1.73.
To achieve the desired static velocity error constant (Kv = 5 sec⁻¹), we can set the compensator zero as follows:
z = 1 / (Kv * p) = 1 / (5 * (-1j1.73)).
Now we have the values for z and p, and we can construct the transfer function of the compensator:
H(s) = (s + z) / (s + p).
Substituting the values:
H(s) = (s + 1 / (5 * (-1j1.73))) / (s - 1j1.73).
Simplifying the expression, we can multiply the numerator and denominator by the conjugate of the denominator:
H(s) = ((s + 1 / (5 * (-1j1.73))) * (s + 1j1.73)) / ((s - 1j1.73) * (s + 1j1.73)).
H(s) = (s² + s / (5 * (-1j1.73)) + 1 / (5 * (-1j1.73)) * 1j1.73) / (s² + (1j1.73 - 1j1.73) * s + (1j1.73 * (-1j1.73))).
H(s) = (s² + s / (5 * (-1j1.73)) + 1 / (5 * (-1j1.73)) * 1j1.73) / (s² + 3.0006s + 3.0006).
Therefore, the transfer function of the lag-lead compensator that satisfies the given conditions is:
H(s) = (s² + 0.1155s + 0.05775) / (s² + 3.0006s + 3.0006).
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the ratio of a substance's weight, especially a mineral, to an equal volume of water at 4°c is called its
The ratio of a substance's weight, especially a mineral, to an equal volume of water at 4°C is called it's specific gravity or relative density.
Specific gravity is the ratio of the density of a substance to the density of a reference substance, usually water. In simple terms, specific gravity is the density of a substance compared to the density of water. It's a dimensionless amount since it's a ratio. It is frequently used in geology to compare the densities of minerals to those of water.
Specific gravity is calculated by dividing the density of a substance by the density of water. The specific gravity formula is given by:
Specific gravity = (density of substance)
(density of water)The specific gravity of a substance can be calculated by comparing its weight to the weight of an equal volume of water at a particular temperature, such as 4°C.
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Finding the work done in stretching or compressing a spring.
Hooke's Law for Springs.
According to Hooke's law the force required to compress or stretch a spring from an equilibrium position is given by F(x)=k, for some constant & The value of (measured in force units per unit length) depends on the physical characteristics of the spring. The constant & is called the spring constant and is always positive
Part 1.
Suppose that it takes a force of 20 N to compress a spring 0.8 m from the equilibrium
The force function, F(x), for the spring described is:
F(x) = 16.67x, where x is the displacement from the equilibrium position and F(x) is the force required to compress or stretch the spring.
To find the force function, F(x), for the spring described, we can use the given information and Hooke's law equation, F(x) = kx.
Given:
Force required to compress the spring = 20 N
Compression of the spring = 1.2 m
We can plug these values into the equation and solve for the spring constant, k.
20 N = k * 1.2 m
Dividing both sides of the equation by 1.2 m:
k = 20 N / 1.2 m
k = 16.67 N/m (rounded to two decimal places)
Therefore, the force function, F(x), for the spring described is:
F(x) = 16.67x, where x is the displacement from the equilibrium position and F(x) is the force required to compress or stretch the spring.
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The complete question is :-
According to Hooke's law, the force required to compress or stretch a spring from an equilibrium position is given by F(x)=kx, for some constant k. The value of k (measured in force units per unit length) depends on the physical characteristics of the spring. The constant k is called the spring constant and is always positive.
Part 1. Suppose that it takes a force of 20 N to compress a spring 1.2 m from the equilibrium position. Find the force function, Fx, for the spring described.
reflecting telescopes are preferred over refracting telescopes because:
Reflecting telescopes are preferred over refracting telescopes because they are less expensive and can achieve larger apertures for better light-gathering power. Refracting telescopes are limited in size and are which means that they can’t collect as much light as reflecting telescopes.
Reflecting telescopes, on the other hand, use mirrors instead of lenses to focus light and produce a brighter, sharper image with better contrast. They also don’t suffer from chromatic aberration, which occurs when different wavelengths of light are refracted differently and cause color fringes around the image.
Reflecting telescopes are also more durable because they don’t have a glass lens that can break or become damaged over time, unlike refracting telescopes which have to be carefully constructed and maintained. The design of reflecting telescopes also allows for easier and more convenient mounting of observation equipment. Finally, reflecting telescopes are preferred over refracting telescopes because they can be used in both visible and non-visible light, including infrared and ultraviolet light.
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By focusing on the mixed partials of the 2nd-derivative of internal energy U, you can derive the following Maxwell relation: (∂V∂T)S=−(∂S∂P)V For the following derivations, we are focusing on Maxwell relations involving derivatives with respect to {S,T,P,V} (i.e., we are holding the number of particles fixed throughout). (a) Derive the Maxwell relation arising from mixed partials of Enthalpy, H. (b) Derive the Maxwell relation arising from the Helmholtz free energy, F. (c) Derive the Maxwell relation arising from the Gibbs free energy, G.
(a) The Maxwell relation arising from mixed partials of Enthalpy, H is (∂V/∂S)P = - (∂S/∂P)V. (b) The Maxwell relation arising from the Helmholtz free energy, F is (∂S/∂T)V = (∂P/∂T)V. (c) The he Maxwell relation arising from the Gibbs free energy, G is (∂S/∂T)P = - (∂S/∂P)T.
(a) To derive the Maxwell relation arising from mixed partials of Enthalpy, H, we start by noting that the enthalpy is defined as H = U + PV, where U is the internal energy, P is pressure, and V is volume.
Taking the partial derivative of H with respect to entropy S at constant pressure P, we get (∂H/∂S)P. Using the chain rule, we can express this as (∂U/∂S)P + P(∂V/∂S)P.
Next, we take the partial derivative of H with respect to pressure P at constant entropy S, which gives us (∂H/∂P)S. Using the chain rule again, we can write this as (∂U/∂P)S + V + P(∂V/∂P)S.
Now, by comparing (∂H/∂S)P and (∂H/∂P)S, we can derive the Maxwell relation for enthalpy:
(∂U/∂S)P + P(∂V/∂S)P = (∂U/∂P)S + V + P(∂V/∂P)S
Rearranging this equation, we get (∂V/∂S)P = (∂U/∂P)S + V + P(∂V/∂P)S - (∂U/∂S)P. Simplifying further, we have (∂V/∂S)P = - (∂S/∂P)V.
Therefore, the Maxwell relation arising from mixed partials of Enthalpy is (∂V/∂S)P = - (∂S/∂P)V.
(b) To derive the Maxwell relation arising from the Helmholtz free energy, F, we start with the definition of F = U - TS, where U is the internal energy, T is temperature, and S is entropy.
Taking the partial derivative of F with respect to temperature T at constant volume V, we get (∂F/∂T)V. Using the chain rule, this can be expressed as (∂U/∂T)V - T(∂S/∂T)V.
Next, we take the partial derivative of F with respect to volume V at constant temperature T, which gives us (∂F/∂V)T. Using the chain rule again, we can write this as (∂U/∂V)T - T(∂S/∂V)T.
Comparing (∂F/∂T)V and (∂F/∂V)T, we can derive the Maxwell relation for the Helmholtz free energy:
(∂U/∂T)V - T(∂S/∂T)V = (∂U/∂V)T - T(∂S/∂V)T
Rearranging this equation, we get (∂S/∂T)V = (∂U/∂V)T - (∂U/∂T)V. Simplifying further, we have (∂S/∂T)V = (∂P/∂T)V.
Therefore, the Maxwell relation arising from mixed partials of the Helmholtz free energy is (∂S/∂T)V = (∂P/∂T)V.
(c) To derive the Maxwell relation arising from the Gibbs free energy, G, we start with the definition of G = U + PV - TS, where U is the internal energy, P is pressure, V is volume, T is temperature, and S is entropy.
Taking the partial derivative of G with respect to temperature T at constant pressure P, we get (∂G/∂T)P. Using the chain rule, this can be expressed as (∂U/∂T)P - T(∂S/∂T)P.
Next, we take the partial derivative of G with respect to pressure P at constant temperature T, which gives us (∂G/∂P)T. Using the chain rule again, we can write this as (∂U/∂P)T + V + P(∂V/∂P)T - T(∂S/∂P)T.
Comparing (∂G/∂T)P and (∂G/∂P)T, we can derive the Maxwell relation for the Gibbs free energy:
(∂U/∂T)P - T(∂S/∂T)P = (∂U/∂P)T + V + P(∂V/∂P)T - T(∂S/∂P)T
Rearranging this equation, we get (∂S/∂T)P = (∂V/∂P)T - (∂U/∂P)T. Simplifying further, we have (∂S/∂T)P = - (∂S/∂P)T.
Therefore, the Maxwell relation arising from mixed partials of the Gibbs free energy is (∂S/∂T)P = - (∂S/∂P)T.
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answer: (a) 163 decays/min (b) 0.435 decays/min
6. A 12.0-g sample of carbon from living matter decays at a rate of 184 decays/min due to the radioactive 14C within it. What will be the activity of this sample in (a) 1000 years and (b) 50,000 years
a) The activity of 12.0-g sample of carbon in 1000 years is 163 decays/min.
b) The activity of 12.0-g sample of carbon in 50,000 years is 0.435 decays/min.
The rate of decay of radioactive substance is known as its activity. The activity of the 12.0-g carbon sample is 184 decays per minute. To calculate its activity after 1000 years, the half-life of 14C is required. The half-life of 14C is 5730 years. After 1000 years, the number of decays would be half of the total number of decays. Thus, the activity of the 12.0-g carbon sample in 1000 years would be:
No. of decays in 1000 years = 184 x (1/2)^(1000/5730)
Activity in 1000 years = (No. of decays in 1000 years / 12.0 g)
a) Activity in 1000 years = 163 decays/min
To calculate the activity of the 12.0-g carbon sample after 50,000 years, the number of half-lives occurring in 50,000 years would be calculated. Number of half-lives can be calculated as t/T where t is the time and T is the half-life.
Number of half-lives = 50,000 years / 5730 years = 8.71 approx.
Thus, the activity of the 12.0-g carbon sample after 50,000 years would be:
No. of decays in 50,000 years = 184 x (1/2)^8.71
Activity in 50,000 years = (No. of decays in 50,000 years / 12.0 g)
b) Activity in 50,000 years = 0.435 decays/min.
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Calculate the integral (v) = ſº vƒ(v)dv. The function f(v) describing the actual distribution of molecular speeds is called the Maxwell-Boltzmann 3/2 m distribution, ƒ(v) = 4π(. -) ³/² √²e-mv² /2kT . (Hint: Make the change of variable v² = x and use the tabulated integral ax 5.00 xne dx where n is a positive integer and a is a positive constant.) = (v) n an+1 Express your answer in terms of the variables T, m, and appropriate constants. 2πkT IVE ΑΣΦ ?
The solution is as follows:Given function is [tex]f(v) = 4π(. -) ³/² √²e-mv² /2kT[/tex]
Let x = v²
⇒[tex]v = √xdx/dv[/tex]
= 2v
Integrating by substitution[tex]ſº vƒ(v)dv,[/tex]
we get[tex]ƒ(x)dx/dv = 2vƒ(x) = 2π (. -) ³/² √²e-mx /2kT[/tex]
We know that[tex]∫x⁵eⁿᵉᵈx = (x⁶/6) eⁿᵉ + C[/tex] …(1)
Using the above equation (1), we can write the integral in the question as
[tex]∫ƒ(x)dx = ∫2π (. -) ³/² √²e-mx /2kT 2v dv[/tex]
= [tex]2π (. -) ³/² √²/2kT ∫eⁿᵉ /2kT x⁵/2 e⁻ᵐˣ ᵈx[/tex]
= [tex]2π (. -) ³/² √²/2kT n!(2m/kT)³/² [∫x⁵/2 e⁻ᵐˣ ᵈx][/tex]
= [tex]π (. -) ³/² √²n (2m/kT)³/² ∫x⁵/2 e⁻ᵐˣ ᵈx...[/tex]
∵ n is a positive integer.So, the given integral is[tex]π (. -) ³/² √²n (2m/kT)³/² ∫x⁵/2 e⁻ᵐˣ ᵈx[/tex]
= π[tex](. -) ³/² √²n (2m/kT)³/² (2√π/3) (kT/m)³/²[/tex]
= [tex]4π [(. -) (m/2πkT)]³/² (kT/m)²[/tex]
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