(a) Field current is calculated as;If = V/ff Rfwhere, V
= 208 V (supply voltage)ff
= 50 Hz (supply frequency)Rf
= 147 Ω (field circuit resistance)Therefore,If
= 208/50*147
= 0.282 A(b) The motor voltage equation is given by,Ea
= KφNwhere,Ea
= V - Ia Raφ is fluxN is the speedK
= 0.7032 V/A rad/sIa
= V1 / Rawhere V1 is the converter output voltage.Rearranging these equations,φ
= (Ea - V1) / KIa
= V1 / RaEa
= KφN + Ia RaV - V1
= KφN + V1 / Ra Ra∴ V1
= (V - KφN Ra ) / (1 + Ra ).
Where,V = 208 VK = 0.7032 V/A rad/sRa
= 0.25 ΩN = 1000 rpm
= 2πN / 60 rad/s≈ 104.67 rad/s Substituting all these values,V1
= (208 - 0.7032 * φ * 104.67 * 0.25) / (1 + 0.25)
= 31.79φHence, Ia
= V1 / Ra
= 31.79/0.25
= 127.16 A The power input to the armature circuit,P
= V1 Ia cos (α)
= 31.79 * 127.16 cos(α)
The load torque TL = 45 Nm
So, α = cos⁻¹ (TL / KIaN)
α = cos⁻¹ (45 / 0.7032 * 127.16 * 104.67)
α = 47.23°(c) The input power factor of armature circuit converters is given as:
PF = cos (α) = cos (47.23°)
= 0.68.
Therefore, the power factor of the armature circuit converters is 0.68.
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3 * 10 ^ 24 molecules of an ideal gas are stored in a 15 litre container at a temperature of 20 deg * C
a) Calculate the absolute pressure in the container at this temperature, rounded to the nearest kilopascal.
(5)
b) The next day, a second measurement is taken which finds that the pressure has fallen by 5% of the previous measurement, while the temperature is now 1.5 deg * C lower. How much of the gas has been removed from the container? Give your answer in moles.
a) The absolute pressure in the container at this temperature is 5.56 * 10⁴ kPa; b) 0.07 moles of gas has been removed from the container.
a) Calculation of absolute pressure: The formula of absolute pressure is given by the ideal gas law formula i.e PV = nRT; where, P = pressure of gas in Pascal (Pa)V = volume of the gas in liters (L)n = number of molecules of gas, R = Universal gas constant which is equal to 8.314 J/K/mol
T = temperature of gas in Kelvin (K)Hence, P = nRT / V, P = (3 * 10²⁴) * 8.314 * (273+20) / (15 * 1000) P
= 5.56 * 10⁷ Pa
≈ 5.56 * 10⁴ kPa
Therefore, the absolute pressure in the container is 5.56 * 10⁴ kPa.
b) Calculation of moles of gas removed: From the ideal gas law PV = nRT, we have; n = PV / RT
Given that the temperature has changed to 1.5 °C lower while the pressure has reduced by 5%, this means that; P₂ = P₁ - 0.05 P₁ = 0.95 P₁ and T₂ = T₁ - 1.5
= 20 - 1.5
= 18.5 °C.
The new pressure (P₂) is given by; P₂ = (0.95 * 5.56 * 10⁷) Pa
= 5.28 * 10⁷ Pa
The new temperature (T₂) in Kelvin is given by T₂ = 18.5 + 273
= 291.5 K
Using the ideal gas law formula again, the number of moles in the gas at initial conditions is given by; n₁ = P₁V / RT₁
Substituting in the values of P₁, V, R and T₁; n₁ = (5.56 * 10⁷ * 15 * 10⁻³) / (8.314 * 293)
= 0.94 mol
Similarly, the number of moles in the gas after the change in temperature and pressure is given by; n₂ = P₂V / RT₂
Substituting in the values of P₁, V, R and T₂; n₂ = (5.28 * 10⁷ * 15 * 10⁻³) / (8.314 * 291.5)
= 0.87 mol
The amount of gas removed is therefore given by the difference between the number of moles in the gas before and after the change; i.e. moles of gas removed = n₁ - n₂
= 0.94 - 0.87
= 0.07 mol
Therefore, 0.07 moles of gas has been removed from the container.
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22. Enthalpy [3P] Consider a process where nitrogen gas with a mass of 2 g and an initial temperature of 27°C undergoes a decrease in pressure by one quarter while the volume stays constant. Determine the enthalpy change of the gas during this process.
The enthalpy change of the gas is 0 J.
According to the first law of thermodynamics, the change in internal energy (ΔU) of a closed system is equal to the heat added to the system (Q) minus the work done by the system (W).
This can be expressed as:
ΔU = Q - W
Since the process in question is isochoric (volume stays constant), the work done by the system is zero. Therefore, the change in internal energy is equal to the heat added to the system. This can be expressed as:
ΔU = Q
Since the nitrogen gas is undergoing a decrease in pressure, it is doing work on the surroundings. This means that the heat added to the system is equal to the work done by the system, but with a negative sign. This can be expressed as:
Q = -W
Plugging in the values, we get:
ΔU = -W = -Q = 0 J
Therefore, the enthalpy change of the gas is 0 J.
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Question 2 In the Davisson-Germer experiment using a Ni crystal, a second-order beam is observed at an angle of 55°. For what accelerating voltage does this occur?
The accelerating voltage for the Davisson-Germer experiment using a Ni crystal is 54.8 V. In the Davisson-Germer experiment, a beam of electrons is incident on a nickel crystal to study their diffraction behavior. This experiment gave a beautiful demonstration of wave-particle duality of electrons.
In the Davisson-Germer experiment, a beam of electrons is incident on a nickel crystal to study their diffraction behavior. This experiment gave a beautiful demonstration of wave-particle duality of electrons. The Ni crystal used in this experiment acts as a diffraction grating, scattering the electrons in various directions to form a diffraction pattern on the detector screen. A second-order beam is observed at an angle of 55°. This means that the electrons in the beam have undergone the second order of diffraction. Using Bragg's law we can relate the angle of diffraction and the interatomic spacing of the crystal.
From this, we can obtain the interatomic spacing of Ni (0.209 nm). Now we can calculate the wavelength of the electron beam by using the de-Broglie relation λ = h/p. where p is the momentum of the electrons and h is the Planck's constant. Using the relation, we get λ = 0.165 nm. Now we can use the relation for accelerating voltage V = h f/ q, where f is the frequency of the electron beam and q is the charge of the electron to obtain the voltage required. Here frequency is given as f = 1/λ. After substituting the values, we get V = 54.8 V. The voltage required to accelerate the electrons in the beam is 54.8 V. Therefore, the accelerating voltage for this experiment is 54.8 V.
Answer: The accelerating voltage for the Davisson-Germer experiment using a Ni crystal is 54.8 V.
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Given that the inputs of two nMOS transistors with W₁/L = 2 and W₂/L = 4 switch simultaneosly. Find the equivalent W/L when the transistors are connected in parallel and series. (4 marks)
The equivalent W/L ratio for the parallel connection is 6, while for the series connection, it is 1.
When transistors are connected in parallel, the total equivalent width (W_eq) is the sum of the individual widths (W) of the transistors, and the equivalent length (L_eq) remains the same.
Given:
Transistor 1: W/L = 2
Transistor 2: W/L = 4
To find the equivalent W/L in parallel, we add up the widths of the transistors:
W_eq = W_1 + W_2 = 2 + 4 = 6
Therefore, the equivalent W/L in parallel is 6/1 = 6.
When transistors are connected in series, the total equivalent length (L_eq) is the sum of the individual lengths (L) of the transistors, and the equivalent width (W_eq) remains the same.
Given:
Transistor 1: W/L = 2
Transistor 2: W/L = 4
To find the equivalent W/L in series, we add up the lengths of the transistors:
L_eq = L_1 + L_2 = 1 + 1 = 2
Therefore, the equivalent W/L in series remains the same: W/L = 2/2 = 1.
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A and B are two reversible Carnot engines which are connected in series working between source temperature of 1500 K and sink temperature of 200 K, respectively. Carnot engine A gets 2000 kJ of heat from the source (maintained at temperature of 1500 K) and rejects heat to second Carnot engine i.e. B. Carnot engine B takes the heat rejected by Carnot engine A and rejects heat to the sink maintained at temperature 200 K. Assuming Carnot engines A and B have same thermal efficiencies, determine: a. Amount of heat rejected by Carnot engine B b. Amount of work done by each Carnot engines i.e. A and B c. Assuming Carnot engines A and B producing same amount of work, calculate the amount of heat received by Carnot B and d. Thermal efficiency of Carnot engines A and B, respectively. c) A flat plate of area = 0.5 m² is pulled at a constant speed of 25 cm/sec placed parallel to another stationary plate located at a distance 0.05 cm. The space between two plates is filled with a fluid of dynamic viscosity =0.004 Ns/m². Calculate the force required to maintain the speed of the plate in the fluid
The force required to maintain the speed of the plate in the fluid is 0.625 N.
a) The amount of heat rejected by Carnot engine B is 1475 kJ.
b) The amount of work done by each Carnot engines i.e. A and B is 125 kJ.
c) The amount of heat received by Carnot B is 125 kJ.
d) The thermal efficiency of Carnot engines A and B, respectively are 83.33% and 41.67% respectively.
Force required to maintain the speed of the plate in the fluid is 0.625 N.
Explanation: Carnot Cycle Formula
The thermal efficiency of Carnot cycle is given by;η = (T1 – T2)/ T1 …….(i)
Where,T1 = temperature of the sourceT2 = temperature of the sink
a) The amount of heat rejected by Carnot engine B is given by;
Q2 = Q1*(T2/T1)Q
1 = 2000 KJQ2
= ?T1
= 1500 KT2
= 200 KQ2
= 2000*(200/1500)
= 267 kJ
Therefore, the amount of heat rejected by Carnot engine B is 267 kJ – 200 kJ = 1475 kJ.
b) The amount of work done by each Carnot engines i.e. A and B is given by;η = 1 – (T2/T1)
Work output = Q1 * η
Work done by engine A,W1 = 2000* (1 – (200/1500)) = 267 kJ
Work done by engine B,W2 = Q2 * η = 1475 * (1 – (200/1500)) = 125 kJ
Therefore, the amount of work done by each Carnot engine i.e. A and B is 125 kJ.
c) The amount of heat received by Carnot B is given by; If both engines produce the same amount of work,
then W1 = W2 = 125 kJ
The amount of heat received by Carnot B, Q2 = W2/η2Q2 = 125/(1 – (200/1500)) = 125 kJ
Therefore, the amount of heat received by Carnot B is 125 kJ.
d) The thermal efficiency of Carnot engines A and B, respectively is given by;η = 1 – (T2/T1)
Carnot engine A,ηA = 1 – (200/1500) = 83.33%
Carnot engine B,ηB = 1 – (200/500) = 41.67%
Therefore, the thermal efficiency of Carnot engines A and B, respectively are 83.33% and 41.67% respectively.
Force required to maintain the speed of the plate in the fluid is given by; F = η*A*(v/d)
Where,η = coefficient of viscosity
A = area = 0.5 m²v = velocity = 25 cm/sec = 0.25 md = distance between plates = 0.05 cm = 0.0005 mη = 0.004 Ns/m²
Therefore, F = 0.004 * 0.5 * 0.25/0.0005 = 0.625 N
Thus, force required to maintain the speed of the plate in the fluid is 0.625 N.
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Lost in the woods by a lake. you carve a block of wood into a cube shape 10 cm wide by 10 cm long by 10 cm high using a part of your body you conveniently identified as 10 cm before you got lost. You mark off the sides of the wood block in 1 cm increments. You want to determine if you'l be able to make a raft of this wood. Wading into the lake with this wood block. you find that 5.5 centimeters of the block stay submerged while the block is floating in water? If the lake water has a density of 1000 kg/m
3
, what is the density of this wood in kg/m
3
to two significant digits?
The density of the wood block in kg/m³ to two significant digits when you are lost in the woods by a lake, is 407 kg/m³.
Here's how to determine the density of the wood:
Volume of the wood block = 10 cm x 10 cm x 10 cm= 1000 cm³
Density = Mass/Volume
Let the mass of the wood be m gm.
To convert m gm to kg, we divide by 1000 i.e m/1000 kg
Volume of wood block in m³ = 1000 cm³ / (100 x 100 x 100) = 0.001 m³
Density = mass / volume 1000 kg/m³ = m / 0.001m³ m = 0.001 m³ x 1000 kg/m³ = 1 kg
So, mass of the wood block is 1 kg
Density of wood = mass of the wood / volume of the wood= 1 kg / 0.00245 m³= 407 kg/m³ (approx).
Therefore, the density of the wood in kg/m³ is 407 kg/m³ to two significant digits.
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What is the Approximate Right Ascension of a full Moon that
occurs in late April
A- 10 Hrs
B-12 Hrs
C- 8 Hrs
D-14 Hrs
Which of the following lists of events in the Moon's monthly
cycle is consecutive
Regarding the consecutive events in the Moon's monthly cycle, the correct answer would be option A- New Moon, First Quarter, Full Moon, Third Quarter.
To determine the approximate right ascension of a full Moon that occurs in late April, we need to consider the position of the Moon in the sky during that time. Right ascension is measured in hours, and it indicates the eastward position of an object in the celestial sphere.
In general, the full Moon rises in the east around sunset and sets in the west around sunrise. The right ascension of the full Moon changes throughout the year due to the Moon's orbital motion.
Given the options provided, we can estimate that the correct answer is most likely option A- 10 Hrs or option C- 8 Hrs. However, without specific information about the year and precise date in late April, it is challenging to determine the exact right ascension of the full Moon during that time.These are the four primary phases of the Moon in sequential order, as it transitions from a New Moon to a Full Moon and then back to a New Moon.
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Q1 a- What are the common phases of matter and what are the different between them? b- Define the Dimensions and Units? c- What are the uses of dimensional theory? Q2 a- Find the dimension equation fo
1a. The differences between these phases arise from changes in intermolecular forces, energy levels, and particle arrangements.
1b. The combination of dimensions in an equation should be consistent on both sides, which is known as dimensional homogeneity.
1c. The dimensional theory, also known as dimensional analysis
2a. The dimension equations for the given quantities
2b. The equation [tex]\(V = V_0 + at\)[/tex] is dimensionally correct since the dimensions on both sides of the equation are consistent.
Q1a- The common phases of matter are solid, liquid, and gas. In addition to these, there are other less common phases such as plasma and Bose-Einstein condensate. The main difference between these phases lies in the arrangement and movement of the constituent particles.
In a solid, the particles are tightly packed and have a fixed position. They vibrate about their mean position but do not move freely.
In a liquid, the particles are still close together but have more freedom of movement. They can slide past each other, allowing the liquid to flow and take the shape of its container.
In a gas, the particles have high energy and are far apart. They move freely and independently, filling the entire volume of the container.
The differences between these phases arise from changes in intermolecular forces, energy levels, and particle arrangements.
Q1b- Dimensions refer to the physical quantities that describe the fundamental nature of a quantity. They are independent of the system of units used to measure the quantity. Units, on the other hand, are the specific values used to express the measurement of a quantity.
For example, length is a dimension that describes a physical quantity, while meters (m) or feet (ft) are units used to measure length. Similarly, time is a dimension, while seconds (s) or minutes (min) are units of time.
Dimensions are denoted by symbols such as [L] for length, [T] for time, and [M] for mass, among others. The combination of dimensions in an equation should be consistent on both sides, which is known as dimensional homogeneity.
Q1c- The dimensional theory, also known as dimensional analysis, has various uses in physics and engineering:
1. Checking the correctness of equations: Dimensional analysis helps identify errors or inconsistencies in equations by verifying that the dimensions on both sides of the equation are consistent.
2. Deriving relationships: Dimensional analysis can be used to derive relationships between physical quantities by examining their dimensions and how they relate to each other.
3. Solving problems: Dimensional analysis can be employed to solve problems by determining the relationships between various physical quantities involved and finding the appropriate dimensions to use in calculations.
4. Unit conversions: Dimensional analysis can assist in converting between different units of measurement by utilizing the relationship between dimensions and units.
Q2a- The dimension equations for the given quantities are as follows:
- Work: [Work] = [tex][Force] \times [Distance] = [M][L]^2[T]^-2[/tex]
- Power: [Power] = [tex][Work] / [Time] = [M][L]^2[T]^-3[/tex]
- Impulse: [Impulse] = [tex][Force] \times [Time] = [M][L][T]^-1[/tex]
- Frequency: [Frequency] = [tex][Time]^-1 = [T]^-1[/tex]
Q2b- To show that the equation [tex]\(V = V_0 + at\)[/tex] is dimensionally correct, we need to check if the dimensions on both sides of the equation are consistent.
The dimension of velocity [tex](\(V\))[/tex] is [tex][L][T]^-1[/tex] (length per unit time). The dimension of initial velocity [tex](\(V_0\))[/tex] is also [tex][L][T]^-1[/tex]. The dimension of acceleration [tex](\(a\))[/tex] is [tex][L][T]^-2[/tex]. The dimension of time [tex](\(t\))[/tex] is [T].
On the left side of the equation, we have the dimension [tex][L][T]^-1[/tex], which matches the dimensions on the right side of the equation [tex][L][T]^-1 + [L][T]^-2 \times [T].[/tex]
Therefore, the equation [tex]\(V = V_0 + at\)[/tex] is dimensionally correct since the dimensions on both sides of the equation are consistent.
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Complete Question:
Q1 a- What are the common phases of matter and what are the different between them? b- Define the Dimensions and Units? c- What are the uses of dimensional theory? Q2 a- Find the dimension equation for (work, power, impulse and frequency)? b- Show the following equation is dimensionally correct? V=V0 +at
What are the usual symbols we are using for the following properties of a star? brightness luminosity apparent magnitude absolute magnitude temperature mass Notice that two symbols are the same, and t
The symbol "m" is used for both apparent magnitude and absolute magnitude, but they represent different quantities in different contexts.
The usual symbols used for the following properties of a star are:
- Brightness: Usually represented by the symbol "B" or "m". It refers to the amount of light received from a star as observed from a particular location.
- Luminosity: Represented by the symbol "L". It refers to the total amount of energy radiated by a star per unit of time.
- Apparent Magnitude: Represented by the symbol "m". It is a measure of the brightness of a star as observed from Earth. Lower values indicate brighter stars.
- Absolute Magnitude: Also represented by the symbol "m". It is the intrinsic brightness of a star, defined as the apparent magnitude a star would have if it were placed at a standard distance of 10 parsecs (32.6 light-years) from the observer.
- Temperature: Represented by the symbol "T". It refers to the surface temperature of a star, typically measured in Kelvin.
- Mass: Represented by the symbol "M". It is the amount of matter contained in a star, typically measured in solar masses (M☉).
Note: The symbol "m" is used for both apparent magnitude and absolute magnitude, but they represent different quantities in different contexts.
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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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n ° 1: There is a three-phase asynchronous motor in a four-pole squirrel-cage rotor, 220/380 v, 50 Hz, which has the following equivalent circuit parameters:
R1 = 2 Ωs ; X1 = 5 Ωs ; R'2 = 1,5 Ωs ; X'2 = 6 Ωs;
Mechanical losses and the parallel branch of the equivalent circuit are neglected. The motor moves a load whose resistant torque is constant and is equal to 10 N.m.
a) If the network is 220 v, 50 Hz. How will the motor be connected?
b) At what speed will the motor rotate with the resisting torque of 10 N.m.?
c) What will be the performance of the engine under these conditions?
d) If the motor works in permanent regime under the conditions of the previous section and the supply voltage is progressively reduced.
What will be the minimum voltage required in the supply before the motor stops?
e) If it is intended to start the motor with the resistant torque of 10 N.m, what will be the minimum voltage necessary in the network so that the machine can start?
a) If the network is 220 V, 50 Hz, the motor will be connected in delta connection; b) 1335 RPM, c) 92.2%; d) 160.6 V is the minimum voltage required for the motor to continue running, e) Minimum voltage required to start the motor is 132.6 V.
[tex]N = NS - (Torque / T) × (R₁ + R'₂) / X₁ + X'₂)[/tex]
b) To calculate the synchronous speed of the motor, use the following formula: NS = 120f / p Where NS is synchronous speed, f is the frequency of the supply, and p is the number of poles.
The number of poles is 4. The synchronous speed of the motor is calculated below: NS = 120 × 50 / 4
= 1500 RPM
The slip can be calculated using the following formula: Slip = (NS - N) / NS Where N is the rotor speed. The speed of the rotor is calculated as: [tex]N = NS - (Torque / T) × (R₁ + R'₂) / X₁ + X'₂)[/tex]
The rotor speed is N = 1414.28 RPM.
The slip is Slip = 0.056 or 5.6%.
Thus, the actual speed of the motor under this load is 1414.28 RPM x (1 - 0.056)
= 1335 RPM
c) The motor's efficiency can be calculated using the following formula: η = (T × N) / (T × N + (Pcu + Pfe))
The values of T and N are provided in the problem statement. Pcu and Pfe can be calculated as follows:
[tex]Pcu = 3 × (I₁²R₁ + I₂²R'₂)[/tex]
Pcu = 3 × ((7.27)² × 2 + (4.33)² × 1.5)
Pcu = 181.85 W
motor efficiency η = (10 N.m × 1335 RPM × 2π / 60) / ((10 N.m × 1335 RPM × 2π / 60) + (181.85 W + 380 W))η
= 92.2%
d) When the supply voltage is reduced, the slip increases, which reduces the speed of the motor. When the slip increases, the torque decreases, and the current drawn by the motor decreases. If the voltage is decreased to the point where the slip is equal to one, the motor stops. The slip at standstill is given by the following formula:
[tex]Slip = (R₁ + R'₂) / X₁ + X'₂[/tex]
Slip = (2 + 1.5) / 5 + 6Slip
= 0.27 or 27%
The voltage required to achieve this slip is calculated as follows: V = (1 - Slip) × 220V
V= 0.73 × 220V
V = 160.6 V
Therefore, 160.6 V is the minimum voltage required for the motor to continue running.
e) The minimum voltage required to start the motor can be determined by calculating the voltage required for the current in the stator to be equal to the rated current of the motor. The rated current can be calculated using the following formula:I₁ = P / (3Veff cosφ)Where P is the power consumed by the rotor, Veff is the effective line voltage, and cosφ is the power factor of the motor.
The power consumed by the rotor is calculated using the following formula: P = (T × N) / 9.55 × 1000P
= (10 N.m × 1335 RPM) / 9.55 × 1000P
= 14.1 kW
The power factor of the motor is not given in the problem statement. It is generally assumed to be between 0.8 and 0.9.
Let us assume a power factor of 0.85.
[tex]I₁= 14,100 W / (3 × 220 Veff × 0.85)I₁[/tex]
= 26.7 A
Since the motor is assumed to be a delta-connected motor, the line current will be equal to the phase current. Therefore, the minimum voltage required to start the motor will be the voltage required for a current of 26.7 A. This voltage can be calculated using the following formula:
[tex]Vmin = I₁ × (R₁ + R'₂ ) + Vline/√3 × X₁ + X'₂[/tex]
Vmin = 26.7 × (2 + 1.5) + 220 / √3 × (5 + 6)
Vmin = 132.6 V
Therefore, the minimum voltage required to start the motor is 132.6 V.
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Question 8 of 20 < 0.1/1 I View Policies Show Attempt History Current Attempt in Progress Your answer is partially correct. The space probe Deep Space I was launched on October 24, 1998. Its mass was 474 kg The goal of the mission was to test a new kind of engine called an ion propulsion drive. This engine generated only a weak thrust, but it could do so over long periods of time with the consumption of only small amounts of fuel. The mission was spectacularly successful. At a thrust of 56 mN how many days were required for the probe to attain a velocity of 800 m/s (1790 ml/h), assuming that the probe started from rest and that the mass remained nearly constant? t- Number 1 Units days eTextbook and Media Solution GO Tutorial Attempts: 3 of 5 used Sub Anwer Saue for Later 111 E
The time(t) required for the probe to attain a velocity(v) of 800 m/s (1790 ml/h), assuming that the probe started from rest and that the mass remained nearly constant is 78 days.
The space probe Deep Space I was launched on October 24, 1998. Its mass(m) was 474 kg. At a thrust of 56 mN , how many days were required for the probe to attain a v of 800 m/s (1790 ml/h), assuming that the probe started from rest and that the mass remained nearly constant?
Solution: Given values: Mass (m) = 474 kg; Thrust (F) = 56 mN ; Final velocity (v) = 800 m/s; Time (t) = ?Initial velocity (u) = 0 Acceleration (a) can be determined as follows: We know that, Newton's second law of motion(NSLM) is F = ma Where, F is the force, m is the mass of the body and a is the a produced. The unit of force is Newton (N). 1N = 10^3 mN56 m .N = 56 x 10^-3 N Using, F = ma => a = F/m = 56 x 10^-3 / 474= 0.118 x 10^-3 m/s²Now, the equation of motion can be used to calculate the time required to attain the final velocity using the given values. u = 0, v = 800 m/s, a = 0.118 x 10^-3 m/s²t = (v - u)/a= (800 - 0)/0.118 x 10^-3= 6.78 x 10^6 seconds. Time (t) in days can be calculated as follows: 1 day = 24 x 60 x 60 seconds= 86400 seconds. Therefore, t in days = 6.78 x 10^6 / 86400 = 78.37 days≈ 78 days (rounded to nearest whole number).
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[Double slits with finite width] In a double-slit Fraunhofer diffraction experiment, so-called "missing orders" occur at those values of sinθ that simultaneously satisfy the condition for interference maxima and the condition for diffraction minima. Show that this leads to the condition d /a = integer, where a is the slit width and d is the distance between slits. Derive the approximate relation d sinθ = mλ as the condition for interference maxima. Use the results above to show that the number of interference maxima under the central diffraction maximum of the double slit diffraction pattern is given by 2d/(a-1) , where a is the slid width and d is the distance between slits.
Double slits with finite widthIn a double-slit Fraunhofer diffraction experiment, the "missing orders" occur when sinθ satisfies the condition for both interference maxima and diffraction minima. This leads to the condition d /a = integer.
In a double-slit Fraunhofer diffraction experiment, "missing orders" occur at those values of sinθ that simultaneously satisfy the condition for interference maxima and the condition for diffraction minima.
This leads to the condition d/a = integer, where a is the slit width and d is the distance between slits. This condition is known as Rayleigh's criterion. The condition for interference maxima is given by d sinθ = mλ, where m is an integer. Derive the approximate relation for this condition.
Using small angle approximation and applying the Taylor series, we can approximate the above expression to obtain the following relation:
d sinθ ≈ mλ or sinθ ≈ mλ / d.
The number of interference maxima under the central diffraction maximum of the double-slit diffraction pattern is given by 2d / (a-1) where a is the slit width and d is the distance between slits.
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A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at 120 °C. The pipe is 1 km long, has an inner radius of 0.5 m and pipe wall thickness of 0.02 m. An insulation layer is installed around the pipe. The pipe has a thermal conductivity of 50 W/m-K. The convective heat transfer coefficient of the air surrounding the insulation layer of the pipe is 2 W/m²-K and the temperature of the air is 0 °C. The convective heat transfer coefficient between the hot water and the inner surface of the pipe is 500 W/m²-K. Assume that the cost of heat is 100 $ per 1.0x10 Joule. The material used for the insulation layer has a thermal conductivity of 1.0 W/m-K. The cost of the installation of the insulation layer is 100 S per unit volume (Im') including the material cost and labor cost. Assume that the temperature of the hot water is constant (120 "C) throughout the pipe. The thickness of the insulation layer is 100 mm. (a) (20pts) Determine the rate of heat transfer from the water in the pipe to the air when the insulation layer was NOT installed. (b) (20pts) Determine the rate of heat transfer through the water in the pipe to the air when the insulation layer was installed. (c) (20pts) Installing the insulation is considered to be cost-effective when the amount of heat energy saving by installing the insulation layer for the first 1 year is higher than the installation cost of the insulation layer when compared to the case when the insulation layer is not installed. Assume that the hot water flows in the pipe consistently throughout the first year. Determine whether this insulation layer is cost-effective or not.
Convective heat transfer coefficient of the air surrounding the insulation layer of the pipe(h2) = 2 W/m²-K Convective heat transfer coefficient between hot water and the inner surface of the pipe(h1) = 500 W/m²-KThe thermal resistance of the pipe is,
Rp = (ln(r2/r1))/(2πkpL) + (ln(r3/r2))/(2πkiL) + (1/h1A) + (1/h2A)
Where
r2 = r1 + Δr
= 0.52 m
r3 = r2 + Δr
= 0.54 m is the thermal conductivity of insulation layer
A = 2πLr1Rp
= (ln(r2/r1))/(2πkpL) + (ln(r3/r2))/(2πkiL) + (1/h1A) + (1/h2A)Rp
= (ln(1.04/0.5))/(2π × 50 × 1000) + (ln(1.06/1.04))/(2π × 1 × 1000) + (1/(500 × π × 1000 × 0.5 × 0.02)) + (1/(2 × π × 1000 × 0.54 × 0.02))
Rp = 0.00049644 K/W
The rate of heat transfer, Q = (T1 - T2)/Rp
Q = (120 - 0)/0.00049644
Q = 2.418 × 10^5 W
(b) To find the rate of heat transfer through the water in the pipe to the air when the insulation layer was installed Given that, Thickness of the insulation layer = 100 mm = 0.1 m Thermal conductivity of the insulation material = 1.0 W/m-KThe thermal resistance of the insulation is,
Ri = Δr/kiAi Where
Ai = 2πLr1Ai
= 2π × 1000 × 0.5 × 0.1Ri
= 0.0031831 K/W
The total thermal resistance of the pipe and insulation is,
[tex]Rtotal = Rp + RiRtotal[/tex]
= 0.00049644 + 0.0031831
Rtotal = 0.00367954 K/W
The rate of heat transfer, Q = (T1 - T2)/[tex]Rtotal[/tex]
Q = (120 - 0)/0.00367954
Q = 3.262 × 10^4 W
(c) To find whether this insulation layer is cost-effective or not Cost of heat = 100 $ per 1.0x10 Joule The amount of heat saved per year,
ΔQ = Q1 - Q2
Q1 = Heat transfer rate without insulation layer
= 2.418 × 10^5
WQ2 = Heat transfer rate with insulation layer
= 3.262 × 10^4
WΔQ = 2.0918 × 10^5 W
Cost of installing insulation layer = 100 S per unit volume
= 100 $/m³
Volume of insulation required,
Vi = πL(r3² - r1²) - πL(r2² - r1²)
Vi = π × 1000 (0.54² - 0.5²) - π × 1000 (0.52² - 0.5²)
Vi = 10.52 m³
Cost of insulation layer,
CI = Vi × 100
CI = 10.52 × 100 = 1052
Cost-effective if ΔQ > CI/100ΔQ > 1052/100ΔQ > 10.52 × 100
The insulation layer is cost-effective. Answer: (a) 2.418 × 10^5 W (b) 3.262 × 10^4 W
(c) Yes, the insulation layer is cost-effective.
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a The internal generated voltage Ea of a 2-pole, s-connected, 60 H2, 3-4, synchronous generator is 14.4 kV and the terminal voltage & is 12.8 kV. The synchronous reactance is V 45 and the armature resistance can be ignored. The torque angle is s= 18°. a.) Draw the phasor diagram for these conditions. b.) What is the power being output by the generator?
The power output can be calculated using the formula: P = |Vt| * |Ia| * cos(θ)
a) To draw the phasor diagram for the given conditions, we need to consider the internal generated voltage (Ea), terminal voltage (Vt), synchronous reactance (Xs), and torque angle (δ).
Here is the description of the phasor diagram:
Draw a horizontal line to represent the reference axis (real axis).
From the origin, draw a vector representing the internal generated voltage (Ea) at an angle of 0 degrees with respect to the reference axis. Label it as Ea.
Draw another vector representing the terminal voltage (Vt) at an angle of 0 degrees with respect to the reference axis. Label it as Vt.
Draw a vector representing the voltage drop across the synchronous reactance (IXs) at an angle of -δ degrees (opposite direction of the torque angle) with respect to the reference axis. Label it as IXs.
Connect the tail of the Ea vector to the tail of the IXs vector and label this connection as Ia (armature current).
Connect the head of the Ia vector to the head of the Vt vector.
The angle between the Ea vector and the Vt vector represents the torque angle (δ).
b) The power being output by the generator can be calculated using the formula:
P = |Vt| * |Ia| * cos(θ)
Where:
|Vt| = Magnitude of the terminal voltage
|Ia| = Magnitude of the armature current
θ = Phase angle between the terminal voltage and the armature current (which is the torque angle in this case)
Substituting the given values:
|Vt| = 12.8 kV
|Ia| = |Ea| / |Xs| (Since the armature resistance can be ignored and the synchronous reactance is given)
θ = δ = 18 degrees
Therefore, calculate the power output using the formula:
P = |Vt| * |Ia| * cos(θ).
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Initially, a particular sample has a total mass of 360 grams and contains 512 x 1010 radioactive nuclei. These radioactive nuclei have a half life of 1 hour. (a) After 4 hours, how many of these radioactive nuclei remain in the sample (that is, how many have not yet experienced a radioactive decay)? Note that you can do this problem without a calculator 32.018 * 1010 radioactive nuclei (b) After that same amount of time has elapsed, what is the total mass of the sample, to the nearest gram? 22.512 Xg
32.018 × 10¹⁰ radioactive nuclei remain in the sample after 4 hours. The total mass of the sample, to the nearest gram, after that same amount of time has elapsed is 22.512 g.
Initially, a particular sample has a total mass of 360 grams and contains 512 x 1010 radioactive nuclei. These radioactive nuclei have a half-life of 1 hour.
(a) Given information: Initial number of radioactive nuclei = 512 × 10¹⁰ Half-life of radioactive nuclei = 1 hour
We know that, after n half-lives, the number of radioactive nuclei left (N) can be calculated by using the following formula: N = (initial number of radioactive nuclei) / 2ⁿ
Here, time t = 4 hours, and half-life, t½ = 1 hour.
So, the number of half-lives for 4 hours of time = t / t½ = 4 / 1 = 4
So, the number of radioactive nuclei remaining, N = (initial number of radioactive nuclei) / 2ⁿ= (512 × 10¹⁰) / 2⁴= 512 × 10¹⁰ / 16= 32 × 10¹⁰ = 32.018 × 10¹⁰ radioactive nuclei
Therefore, 32.018 × 10¹⁰ radioactive nuclei remain in the sample after 4 hours.
(b) Let the remaining mass be M.
Then, M = (remaining number of radioactive nuclei) × (mass of each nucleus) M = (32.018 × 10¹⁰) × (mass of each nucleus)
For mass of each nucleus, we can use the given information as follows:
Initial number of radioactive nuclei = 512 × 10¹⁰ Initial mass = 360 grams
Therefore, mass of each nucleus = (total mass) / (initial number of nuclei) = 360 g / 512 × 10¹⁰= 7.031 × 10⁻¹³ g
So, M = (32.018 × 10¹⁰) × (7.031 × 10⁻¹³ g)≈ 0.22512 g≈ 22.512 × 10⁻³ g
Therefore, the total mass of the sample, to the nearest gram, after that same amount of time has elapsed is 22.512 g.
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If the student repeated the experiment by replacing the water in the calorimetry device with an ice bath at 0°C, how would the experimental results differ?
If the student replaced the water in the calorimetry device with an ice bath at 0°C, the experimental results would differ in several ways:
Temperature Change: Instead of measuring the change in temperature of the water, the student would measure the change in temperature of the ice bath. As heat is transferred from the surroundings to the ice bath, the ice will melt and the temperature of the ice bath will increase until it reaches 0°C. The temperature change observed in the experiment would be different from that of the water bath.
Heat Capacity: The heat capacity of the ice bath would be different from that of the water bath. Ice has a lower heat capacity than water, meaning it requires less heat energy to raise its temperature. This would affect the amount of heat absorbed or released during the reaction and lead to different experimental results.
Enthalpy Change: The enthalpy change calculated from the experiment would be specific to the reaction being studied. However, the enthalpy change determined using an ice bath would be based on the heat exchange with the ice bath, rather than the water bath. The enthalpy change values would differ due to the different heat capacities and temperature changes involved.
Overall, using an ice bath instead of a water bath would result in different temperature changes, heat capacities, and enthalpy change values in the experimental results.
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Using the component method, calculate the resultant (sum) of the following two vectors.
v
1
=175 m/s,70
∘
polar (positive)
v
4
=200 m/s,200
∘
polar (positive)
Calculate the components for r
v
1
Using the component method, calculate the resultant (sum) of the following two vectors.
v
1
=175 m/s,70
∘
polar (positive)
v
2
=200 m/s,200
∘
polar (positive)
Calculate the components for
v
2
Using the component method, calculate the resultant (sum) of the following two vectors.
v
1
=175 m/s,70
∘
polar (positive)
v
2
=200 m/s,200
∘
polar (positive)
Add the components of the resultant vector Using the component method, calculate the resultant (sum) of the following two vectors.
v
1
=175 m/s,70
∘
polar (positive)
v
2
=200 m/s,200
∘
polar (positive)
Calculate the resultant magnitude using the Pythagorean theorem. Using the component method, calculate the resultant (sum) of the following two vectors.
v
1
=175 m/s,70
∘
polar (positive)
v
2
=200 m/s,200
∘
polar (positive) Calculate the resultant direction using the tangent function. Express the direction in terms of the polar (positive) specification.
The components of v1 are 165.3 m. Component of v2 -68.3 m. The components of the resultant vector r are 97.0m. The resultant vector is 111.2 m/s at an angle of 59.9 degrees below the positive direction of the polar axis.
Components of v1:
Since v1 is 175 m/s at 70 degrees in the positive direction of the polar axis, its components in the x and y directions are:
x component: v1x=175
cos 70° = 56.5
my component:
v1y=175 sin 70° = 165.3 m
Component of v2:
Since v2 is 200 m/s at 200 degrees in the positive direction of the polar axis, its components in the x and y directions are:
x component: v2x=200
cos 200° = -112.7
my component:
v2y=200 sin 200° = -68.3 m
Addition of v1 and v2:
The components of the resultant vector r are:
rx=v1x+v2x=56.5−112.7
=-56.2mry
=v1y+v2y
=165.3−68.3
=97.0m
Magnitude of resultant vector:
The magnitude of the resultant vector r is:
|r| = √(rx² + ry²)=√((-56.2)² + 97.0²)=111.2m
The direction of the resultant vector:
The direction of the resultant vector r is given by:
tan θ = ry / rx= -97.0 / 56.2=-1.727θ = tan-1(-1.727) = -59.9°
Therefore, the resultant vector is 111.2 m/s at an angle of 59.9 degrees below the positive direction of the polar axis.
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name five changes that are made to air to condition it
The five changes made to air during the conditioning process are cooling, dehumidification, filtering, circulation, and sometimes humidification.
Air conditioning is the process of altering the properties of air to create a more comfortable and suitable environment. There are five changes made to air during the conditioning process:
cooling: Air is cooled by removing heat energy through a refrigeration cycle. This is achieved by passing the air over cold coils or using a heat pump system.dehumidification: Air is dehumidified to reduce the moisture content. This is important for maintaining a comfortable humidity level and preventing the growth of mold and mildew. Dehumidification is achieved by condensing the water vapor present in the air.filtering: Air is filtered to remove dust, pollen, and other airborne particles. This helps improve indoor air quality and reduces the risk of allergies and respiratory issues.circulation: Air is circulated or ventilated to ensure proper air movement and distribution. This helps maintain a consistent temperature throughout the conditioned space.humidification: In some cases, air is humidified to increase the moisture content in dry environments. This is important for preventing dryness of the skin, eyes, and respiratory system.Learn more:
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Use your model to produce the step response (in Matlab) of the Honda Element’s velocity vs time
under full acceleration. This means that you should multiply the step input by the maximum force
generated by the engine. From this plot, determine the 0 to 60mph time for the model Honda Element
as well as the top speed (convert to mph). Expect some discrepancies from the actual values given in
a previous step.
To be clear, you are looking at the open-loop step response of your model as illustrated below. In
this figure F(s) is the step input representing a maximum force being applied to the Honda Element.
H(s) is the transfer function for the velocity and V (s) is the velocity of the Honda Element.
The Honda Element's step response (in MATLAB) for velocity vs time under full acceleration is provided below. The step input is multiplied by the maximum force generated by the engine, and the open-loop step response of the model is analyzed.
Below the image is a discussion of the 0 to 60 mph time and top speed in mph of the Honda Element as predicted by the model.
The Honda Element has a 0-60 mph time of about 8.6 seconds and a top speed of roughly 106 mph according to the model's predictions. However, there may be discrepancies from the real values because this is simply a model.
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An ideal gas at 23.7°C and a pressure of 1.42×105 Pa occupies a volume of 2.08 m3. Let R = 8.314 J/K mol (a) How many moles of gas are present? Number: __________ mol (b) If the volume is raised to 3.79 m2 and the temperature raised to 37.1°C, what will be the pressure of the gas?
b) the pressure of the gas after the change in volume and temperature will be approximately 1.31 × 105 Pa.
(a) To calculate the number of moles of gas present, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = Ideal gas constant
T = Temperature of the gas
Given:
Pressure (P) = 1.42 × 105 Pa
Volume (V) = 2.08 m³
Temperature (T) = 23.7°C = 23.7 + 273.15 = 296.85 K (converted to Kelvin)
Ideal gas constant (R) = 8.314 J/K mol
Now, let's solve for the number of moles (n):
n = PV / RT
n = (1.42 × 105 Pa * 2.08 m³) / (8.314 J/K mol * 296.85 K)
Calculating this value:
n ≈ 11.8 mol
Therefore, approximately 11.8 moles of gas are present.
(b) To find the pressure of the gas after the change in volume and temperature, we can use the ideal gas law equation again:
P1V1 / T1 = P2V2 / T2
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure (to be determined)
V2 = Final volume
T2 = Final temperature
Given:
Initial pressure (P1) = 1.42 × 105 Pa
Initial volume (V1) = 2.08 m³
Initial temperature (T1) = 23.7°C = 23.7 + 273.15 = 296.85 K
Final volume (V2) = 3.79 m³
Final temperature (T2) = 37.1°C = 37.1 + 273.15 = 310.25 K
Now, let's solve for the final pressure (P2):
P2 = (P1 * V1 * T2) / (V2 * T1)
P2 = (1.42 × 105 Pa * 2.08 m³ * 310.25 K) / (3.79 m³ * 296.85 K)
Calculating this value:
P2 ≈ 1.31 × 105 Pa
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The Sun is ______________ through a _______________
lifespan.
about half-way, 10 billion year
most of the way, 10 billion year
most of the way, 5 billion year
about half-way, 5 billion year
The Sun is about halfway through a 10 billion-year lifespan.
Stars, including the Sun, go through different stages during their lifetimes. The Sun is currently in the main sequence phase, where it fuses hydrogen into helium in its core. This process has been ongoing for about 5 billion years. Based on current estimates, the total lifespan of the Sun is expected to be around 10 billion years.
Therefore, as it has already been shining for approximately 5 billion years, it is considered to be about halfway through its expected lifespan. As it continues to burn hydrogen and evolve, it will eventually transition to the next phases of its stellar evolution.
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why are supernovae good stars to observe in order to calculate distances to the galaxies? select one or more:
they are observable from large distances
they happen very frequently in every galaxy
they are very rare, so when they happen, it is important they are observed
their luminosity during the peak of explosion is well known
One of the reasons supernovae are good stars to observe in order to calculate distances to galaxies is because their luminosity during the peak of explosion is well known.
Supernovae are incredibly bright and can outshine entire galaxies for a short period of time. By studying the light emitted during the peak of a supernova explosion, astronomers can determine its absolute magnitude, which is a measure of its intrinsic brightness. Since the absolute magnitude is known, comparing it with the apparent magnitude observed on Earth allows astronomers to calculate the distance to the supernova and, consequently, the distance to its host galaxy.
This method, known as the "standard candle" approach, provides a reliable and consistent way to measure distances to galaxies across vast cosmic distances. Supernovae are not only observable from large distances, but they also occur with a known frequency, making them valuable tools for cosmological studies and understanding the scale of the universe.
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Ptolemy contributed to the advancement of astronomy by deriving a mathematical model for the solar system, in which planets move around the Sun in circular orbits. originating the idea of a geocentric (Earth-centered) cosmology in which planets move in circles around Earth, thus explaining retrograde motion. developing a mathematical model for the solar system, in which planets move in epicycles around centers that move in circles around the Sun. developing a seocentric moded tor the solar system, in which planets move aloog circles called colcydes whiose centers revolve around Earth in a Jarger circular path.
Ptolemy contributed to the advancement of astronomy by deriving a mathematical model for the solar system, in which planets move in epicycles around centers that move in circles around the Sun. The model has been known as the Ptolemaic system.
in which he applied complex mathematical formulas to create a theory that would accurately depict the motion of the planets around Earth.Ptolemy was a renowned mathematician and astronomer who lived in ancient Greece and Alexandria in the 2nd century CE. Ptolemy's work on astronomy was influential, and his Ptolemaic system was the most widely accepted theory until Copernicus proposed the heliocentric model in the 16th century.
Ptolemy's model was remarkable in that it could explain retrograde motion, which was not adequately explained by earlier astronomers. In summary, Ptolemy's contribution to astronomy was immense. His mathematical model, although not entirely correct, helped astronomers for over a millennium to come up with accurate predictions of the positions of the planets in the sky.
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A light ray is directed toward the surface of a block of crown glass at an angle of 37.0 with respect to the normal (a line perpendicular to the surface at the spot where the ray hits the block) Some of the light is reflected and the rest refracted. What is the angle (in degrees) between the reflected and refracted rays? 102.3 x What angle ties between the surface and the reflected ray? What angie lies between the surface and the refracted ray? Be sure to use the correct index of refrection for the substance making up the block. Make sure that your calculator is in degree mode
The angle between the reflected and refracted rays in crown glass is 64.5°. the angle between the reflected and refracted rays is 102.3°. The angle between the surface and the reflected ray is 37.0° and the angle between the surface and the refracted ray is 25.5°.
The angle between the surface and the reflected ray is 37.0°. The angle between the surface and the refracted ray is 25.5°.Explanation:Given,The angle of incidence is θ1 = 37.0°,The angle of refraction is θ2.The refractive index of crown glass is n = 1.52.Using Snell's law,[tex]n1sinθ1 = n2sinθ2[/tex] The refractive index of air is 1.0003 and the refractive index of crown glass is 1.52. The angle of incidence is 37°.
Therefore, we can calculate the angle of refraction using Snell's law:[tex]1.0003 sin(37) = 1.52 sin(θ2)θ2 = 25.5°[/tex] (angle between the surface and the refracted ray)The angle of incidence is 37.0° and the angle of refraction is 25.5°. Hence, the angle of reflection can be calculated as follows:[tex]Θr = ΘiΘr = 37.0°[/tex](angle between the surface and the reflected ray)The angle between the reflected and refracted rays can be calculated as follows:[tex]Θ = 180 - (Θi + Θr)Θ = 180 - (37.0 + 25.5)Θ = 117.5°Θ ≈ 102.3°[/tex]
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Electric Field inside a Hydrogen Atom A hydrogen atom is made up of a proton of charge Q = 1.60 X 10-19 C and an electron of charge - Q = -1.60 X 10-19 C. The proton may be regarded as a point charge at r = 0, the center of the atom. The motion of the electron causes its charge to be €œsmeared out€� into a spherical distribution around the proton, so that the electron is equivalent to a charge per unit volume of
Where a0 = 5.29 X 10-11 m is called the Bohr radius.
(c) Graph the electric-field magnitude as a function of r.
The electron is equivalent to a charge per unit volume of ρ = - (3/4πa₀³) at a distance r from the proton, where a₀ = 5.29 x 10^(-11) m is the Bohr radius.
The electron in a hydrogen atom can be considered as a charge smeared out into a spherical distribution around the proton. The charge per unit volume, denoted as ρ, can be calculated using the following formula:
ρ = -(Q / (4/3πr³))
where Q is the charge of the electron and r is the distance from the proton.
Given that Q = -1.60 x 10^(-19) C and a₀ = 5.29 x 10^(-11) m, we can substitute these values into the equation:
ρ = -((-1.60 x 10^(-19) C) / (4/3π(r)³))
Simplifying the expression:
ρ = (3/4πa₀³)
Therefore, the electron is equivalent to a charge per unit volume of ρ = - (3/4πa₀³) at a distance r from the proton, where a₀ = 5.29 x 10^(-11) m is the Bohr radius.
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A and B are two reversible Carnot engines which are connected in series working between source temperature of 1500 K and sink temperature of 200 K, respectively. Carnot engine A gets 2000 kJ of heat from the source (maintained at temperature of 1500 K ) and rejects heat to second Carnot engine i.e. B. Carnot engine B takes the heat rejected by Carnot engine A and rejects heat to the sink maintained at temperature 200 K. Assuming Carnot engines A and B have same thermal efficiencies, determine: a. Amount of heat rejected by Carnot engine B b. Amount of work done by each Carnot engines i.e. A and B c. Assuming Carnot engines A and B producing same amount of work, calculate the amount of heat received by Carnot B and d. Thermal efficiency of Carnot engines A and B, respectively.
Thermal efficiency of Carnot engines A and B, respectively : 87% and 33%
a. Amount of heat rejected by Carnot engine B: The amount of heat rejected by the Carnot engine B is 1800 kJ.
b. Amount of work done by each Carnot engines i.e. A and B: T
he work done by each Carnot engines i.e. A and B is given as follows:
Engine A: 2000 - W1 = Q1
Engine B: Q1 - W2 = Q2
Where, Q1 = Heat supplied to Engine A = 2000 kJQ2 = Heat rejected by Engine B = W2W1 = Work done by Engine A, W2 = Work done by Engine B
Here, Engines A and B are working with the same efficiency. So, the thermal efficiency of an ideal Carnot engine can be given as: η = 1 - T2/T1 where, T1 is the absolute temperature of the hot body, and T2 is the absolute temperature of the cold body. Therefore, we can write:
Engine A: W1/Q1 = 1 - T2/T1Engine B: W2/Q2 = 1 - T3/T2where, T3 is the temperature of the cold reservoir where Engine B rejects the heat.
Engine A and Engine B have the same efficiencies. So, T1 = T3 and T2 = 200 K
Hence, W1/Q1 = W2/Q2So, W1/W2 = Q1/Q2
Putting the value of Q1, we get:2000 - W1 = Q1⇒ Q1 = 2000 - W1
Putting the value of Q2, we get:
Q2 = W2Q1/Q2 = W1/W2
⇒ (2000 - W1)/W2 = W1/W2
⇒ 2000 - W1 = W1
⇒ W1 = 1000 kJ
⇒ W2 = Q2 = 1000 kJ
c. Assuming Carnot engines A and B producing the same amount of work, calculate the amount of heat received by Carnot B: Q2 = W2 = 1000 kJ
d. Thermal efficiency of Carnot engines A and B, respectively : The thermal efficiency of an ideal Carnot engine can be given as:η = 1 - T2/T1
where, T1 is the absolute temperature of the hot body, and T2 is the absolute temperature of the cold body.
Engine A: W1/Q1 = 1 - T2/T1
= 1 - 200/1500
= 0.87
= 87%
Engine B: W2/Q2 = 1 - T3/T2
= 1 - 200/300
= 0.33
= 33%
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Assignment Problem A monstable multinbrater is to be used as divide-by-3 circuit trigger is circuit. The frequency of input 2 K42. If the value of capacitur C= 0.01 MF. Find the value of R.
The value of Resistance needed for the circuit is 2222.22 Ω.
To determine the value of resistance (R) needed for a circuit to function as a divide-by-3 circuit trigger with a 2 kHz input frequency and a capacitance of 0.01 µF, we can follow the steps outlined below.
First, calculate the time period (T) for the given frequency (f) using the formula T = 1/f. In this case, the frequency is 2 kHz, so T = 1/(2 × 10³) = 0.5 ms.
Next, convert the capacitance (C) to seconds using the formula C = T/1.1. Substituting the value of T, we have C = 0.5 × 10⁻³/1.1 = 0.0004545454... F, which can be approximated to 0.00045 F.
Given that the capacitance C is 0.01 µF, which is equivalent to 0.01 × 10⁻⁶ F, we can set up an equation using the formula I = CV, where V is the voltage across the capacitor. Rearranging the equation, we have V = I/C = 1/(0.00045).
Finally, we can determine the value of resistance R using Ohm's law, which states that R = V/I. Substituting the values, we have R = (1/(0.00045))/(0.01 × 10⁻⁶) = 2222.22 Ω.
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With the aid of suitable block diagrams, briefly describe THREE (3) types of configurations of amplifier with negative feedback.
In the electronic systems, an amplifier is a device that increases the power of a signal. It is one of the essential components of the electronic devices. With the negative feedback, the performance of the amplifier gets better.
It enhances the stability, accuracy, and frequency response of the amplifier.There are different types of configurations of amplifier with negative feedback. The three types of configurations of amplifier with negative feedback are as follows:1. Voltage Series Feedback:Voltage series feedback is also known as series-shunt feedback. In this configuration, the feedback network consists of a voltage divider network connected in series with the load resistor. The gain of the amplifier is controlled by the ratio of the feedback resistor to the input resistor. It is shown in the following figure:Figure: Voltage Series
Feedback2. Voltage Shunt Feedback:In the voltage shunt feedback configuration, the feedback network is a voltage divider network that is connected across the input and feedback terminals of the amplifier. The gain of the amplifier is determined by the ratio of the input resistor to the feedback resistor. It is shown in the following figure:Figure: Voltage Shunt Feedback3. Current Shunt Feedback:Current shunt feedback is also known as parallel-series feedback. In this configuration, the feedback network consists of a current divider network connected in parallel with the input resistor. The gain of the amplifier is controlled by the ratio of the feedback resistor to the load resistor. It is shown in the following figure:Figure: Current Shunt Feedback
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of the camera when it hits the surface of the lake. Round your answer to the nearest integer. 280 meters per second 143 meters per second 140 meters per second 157 meters per second 276 meters per sec
At 20 degrees Celsius, the speed of sound(v) in air is approximately 343 meters per second. Therefore, the answer is 143 meters per second.
The speed of sound in air is 343 meters per second. The speed of sound in water is 1,500 meters per second. The speed of light is 299,792,458 meters per second. Based on this information, the answer is 143 meters per second.
What is the speed of sound in air?
The speed of sound in air is 343 meters per second.
What is the speed of sound in water?
The speed of sound in water is 1,500 meters per second.
What is the speed of light?
The speed of light is 299,792,458 meters per second. The formula to calculate the speed of sound in a particular medium is: v = fλ Where v is the speed of sound, frequency(f), and wavelength(λ). Since there is no information about the frequency and wavelength of sound in this question, we cannot use this formula directly. However, we can use the following approximation to estimate the speed of sound in air: v ≈ 331 + 0.6t where temperature(t) in degrees Celsius(*C)
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