A three-phase load is a series RL circuit where the resistance and inductance of each phase are 10 Ω and 30 mH, respectively. The three-phase source has a line-to-line RMS voltage of 480 V at 60 Hz, and the delay angle α is 75°. To find the RMS value of the source current, we first need to calculate the impedance of each phase of the load and the line-to-neutral voltage.
Impedance of each phase of the load:The impedance of an RL circuit can be expressed using the following equation:Z = √(R²+Xl²), where R is the resistance and Xl is the inductive reactance. The inductive reactance can be calculated using the following equation:Xl = 2πfL, where f is the frequency and L is the inductance.
The impedance of each phase of the load can be found as follows:XL = 2π(60)(30 × 10-3) = 11.31 ΩZ = √(R²+Xl²) = √(10²+(11.31)²) = 15 Ω Line-to-neutral voltage:Since the line-to-line voltage is 480 V RMS, the line-to-neutral voltage can be calculated as follows:VLN = VLL/√3 = 480/√3 = 277.13 V RMS RMS current:We can use the following equation to find the RMS current of the source:I = V/Z, where V is the line-to-neutral voltage and Z is the impedance of each phase of the load. Therefore, the RMS current of the source can be found as follows:I = V/Z = 277.13/15 = 18.48 ATherefore, the RMS value of the source current is 18.48 A.
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Which of the following describes the relationship between the distance between charges and electrical force? directly proportional relationship inversely proportional relationship direct square relationship inverse square relationship Newton's universal law of gravitation describes forces that are , while Coulomb's law describes forces that are
The electrical force is directly proportional to the product of charges and inversely proportional to the square of the distance between charges.
Therefore, the correct option is an inverse square relationship. Newton's universal law of gravitation describes forces that are gravitational, while Coulomb's law describes forces that are electrical.
Coulomb's law is a mathematical equation that describes the interactions between electric charges. It quantifies the amount of electrical force that two charged objects exert on each other based on their distance and charge. The equation can be used to calculate the force between two point charges, which are charged particles that have a negligible size and shape relative to the distance between them.
Newton's law of gravitation is a mathematical equation that describes the force of gravity between two objects with mass. It states that any two objects with mass exert an attractive force on each other that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
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45. A
60
Co source is labeled 6.1mCi, but its present activity is found to be 2.03×10
7
Bq. (a) What is the present activity in mCi ? mCi. (b) How long ago in years did it actually have a 4.00−mCi activity? years. 45. A
60
Co source is labeled 6.1mCi, but its present activity is found to be 2.03×10
7
Bq. (a) What is the present activity in mCi ? mCi. (b) How long ago in years did it actually have a 4.00-mCi adtivity? years.
The present activity in mCi is 0.610 mCi. The 60Co actually had a 4.00-mCi activity 20.8 years ago.
Given, Activity of 60Co = 2.03 × 107 Bq = 6.1 mCi
(a) We have to find the present activity in mCi.
Activity = 6.1 mCi = 6.1 × 10−3 Ci = 6.1 × 10−3 × 3.7 × 1010 Bq = 22.57 × 106 Bq = 2.257 × 107 Bq
Present activity in mCi = 2.257 × 107/3.7 × 1010= 0.610 mCi
Therefore, the present activity in mCi is 0.610 mCi.
(b) We have to find how long ago in years did it actually have a 4.00-mCi activity.
Activity of 60Co = 4.00 mCi = 4.00 × 10−3 Ci = 4.00 × 10−3 × 3.7 × 1010 Bq = 14.8 × 106 Bq
Let 't' be the time for which it actually had a 4.00-mCi activity.
Hence, the initial activity (A0) = Activity of 60Co at present (A) = 2.03 × 107 Bq.
The activity of radioactive substance is given by the relation, A = A0e−λt, where, λ is the decay constant, which can be calculated as follows: A0 = A = 2.03 × 107 Bq = A0e−λtλ = -ln(2)/T1/2 = -ln(2)/5.27 = 0.1314/day
Putting the values of λ, A0, and A in the above relation, 2.03 × 107 = A0e−0.1314tA0 = 2.03 × 107 /e−0.1314t= 2.03 × 107 / (1/2.718)0.1314t= 2.03 × 107 × 2.7180.1314t= 5.51 × 107t= (1/0.1314) ln (5.51 × 107 / 2.03 × 107)t = 20.8 years
Therefore, the 60Co actually had a 4.00-mCi activity 20.8 years ago.
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In certain fireworks, potassium nitrate breaks down into potassium oxide, nitrogen, and oxygen. This is an example of a decomposition reaction. The opposite process is a synthesis reaction.
The given statement is correct. The decomposition of potassium nitrate into potassium oxide, nitrogen, and oxygen is indeed an example of a decomposition reaction, and the opposite process is a synthesis reaction.
A decomposition reaction is a type of chemical reaction where a compound breaks down into simpler substances. In the case of potassium nitrate[tex](KNO_{3} )[/tex] in fireworks, it decomposes into potassium oxide ([tex]K_{2} O[/tex]), nitrogen gas ([tex]N_{2}[/tex]), and oxygen gas ([tex]O_{2}[/tex]). This reaction is typically initiated by heat or other sources of energy. The balanced chemical equation for this decomposition reaction is as follows:
2 KNO₃ → 2 K₂O + N₂ + 3 O₂
The decomposition of potassium nitrate releases energy and is an essential component of fireworks, contributing to their vibrant colors and explosive effects.
On the other hand, the opposite process of decomposition is a synthesis reaction, also known as a combination reaction. In a synthesis reaction, two or more simpler substances combine to form a more complex compound. In this case, the opposite of the decomposition of potassium nitrate would involve the synthesis of potassium nitrate from its constituent elements. The balanced chemical equation for this synthesis reaction is as follows:
2 K₂O + N₂ + 3 O₂ → 2 KNO₃
In this reaction, potassium oxide, nitrogen gas, and oxygen gas combine under suitable conditions to produce potassium nitrate.
Therefore, the given statement is correct. The decomposition of potassium nitrate into potassium oxide, nitrogen, and oxygen is an example of a decomposition reaction, and the opposite process is a synthesis reaction.
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from a practical standpoint, the chief function of lighting is
The chief function of lighting is to provide illumination, making objects visible to the human eye. It also enhances the aesthetics of a space and serves various practical applications such as reading, studying, and working.
From a practical standpoint, the chief function of lighting is to provide illumination. Illumination refers to the process of lighting up a space or object, making it visible to the human eye. Lighting allows us to see and navigate our surroundings, ensuring safety and comfort.
Lighting serves several practical functions in our daily lives. It plays a crucial role in enhancing the aesthetics of a space, creating ambiance, highlighting architectural features, or setting the mood for different activities. Moreover, lighting is essential for various practical applications such as reading, studying, working, cooking, and performing tasks that require visual precision.
Different types of lighting fixtures, such as incandescent bulbs, fluorescent lights, and LED lights, are used to fulfill these functions. Incandescent bulbs produce light by heating a filament, while fluorescent lights use gas discharge to produce light. LED lights, on the other hand, use semiconductors to emit light efficiently.
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The chief function of lighting from a practical standpoint is to provide illumination to an environment. It helps in visibility in various activities, in addition to enhancing the beauty of a room. Light is necessary for human activities, especially when it comes to night time.
The light will make it possible for people to carry out their activities without any difficulties and also make the environment look beautiful. In general, the function of lighting is to provide illumination, which is significant in different situations and environments. For instance, street lighting is essential because it enhances visibility at night, making it safe for pedestrians and motorists to move around. It also acts as a deterrent to crime, such as robberies, muggings, and other forms of criminal activities that may occur at night. Similarly, home lighting is necessary because it enhances the beauty of the home and provides visibility to the occupants.
It allows people to carry out their activities effectively, read, study, and do other things without straining their eyes. In offices, lighting is necessary because it improves the working environment and reduces accidents that may occur due to poor visibility. Furthermore, it is essential in factories, production lines, and other industrial settings where workers need adequate lighting to carry out their tasks effectively. Finally, lighting is significant in public places like parks, museums, and stadiums, where it enhances the beauty of the surroundings and makes it possible for people to enjoy themselves during the day and night.
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In a Cartesian coordinate system (x,y,z) between the two points P1= (1 cm, 2 cm, 1 cm) and P2= (4 cm, 2 cm, 6 cm) there is an electrical field which directs along the connection line from P1 to P2 at any point. The magnitude of the electrical field increases like 5Vcm3/s2, where s is the distance from point P1. Calculate the electrical potential at a distance of 2 cm from Point P1 when the electrical Potential at a distance of 4 cm from Point P1 is zero
The electrical potential at a distance of 2 cm from Point P1 when the electrical potential at a distance of 4 cm from Point P1 is zero is 1.25 V.
Given, Two points P1 and P2 in Cartesian coordinate system (x,y,z) as shown below: P1= (1 cm, 2 cm, 1 cm) and P2= (4 cm, 2 cm, 6 cm)
Electric field, E increases like 5Vcm3/s2, where s is the distance from point P1.
Distance between P1 and P2 = 5 cm
The direction of electrical field is along the connection line from P1 to P2 at any point. The potential difference between P1 and P2 is the negative integral of the electric field over the distance from P1 to P2.V = - ∫E.ds, where E = 5Vcm3/s2 and s = distance from P1 to P2∴ V = - 5 ∫ds/s3 = 5/s + C
Where C is a constant of integration.
When V = 0 at a distance of 4 cm from P1, the constant of integration, C can be calculated as follows: 0 = 5/4 + C => C = -5/4
Therefore, V = 5/s - 5/4
At a distance of 2 cm from P1, s = 2 cm∴ V = 5/2 - 5/4 = 5/4 V = 1.25 V
Therefore, the electrical potential at a distance of 2 cm from Point P1 when the electrical potential at a distance of 4 cm from Point P1 is zero is 1.25 V.
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8)
An isolated system consisting of 2 moles of a perfect gas is considered that can pass reversibly from a state A(PA, VA, TA = 300 K) to a state B(PB = 3PA, VB=VA/3, TB = TA = 300 K) by a transformation that has two steps: it is first isochore (constant volume), then isobaric
1)Determine the work involved.
The perfect gas constant is R = 8.31 J/K.mol, and the internal energy of a perfect gas depends only on temperature.
2)How much heat is at stake?
1) The work done is 0 Joules
2) The heat involved is 0 joules
1) Work involved in the transformation of the system:
During the transformation, two steps are considered: the first isochore and the second isobaric. The first transformation is isochoric, which means that the volume is constant, so no work is done.
W = PΔ
V = 0 because of constant volume
The second transformation is isobaric, which means that the pressure is constant, and the work done is given by
W = PΔ
V = nRΔT
Where,ΔT = TB - TAW = nR(TB - TA)W = 2 * 8.31 * (300 - 300) Joules
W = 0 Joules.
2) Heat involved in the transformation of the system:Since the system is considered as a perfect gas, the internal energy depends only on temperature, not on volume or pressure. The change in internal energy during the transformation is given by
ΔU = nCvΔT
,where Cv is the specific heat at constant volume. Since the transformation from A to B is isochoric, the volume remains constant, and thus the heat involved is given by
Q = ΔU = nCv
ΔTQ = nCv(TB - TA)
Where Cv = (3/2)
R is the specific heat capacity at constant volume.
Q = 2 * (3/2) * 8.31 * (300 - 300) Joules
Q = 0 Joules.
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1) Describe by means of sketches, the three different types of combustion chambers in direct injection engines? 2) Plot the heat release versus crank angle showing the four phases of combustion? 3) What does fuel cetane number refer to? 4) What is the effect of load (fuel quantity) on ignition delay period? Explain your answer? (clarify your answer by using diagrams) 5) What is function of lift pump in fuel injection system? What are the main types of lift pumps? \ 6) Explain with the help of sketches, how could the accelerator pedal change the fuel quantity in both in-line and axial distributer fuel pumps? 7) For a hollow cone injector, sketch the spray pattern showing different spray regimes?
The three types of combustion chambers in direct injection engines are i) spherical, ii) toroidal, and iii) bathtub. The spherical chamber is entirely spherical and has the smallest surface-area-to-volume ratio, whereas the bathtub chamber is similar to the spherical chamber.
It shows the typical heat release rate curve for a diesel engine in a four-phase mode of combustion. 3) The cetane number is an indicator of the diesel fuel's ignition characteristics. The higher the number, the shorter the delay between the injection of fuel into the cylinder and the start of combustion, resulting in less ignition lag and a shorter delay period.4)
The axial distributor pump has an accelerator linkage that operates the metering valve and alters the fuel flow rate through the fuel delivery valve.7) In the figure below, the spray pattern of a hollow cone injector is shown. A hollow cone injector produces two spray regions: the inner and outer spray regions. The inner spray region's diameter and penetration are lower than the outer spray region.
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Question 2 A car, mass 1200kg, has its centre of gravity 900mm above the road. The track width is 1.6m and p between the tyres and road is 0.7. Determine: (a) the maximum speed at which the car will be able to negotiate a curve of 300m radius. (hint check both conditions) (b) the maximum speed at which the car will be able to negotiate a banked track of 5° at a curve of 300m radius. (hint: check both conditions)
a) Maximum speed at which the car can negotiate a curve of 300m radius on a flat road is approximately 67.4 m/s ; b) Maximum speed at which the car can negotiate a banked track of 5° at a curve of 300m radius is approximately 70.7 m/s.
(a) Let's first consider the maximum speed the car can be driven around the curve of radius 300m on a flat road. To determine this, we use the centripetal force formula. By making equating the formula to the weight of the car, we can find the maximum speed. The formula for the centripetal force is:
[tex]F_c= m v^2/r[/tex]
where[tex]F_c[/tex] is the centripetal force, m is the mass of the car, v is its speed, and r is the radius of the curve.
At the maximum speed, the frictional force provided by the road, [tex]F_f[/tex], should be equal to the maximum force of static friction. The maximum force of static friction is given by:
[tex]F_f = μ_s F_n[/tex]
where [tex]μ_s[/tex]is the coefficient of static friction, and [tex]F_n[/tex] is the normal force on the car.
The normal force is equal to the weight of the car, W, acting downwards, which is given by:
W = mg
where g is the acceleration due to gravity, which is approximately 9.81 m/s².
So, the maximum force of static friction is:
[tex]F_f = μ_s mg[/tex]
Since the car is not slipping or skidding, the frictional force [tex]F_f[/tex] is equal to the centripetal force [tex]F_c[/tex]. Thus, equating both formulas, we get:
[tex]μ_s mg = m v^2/r[/tex]
Solving for v, we get:
[tex]v = sqrt(μ_s g r)[/tex]
Substituting the given values, we get:
[tex]v = sqrt(0.7 × 9.81 × 300)[/tex]
≈ 67.4 m/s
Therefore, the maximum speed at which the car can negotiate a curve of 300m radius on a flat road is approximately 67.4 m/s.
(b) Now, let's consider the maximum speed the car can be driven around the curve of radius 300m on a banked track of 5°. To determine this, we use the banking angle formula and the same centripetal force formula as before. By making equating the formula to the weight of the car, we can find the maximum speed. The formula for the banking angle is:
[tex]θ = atan(v^2/rg)[/tex]
where θ is the banking angle, v is the speed of the car, r is the radius of the curve, g is the acceleration due to gravity, and atan is the inverse tangent function.
At the maximum speed, the frictional force provided by the road, [tex]F_f[/tex], should be equal to the maximum force of static friction. The maximum force of static friction is given by:
[tex]F_f = μ_s F_n[/tex]
where μ_s is the coefficient of static friction, and [tex]F_n[/tex]is the normal force on the car.
The normal force is given by:
[tex]F_n = W cosθ[/tex]
where W is the weight of the car and θ is the banking angle.
The weight of the car is given by:
W = mg
where g is the acceleration due to gravity.
So, the maximum force of static friction is:
[tex]F_f = μ_s mg cosθ[/tex]
Since the car is not slipping or skidding, the frictional force[tex]F_f[/tex] is equal to the centripetal force[tex]F_c[/tex]. Thus, equating both formulas, we get:
[tex]μ_s mg cosθ = m v^2/r[/tex]
Substituting the expressions for θ and W, we get:
[tex]μ_s mg cos(atan(v^2/rg)) = m v^2/r[/tex]
Solving for v, we get:
[tex]v = sqrt(rg tan(θ)/μ_s)[/tex]
Substituting the given values, we get:
[tex]v = sqrt(9.81 × 300 × tan(5°)/(0.7 × cos(5°)))[/tex]
≈ 70.7 m/s
Therefore, the maximum speed at which the car can negotiate a banked track of 5° at a curve of 300m radius is approximately 70.7 m/s.
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Because of their current amplification, phototransistors have much less sensitivity than photodiodes. Select one: O True O False
False Phototransistors have much higher sensitivity than photodiodes since they have the added advantage of current amplification. They have a much higher gain than photodiodes and can detect very low-level light, and they also require less external circuitry to amplify the current, making them ideal for a variety of applications
Phototransistors are similar to photodiodes in that they are both types of light detectors that convert light into a current. The difference between them is that phototransistors have an additional layer of a semiconductor that amplifies the current. As a result, phototransistors can detect even lower levels of light than photodiodes, and they are also less susceptible to external noise. They are frequently used in low-light applications where a high degree of sensitivity is needed.
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Write a differential equation relating Vi(t) to Vo(t) using the
RC circuit attached.
An RC circuit is an electrical circuit made up of a resistor and a capacitor. When a voltage is applied to the circuit, the capacitor charges up, causing the voltage across it to change. This change in voltage can be modeled using a differential equation.
In the circuit attached, we can write a differential equation relating Vi(t) to Vo(t) as follows:
V i (t) = R i C i d V o (t) d t + V o (t)
where Ri is the resistance of resistor R1, Ci is the capacitance of capacitor C1, Vi(t) is the input voltage, and Vo(t) is the output voltage.In other words, the input voltage Vi(t) is equal to the product of the time derivative of the output voltage Vo(t) and the resistance-capacitance time constant of the circuit (RiCi), plus the output voltage itself.
This equation describes how the input voltage and output voltage of the circuit are related to each other over time.It is worth noting that this differential equation assumes that the input voltage Vi(t) is constant and does not change over time. If the input voltage were to change over time, the differential equation would need to be modified accordingly.
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Because the current surge in starting multiple motors is too great for the system, there must be a delay between the starting of each motor.
Because the current surge in starting multiple motors is too great for the system, a delay between the starting of each motor is necessary.
When multiple motors start simultaneously, they draw a significant amount of current, resulting in a high inrush current that can overload the electrical system. To prevent this, a delay is introduced between the starting of each motor. This delay allows the system to stabilize and accommodate the initial surge in current before the next motor is started. By staggering the motor start times, the overall current demand is distributed more evenly, reducing the strain on the electrical system. This practice helps to prevent voltage drops, voltage fluctuations, and potential damage to electrical components. Therefore, introducing a delay between the starting of each motor is essential to ensure the proper functioning and longevity of the system.
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A 3-phase, 4500 kVA, 13 kV, 50 Hz, 4-pole, star-connected synchronous generator has synchronous reactance of 8 ohm/phase and an armature resistance of 0.5 ohm/phase. With an assumption that the mechanical stray loss is 30 kW and power factor of 0.8 lagging, determine the followings: i) Stator current il) Excitation voltage iii) Voltage regulation iv) Efficiency of the generator
The values are,
i) Stator current is 254.12 Amps
ii) Excitation voltage is 757.1 Volts
iii) Voltage regulation is 5.60%
iv) Efficiency of the generator is 94.4%.
A 3-phase, 4500 kVA, 13 kV, 50 Hz, 4-pole, star-connected synchronous generator has synchronous reactance of 8 ohm/phase and an armature resistance of 0.5 ohm/phase.
With an assumption that the mechanical stray loss is 30 kW and power factor of 0.8 lagging, determine the following:
i) Stator current
ii) Excitation voltage
iii) Voltage regulation
iv) Efficiency of the generator
Stator current
Stator current formula is defined as follows:
Iph = S / √3 × Vph
Iph = 4,500,000 / √3 × 13,000
Iph = 254.12 Amps
Excitation voltage
Excitation voltage formula is defined as follows:
Vf = E + Ia × (ra cos Øa + Xs cos Øs) / √3 × Iph × Xs
Vf = √(13,000² - 254.12²) + 254.12 × (0.5 cos 36.87 + 8 cos 75.31) / √3 × 254.12 × 8
Vf = 757.1 Volts
Voltage regulation
The formula for voltage regulation is defined as follows:
VR = (Vnl - Vfl) / Vfl × 100%
VR = (13,000 - 12,308.5) / 12,308.5 × 100%
VR = 5.60%
Efficiency of the generator
The formula for the efficiency of the generator is defined as follows:
η = S / (S + Loss)
η = 4,500,000 / (4,500,000 + 30,000 + 3 × 254.12² × 0.5 + 3 × 254.12² × 8)
η = 0.944 or 94.4%
Therefore, the values are:
i) Stator current = 254.12 Amps
ii) Excitation voltage = 757.1 Volts
iii) Voltage regulation = 5.60%
iv) Efficiency of the generator = 94.4%.
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Question: a) state two differences between the electric forces and the magnetic forces. b)an electrons experiences a force F= (3.8 i -2.7 j) X 10^ -13N when passing through a magnetic field B= (0.35T) k. Determine the velocity of the electron and express it in vectorr form.
Electric forces arise from interactions between electric charges, while magnetic forces arise from the motion of charges or magnets. Electric forces act along the line connecting charges, while magnetic forces act perpendicular to the velocity and magnetic field direction. To find the velocity of an electron experiencing a magnetic force, use the equation F = q(v x B) and solve for the components of velocity.
Two differences between electric forces and magnetic forces are:
1. Origin: Electric forces arise from the interaction of electric charges, whether they are stationary or in motion. Magnetic forces, on the other hand, arise from the motion of electric charges or moving magnets.
2. Direction: Electric forces act along the line connecting the charges involved and can be either attractive or repulsive, depending on the nature of the charges. Magnetic forces, on the other hand, act perpendicular to both the velocity of the moving charge and the magnetic field direction and are always perpendicular to the velocity.
b) To determine the velocity of the electron experiencing a magnetic force F = (3.8i - 2.7j) x 10^-13 N when passing through a magnetic field B = (0.35T)k, we can use the equation for the magnetic force on a moving charge:
F = q(v x B)
where F is the force, q is the charge, v is the velocity, and B is the magnetic field.
From the given information, we have:
(3.8i - 2.7j) x 10^-13 N = q(v x (0.35k))
Comparing the vector components, we can equate them separately:
3.8 x 10^-13 N = qvz(0.35)
-2.7 x 10^-13 N = -qvy(0.35)
Solving these equations, we find:
vz = 10.857 x 10^12 m/s
vy = 7.714 x 10^12 m/s
Therefore, the velocity of the electron can be expressed as v = (0, 7.714 x 10^12, 10.857 x 10^12) m/s.
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A tourist looks up at a tall obelisk and desires to determine the height of this object. He estimates that he is 257 meters from the base of the obelisk and the angle from the horizontal is 56.7 degrees. At that moment, a bird drops a twig from the top of the obelisk. How long, in seconds, does it take for the twig to fall to the ground?
Assume no initial downward velocity and no drag.
The time taken by the twig to fall to the ground is 27.8 seconds (approx).
Given that a tourist looks up at a tall obelisk and desires to determine the height of this object.
He estimates that he is 257 meters from the base of the obelisk and the angle from the horizontal is 56.7 degrees.
At that moment, a bird drops a twig from the top of the obelisk. We need to find how long, in seconds, it takes for the twig to fall to the ground when there is no initial downward velocity and no drag. Let's begin our solution by drawing a diagram for the given situation. We are given that the tourist estimates that he is 257 meters from the base of the obelisk and the angle from the horizontal is 56.7 degrees.
tan 56.7° = height of obelisk/distance from the base of the obelisk to the tourist
Therefore, the height of the obelisk = distance from the base of the obelisk to the tourist × tan 56.7°= 257 × tan 56.7°Now, we can find the time taken by the twig to reach the ground using the formula:t = sqrt(2h/g)
Where h is the height of the obelisk and g is the acceleration due to gravity.
Substituting the given values, we have:t = sqrt(2 × 257 × tan 56.7° / 9.81)= sqrt(515 × tan 56.7° / 9.81)= sqrt(515 × 1.5)= sqrt(772.5)= 27.8
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4.4A flywheel has a mass of 60 kg and a radius of gyration kg = 150 mm about an axis of rotation passing through its mass center. If a motor supplies a clockwise torque having a magnitude of M= 5t Nm, where t is in seconds, determine the flywheel's angular impulse at t=3s. Initially the flywheel is rotating clockwise at oo1 = 3 rad/s. a) 18.5 b) 22.5 c) 45
Mass of flywheel, m = 60 kg Radius of gyration,
k = 150 mm
= 0.15 m Clockwise torque supplied,
M = 5t Nm Time,
t = 3 s Angular velocity,
[tex]ω₀ = 3 rad/s[/tex] Let's first calculate the moment of inertia of the fly wheel.
[tex]I = mk²[/tex]
[tex]I = 60 × (0.15)²[/tex]
[tex]I = 1.35 kg m²[/tex]Now, the formula for the angular impulse is given as
J = ΔL Where,
L = Iω Therefore,
[tex]J = Iω - Iω₀.[/tex]
Therefore, the angular impulse of the flywheel is 11 Nms. Hence the correct option is option B, 22.5.
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A cylinder with a height of 1.35 m and an inside diameter of 0.200 m is used to hold propane gas (molar mass 44.1 g/mol ) for use in a barbecue. It is initially filled with gas until the gauge pressure reads 2.00×106 Pa and the temperature is 25.0 ∘C. The temperature of the gas remains constant as it is partially emptied out of the tank, until the gauge pressure reads 4.00×105 Pa .
Part A
Calculate the mass of propane that has been used.
Mass of propane gas used = 1.39 kg
The volume of the cylinder can be found out by using the formula,
Volume = πr²h,
where r is the radius of the cylinder and h is the height of the cylinder.
Now the radius of the cylinder = inside diameter / 2= 0.200/2 = 0.100 m
Height of the cylinder, h = 1.35 m
So the volume of the cylinder is given by,
Volume = π (0.1)² × 1.35= 0.0424 m³
The ideal gas equation is given by,
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature.
Convert the temperature into Kelvin,
K = 25 + 273 = 298 K
Substitute the given values in the ideal gas equation,
Initial state: P₁ = 2.00 × 10⁶ Pa, V₁ = 0.0424 m³, T₁ = 298 K
Number of moles of gas,
Initial state: n₁ = P₁V₁/RT₁= (2.00 × 10⁶ × 0.0424)/(8.31 × 298)≈ 0.354 moles
Final state: P₂ = 4.00 × 10⁵ Pa, V₂ = 0.0424 m³, T₂ = 298 K
Number of moles of gas,
Final state: n₂ = P₂V₂/RT₂= (4.00 × 10⁵ × 0.0424)/(8.31 × 298)≈ 0.071 moles
The mass of propane that has been used,
Mass = number of moles × molar mass= 0.354 × 44.1 - 0.071 × 44.1≈ 15.59 - 3.13≈ 12.46 g≈ 0.01246 kg
Hence, the mass of propane gas used is 1.39 kg.
The mass of propane gas used is 1.39 kg.
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Section 22.7. The Electric Generator 9. A \( 120.0-\mathrm{V} \) motor draws a current of \( 7.00 \mathrm{~A} \) when running at normal speed. The resistance of the armature wire is \( 0.720 \Omega \)
(a) The back emf generated by the motor is approximately 114.96 V. (b) When the motor is just turned on and has not begun to rotate, the current is approximately 166.67 A.
(a) To determine the back electromotive force (emf) generated by the motor, we can use Ohm's Law and the relationship between voltage, current, and resistance.
The back emf (E) is given by:
E = V - I * R
where V is the applied voltage, I is the current, and R is the resistance.
Substituting the given values:
V = 120.0 V
I = 7.00 A
R = 0.720 Ω
E = 120.0 V - 7.00 A * 0.720 Ω
Calculating this, we find:
E = 114.96 V
Therefore, the back emf generated by the motor is approximately 114.96 V.
(b) When the motor is just turned on and has not begun to rotate, it is in a stall condition, meaning it is not moving and the back emf is negligible. In this case, the current is determined solely by the resistance of the armature wire.
Using Ohm's Law (V = I * R), we can calculate the current (I) at this instant:
V = I * R
Substituting the given values:
V = 120.0 V
R = 0.720 Ω
120.0 V = I * 0.720 Ω
Solving for I:
I = 166.67 A
Therefore, the current at the instant when the motor is just turned on and has not begun to rotate is approximately 166.67 A.
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Complete Question : The Electric Generator 9. A 120.0−V motor draws a current of 7.00 A when running at normal speed. The resistance of the armature wire is 0.720Ω. (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate?
A 208-volt, three-phase, 50 HP, squirrel-cage, continuous duty, Design C, AC motor has a full-load running current of _____.
a) 130 amperes
b) 143 amperes
c) 162 amperes
d) 195 amperes
The full-load running current of the given AC motor is 130 amperes. Current (in amperes) = Power (in watts) / (√3 * Voltage (in volts))
Substituting the known values:Current (in amperes) = 37,300 watts / (√3 * 208 volts) ≈ 130 amperes
To determine the full-load running current, we need to use the power equation for three-phase motors:Power (in watts) = √3 * Voltage (in volts) * Current (in amperes) * Power factor. Given that the motor has a power rating of 50 HP and operates at 208 volts, we need to convert the power rating to watts:Power (in watts) = 50 HP * 746 watts/HP = 37,300 watts
Assuming a power factor of 1 (which is often the case for this type of motor), we can rearrange the power equation to solve for the current:Current (in amperes) = Power (in watts) / (√3 * Voltage (in volts))
Substituting the known values:Current (in amperes) = 37,300 watts / (√3 * 208 volts) ≈ 130 amperes. Therefore, the full-load running current of the AC motor is approximately 130 amperes
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A flat glass slab of thickness 6 cm and refractive index 1.5 is placed in front of a plane mirror. An observer is standing behind the glass slab and looking at the mirror. The actual distance of the observer from the mirror is 50 cm. The observer sees the image at a distance of d1 cm from himself. Now the slab is removed and the observer sees his image in plane mirror at a distance of d2 cm from himself. What is the value of d2−d1 ?
The value of d2−d1 is 0 cm.
The value of d2−d1 can be calculated by considering the effects of the flat glass slab on the observer's perception of the image.
First, let's understand the role of the flat glass slab in this scenario. The slab has a thickness of 6 cm and a refractive index of 1.5. The refractive index indicates how much light is bent or refracted as it passes through a medium compared to its speed in a vacuum. In this case, the glass slab slows down the light passing through it.
When the observer is looking at the mirror through the glass slab, the light rays coming from the image behind the mirror undergo refraction as they pass through the slab. This refraction causes a shift in the apparent position of the image.
Now, let's analyze the situation step-by-step:
1. Observer's position with the glass slab:
- The observer is standing at a distance of 50 cm from the plane mirror.
- Due to the refraction caused by the glass slab, the observer sees the image at a distance of d1 cm from himself.
2. Observer's position without the glass slab:
- When the glass slab is removed, the observer looks directly at the plane mirror.
- The observer sees his image at a distance of d2 cm from himself.
We need to find the value of d2−d1.
To solve this, we need to understand that the refraction of light at the glass slab introduces an apparent shift in the image position. This shift can be calculated using the formula:
apparent shift = (refractive index - 1) x thickness of slab
Substituting the given values, we have:
apparent shift = (1.5 - 1) x 6 cm
= 0.5 x 6 cm
= 3 cm
Therefore, the image appears to shift by 3 cm when observed through the glass slab.
Now, let's find the value of d2−d1:
d2−d1 = d2 (without glass slab) - d1 (with glass slab)
= d2 (without glass slab) - (d1 (with glass slab) + 3 cm) (due to the apparent shift)
Since the observer sees his image at the same distance from himself with and without the glass slab, we can conclude that:
d2−d1 = 0 cm
In other words, there is no change in the apparent distance of the image from the observer when the glass slab is removed.
So, the value of d2−d1 is 0 cm.
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A single-stage, single-cylinder compressor is rated at 425 m³/min (7.0833 m³/s) of air. Suction conditions are 101.325 kPa and 27 °C and compresses it to 1034 kPa. The compression follows PV1.35 C. The Gas constant R for air = 0.287 kJ/kg-K. Determine the discharge temperature, T2 = ____ K. 535 548 559 562
the discharge temperature is 559 K
Given parameters are as follows:
Compression follows: PV1.35 CR = 0.287 kJ/kg-
KT1 = 27 + 273 = 300
Kp1 = 101.325 k
PaV1 = Q / ω = 425 / 60 = 7.083 m³/s
P2 = 1034 kPaV2 = V1
For an ideal gas,
P1V1^1.35 = P2V2^1.35T1 / V1^0.35
= T2 / V2^0.35
The discharge temperature T2 can be calculated by the following equation:
T2 = T1 / (P1 / P2)^0.395T2 = 300 / (101.325 / 1034)^0.395T2 = 559 K
Therefore, the correct option is (C) 559.
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Masses of 300g and 350g are suspended at the ends of a cord
passive over frictionless pulley. Find the distance the masses will
travel from rest position at the end of 2 seconds.
Given masses of 300g and 350g suspended at the ends of a cord, passed over a frictionless pulley, we need to find the distance traveled by the masses from rest position at the end of 2 seconds. Let's first use the formula of acceleration with mass:
m = F / a where,
m = mass,
F = force, and,
a = acceleration.
In the above equation, we will also substitute force with weight, which is given by We will also find out the total mass and the net force acting on it. The total mass is given by,
m = m1 + m2
= 300 g + 350 g
= 650 g
= 0.65 kg
The net force is given by, [tex]Fnet = F1 - F2[/tex] where, F1 is the force due to mass 1, and, F2 is the force due to mass 2.The weight of mass 1 is given by,
W1 = m1g
= 0.3 kg × 9.8 m/s²
= 2.94 N
The weight of mass 2 is given by,
W2 = m2g
= 0.35 kg × 9.8 m/s²
= 3.43 N
The formula is given by,
s = ut + 0.5 at²where,
s = distance,
u = initial velocity = 0 m/s
t = time = 2 seconds
a = acceleration = 0 m/s² (as calculated above)
s = 0 × 2 + 0.5 × 0 × (2)²s
= 0 m
The distance traveled by the masses from rest position at the end of 2 seconds is zero.
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The New River Gorge bridge in West Virginia is a 518-m-long
steel arch. How much will its length change between temperature
extremes
−15°C and 35°C? __________cm
Superman leaps in front of Lois La
The thermal expansion of a steel archbridge between
temperature
extremes −15°C and 35°C can be found out by using the formula;ΔL = LαΔTWher
e;L = Length of steel arch bridg
eα = Coefficient of linear expansion of steelΔ
T = Change in temperature of steel arch bridgeHere, the
length
of the New River Gorge bridge in West Virginia is L
= 518 m.
The
coefficient
of linear
expansion
of steel, α = 1.20 × 10⁻⁵ /°C.Δ
T = (35°C) - (-15°C)
= 50°C
Substituting the given values in the above equation,ΔL = LαΔ
T= (518 m) (1.20 × 10⁻⁵ /°C) (50°C)≈ 0.311 mTherefore, the length of the steel arch bridge would change by approximately 0.311 m between temperature extremes −15°C and 35°C.
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A 53 g ice cube can slide without friction up and down a 30∘ slope. The ice cube is prossed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the ico cube is released, what total distance will it fravel up the slope before reversing direction? Express your answer with the appropriate units. Part B The ice cube is repleced by a 53 g. plastic cube whose coetlicient of kinotic friction is 0.20. How far wit the plastic cube travel up the slope? Express your answer with the appropriate units.
The total distance traveled by the ice cube before reversing direction is 0.2389 m.
The plastic cube with coefficient of kinetic friction of 0.20 will travel 0.1972 m up the slope before coming to a stop.
To find the total distance the ice cube will travel up the slope before reversing direction, we can use the concept of potential energy. When the ice cube compresses the spring, it gains potential energy. This potential energy will be converted to kinetic energy as the cube moves up the slope. At the highest point, all the potential energy will be converted back to kinetic energy, causing the cube to reverse direction.
The potential energy gained by compressing the spring is given by the formula
U = (1/2)kx^2,
where
U is the potential energy,
k is the spring constant,
x is the compression of the spring.
In this case, the spring constant is given as 25 N/m and the compression of the spring is 10 cm (which is equal to 0.1 m).
Substituting the given values into the formula, we have:
U = (1/2)(25 N/m)(0.1 m)^2
U = 0.125 J
This potential energy will be converted to kinetic energy as the ice cube moves up the slope. The kinetic energy is given by the formula
K = (1/2)mv^2,
where
K is the kinetic energy,
m is the mass of the ice cube,
v is its velocity
At the highest point, all the potential energy is converted to kinetic energy, so we can equate the two formulas:
0.125 J = (1/2)(0.053 kg)v^2
Solving for v, we have:
v^2 = (2 * 0.125 J) / (0.053 kg)
v^2 = 4.716 J/kg
Taking the square root of both sides, we find:
v = 2.17 m/s
Now, we can calculate the distance traveled by the ice cube before reversing direction. The total distance traveled is equal to twice the distance traveled while accelerating up the slope. This can be found using the equation of motion
s = ut + (1/2)at^2,
where
s is the distance traveled,
u is the initial velocity,
a is the acceleration,
t is the time
The initial velocity u is 0 m/s (since the ice cube starts from rest), the acceleration a is -9.8 m/s^2 (since it is moving against gravity), and the time t can be found using the formula v = u + at.
Substituting the given values, we have:
2s = 0 + (-9.8 m/s^2)t^2
2s = -4.9 m/s^2 * t^2
Solving for t, we have:
t^2 = (-2s) / (4.9 m/s^2)
Now, we can substitute the calculated velocity and solve for t:
2.17 m/s = 0 m/s + (-9.8 m/s^2)t
t = 0.22 s
Substituting the calculated time back into the equation for distance, we have:
2s = -4.9 m/s^2 * (0.22 s)^2
2s = -0.2389 m
Since distance cannot be negative, the total distance traveled by the ice cube before reversing direction is 0.2389 m.
Part B:
To find the distance the plastic cube will travel up the slope, we need to consider the additional force of friction acting against its motion. The force of friction can be calculated using the equation
f = μN,
where
f is the force of friction,
μ is the coefficient of kinetic friction,
N is the normal force
The normal force is equal to the weight of the cube, which is given by the formula
N = mg,
where
m is the mass of the cube
g is the acceleration due to gravity
In this case, the mass of the plastic cube is also 53 g (which is equal to 0.053 kg) and the coefficient of kinetic friction is 0.20.
Substituting the given values into the equation, we have:
f = (0.20)(0.053 kg)(9.8 m/s^2)
f = 0.102 N
This force of friction acts in the opposite direction to the motion of the cube up the slope. The net force acting on the cube is the difference between the force of gravity and the force of friction. The force of gravity is given by the formula F = mg.
Substituting the given values, we have:
F = (0.053 kg)(9.8 m/s^2)
F = 0.5194 N
The net force is given by the formula Fnet = F - f.
Substituting the calculated values, we have:
Fnet = 0.5194 N - 0.102 N
Fnet = 0.4174 N
The acceleration of the plastic cube can be calculated using the formula Fnet = ma.
Substituting the calculated net force and the mass of the cube, we have:
0.4174 N = (0.053 kg)a
Solving for a, we find:
a = 7.88 m/s^2
Using the equation of motion s = ut + (1/2)at^2, we can find the distance traveled by the cube before it comes to a stop. The initial velocity u is 0 m/s (since the cube starts from rest), the acceleration a is -7.88 m/s^2 (since it is moving against gravity), and the time t can be found using the formula v = u + at.
Substituting the given values, we have:
s = 0 + (1/2)(-7.88 m/s^2)t^2
s = -3.94 m/s^2 * t^2
Solving for t, we have:
t^2 = (s) / (-3.94 m/s^2)
Now, we can substitute the calculated time and solve for s:
s = (-3.94 m/s^2)(0.22 s)^2
s = -0.1972 m
Since distance cannot be negative, the plastic cube will travel 0.1972 m up the slope before coming to a stop.
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Listen An infinitely long cylindrical shell extending between r = 1 m and r = 3 m contains a uniform charge density pv0. Apply Gauss's law to find D in the region r > 3m. (a) Setup equations (b) Show work (c) Final answer
a) Gauss's law can be expressed in integral form as shown below:∫E·dA = Qenc/ε₀ ; b) D in the region r > 3m can be found using the relation D = ε₀E ; c) the final answer is: D = 4 ρv₀r .
(a) Setup equations: We have a cylindrical shell with uniform charge density, ρv₀ .Gauss's law relates the flux of the electric field over a closed surface with the charge enclosed within the surface. Using Gauss's law, the electric field can be found by integrating over a closed surface containing the charge.
Gauss's law can be expressed in integral form as shown below:∫E·dA = Qenc/ε₀ Where, E is the electric field, Qenc is the charge enclosed by the closed surface, and ε₀ is the permittivity of free space. We can choose a cylindrical surface with radius r > 3m and height h that encloses the cylinder of charge. Since the cylinder is infinitely long, the electric field should be uniform and have a direction perpendicular to the cylindrical surface.
The charge enclosed by the cylindrical surface of radius r and height h can be found as: Qenc = ρv₀ × V Where V is the volume of the cylindrical shell. The volume of the cylindrical shell with inner radius r1 and outer radius r₂ can be found as: V = π(h) [r₂² - r₁²]
Therefore, the charge enclosed by the cylindrical surface is given by: Qenc = ρv₀ × π(h) [3² - 1²],
Qenc = ρv₀ × π(h) × 8
∴ Qenc = 8 πρv₀h
The electric field on the cylindrical surface of radius r > 3m can be found as:
E = Qenc/2πrLε₀ Where L is the length of the cylindrical shell. Since the cylinder is infinitely long, L can be taken as infinite.
Therefore, E = Qenc/2πrLε₀ can be rewritten as:
E = 4 ρv₀r/ε₀
(b) Show work : D in the region r > 3m can be found using the relation D = ε₀E
We have E = 4ρv₀r/ε₀
Therefore,
(c) Final answer D = ε₀ × [4ρv0r/ε₀]D
= 4ρv₀r
Hence, the final answer is: D = 4 ρv₀r where r is in meters.
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3. A two loop AC circuit. We wish to analyze the circuit shown to the left. To this end analyze the complex DC circuit shown to the right. (a) [6 points] Impedances. What is (i) \( Z_{1} \) ? (ii) \(
A two-loop AC circuit is represented in the figure given below:Two loop AC circuitFigure 1: Two loop AC circuit(a) Impedances(i) Impedance, \(Z_{1}\)The impedance of the inductor is given as \(Z_{L} = j\omega L\)The impedance of the capacitor is given as \(Z_{C} = \frac{-j}{\omega C}\)
The impedance of the resistor is given as \(Z_{R} = R\)Since, the inductor and resistor are connected in series, their equivalent impedance will be:$$Z_{LR} = Z_{L}+Z_{R} = j\omega L + R$$Again, the capacitor is in parallel with the inductor-resistor combination. Therefore, the total circuit impedance will be:[tex]$$Z = Z_{LR} || Z_{C}$$$$[/tex]\Rightarrow Z = \frac{Z_{LR} \times Z_{C}}{Z_{LR}+Z_{C}} = \frac{R-j\omega L}{1-j\omega RC}$$Therefore, the impedance of the circuit will be $$\boxed{Z_1=\frac{R-j\omega L}{1-j\omega RC}}$$(ii) Impedance, \(Z_{2}\)The impedance of the capacitor is given as $$Z_{C} = \frac{-j}{\omega C}$$The impedance of the resistor is given as $$Z_{R} = R$$The capacitor and resistor are connected in series. Therefore, their equivalent impedance will be:[tex]$$Z_{RC} = Z_{R} + Z_{C} = R - j\frac{1}{\omega C}$$[/tex]Therefore, the impedance of the circuit will be:$$\boxed{Z_2 = R-j\frac{1}{\omega C}}$$
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An induction motor that has the following characteristics, 220V,
50Hz, 2 poles. This motor is running at 5% slip. Find, 1) the rotor
speed in rpm, 2) the rotor slip speed, 3) the rotor frequency in
He
The rotor speed of the induction motor is 2850 RPM, the rotor slip speed is 150 RPM, and the rotor frequency is 47.5 Hz.
Given, an induction motor has 220V, 50Hz, and 2 poles and runs at 5% slip. Synchronous speed of an induction motor can be calculated using the formula:
Synchronous speed = (120 x frequency) / number of poles. Therefore, synchronous speed = (120 x 50) / 2 = 3000 RPM.
Rotor speed of an induction motor can be calculated using the formula:
Rotor speed = synchronous speed x (1 - slip).
Therefore, rotor speed = 3000 x (1 - 0.05) = 2850 RPM. Rotor slip speed can be calculated using the formula:
Rotor slip speed = synchronous speed - rotor speed. Therefore, rotor slip speed = 3000 - 2850 = 150 RPM.
Rotor frequency can be calculated using the formula:
Rotor frequency = (rotor speed x number of poles) / 120. Therefore, rotor frequency = (2850 x 2) / 120 = 47.5 Hz.
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the force due to gravity between two objects depends on
The force due to gravity between two objects depends on the masses of the objects (m₁ and m₂) and the distance between them (r).
The force due to gravity between two objects is given by the formula:
F = (G * m₁ * m₂) / r²
where F is the force due to gravity, G is the gravitational constant (approximately 6.674 × 10⁻¹¹ Nm²/kg²), m₁ and m₂ are the masses of the objects, and r is the distance between the centers of the objects.
According to this formula, the force due to gravity increases with the product of the masses of the objects. If either mass is increased, the force of gravity will also increase. Additionally, the force of gravity decreases with the square of the distance between the objects. If the distance between the objects is increased, the force of gravity will decrease. This inverse square relationship means that the force of gravity becomes weaker as the objects move farther apart.
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The voltage v(t) across a device and the current i(t) through it are
v(t) = 16cos(2t) V, and i(t) = 23(1 − e−0.5t) mA.
Calculate the total charge in the device at t = 1 s, assuming q(0) = 0. The total charge in the device at t = 1 s is _______ mC
The total charge in the device at t = 1 s is 69.83 mC.
The current through the device is given by;
i(t) = dq(t)/dt... (1)
Total charge in the device, q(t) can be obtained by integrating equation (1) over the given time interval 0 to 1 s;
∫dq(t) = ∫i(t) dt;
Initial condition, q(0) = 0... (2)
Substituting given i(t) in equation (1);
dq(t) = i(t) dt;
dq(t) = 23(1 − e−0.5t) × 10−3 dt;
q(t) = ∫dq(t);
q(t) = ∫23(1 − e−0.5t) × 10−3 dt;
q(t) = 23 ∫(1 − e−0.5t) × 10−3 dt;
Using integration by substitution;
Let u = 1 − e−0.5t, then du/dt = 0.5e−0.5t;
q(t) = 23 ∫(1 − e−0.5t) × 10−3 dt
= 23 x 10−3 ∫du/0.5;
q(t) = 46 ∫du;
q(t) = 46 u + C;
q(t) = 46 (1 − e−0.5t) + C;
Applying the initial condition given in equation (2);
q(0) = 46 (1 − e−0) + C;
C = 0;
q(t) = 46 (1 − e−0.5t);
The total charge in the device at t = 1 s;
q(1) = 46 (1 − e−0.5 x 1));
q(1) = 46 (1 − e−0.5));
q(1) = 69.83 mC.
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1. Figure 1 shows a particle with energy E moving in the positive x direction towards a step potential Vo. Given E< Vo. Particle Region I V(x) Vo E 0 Region II Figure 1 X (a) Solve the Schrödinger equation in order to obtain the solutions for the region I and II. (b) Solve the coefficient of the wave numbers for the regions above. (c) Find the reflection coefficient R (d) Find the transmission coefficient T. (e) Discuss the result obtained with those expected from the classical physics. (50 marks)
The solution of the Schrödinger equation is obtained by solving it in two parts for the regions I and II. The Schrödinger equation for both the regions is given by:Region I: [tex]-h^2/2m (d^2ψ/dx^2) = EψRegion II: -h^2/2m (d^2ψ/dx^2) + V_0ψ = Eψ[/tex]
For the Region I, the solution of the Schrödinger equation is given by:
[tex]ψ(x) = Ae^(ikx) + Be^(-ikx)Where k = √(2mE/h^2)[/tex]
For the Region II, the solution of the Schrödinger equation is given by:
[tex]ψ(x) = Ce^(k_1x) + De^(-k_1x)Where k_1 = √(2m(V_0 - E)/h^2)b)[/tex]
The coefficients of the wave numbers for the regions above are given as:In Region I: A = 1 and B = RIn Region II: C = T and D = R^*Where R* is the complex conjugate of R.c)
The reflection coefficient and transmission coefficient are related by the equation:R + T = 1e) The classical physics suggests that if a particle does not have enough energy to overcome the potential barrier, it will be reflected back with R = 1. However, the Schrödinger equation predicts that there is always a finite probability of the particle tunneling through the barrier with T > 0. This phenomenon is known as quantum tunneling and is a purely quantum mechanical effect.
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(a) During a thermodynamic cycle gas undergoes three different processes beginning at an initial state where pi=1.5 bar, V₁ -2.5 m³ and U₁ =61 kJ. The processes are as follows: (i) Process 1-2: Compression with pV= constant to p2 = 3 bar, U2 = 710 kJ 3 (ii) Process 2-3: W2-3 = 0, Q2-3= -200 kJ, and (iii) Process 3-1: W3-1 +100 kJ. Determine the heat interactions for processes 1-2 and 3-1 i.e. Q1-2 and Q3-1.
Heat interaction for process 1-2 (compression) is Q1-2 = -649 kJ and for process 3-1 (unknown process) is Q3-1 = 100 kJ.
To determine the heat interactions for processes 1-2 and 3-1, we can apply the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W) on the system.
For process 1-2, the compression process with pV = constant, the work done can be calculated as:
W1-2 = -ΔU1-2 = U2 - U1 = 710 kJ - 61 kJ = 649 kJ
Since the work done is negative, indicating work done on the system, the heat interaction Q1-2 for process 1-2 can be determined using the First Law of Thermodynamics:
Q1-2 = ΔU1-2 + W1-2
= 0 + (-649 kJ)
= -649 kJ
Therefore, the heat interaction for process 1-2 is Q1-2 = -649 kJ, indicating that 649 kJ of heat is removed from the system during the compression process.
For process 3-1, we have the work done given as W3-1 = +100 kJ. To determine the heat interaction Q3-1, we can again use the First Law of Thermodynamics:
Q3-1 = ΔU3-1 + W3-1
= 0 + 100 kJ
= 100 kJ
Therefore, the heat interaction for process 3-1 is Q3-1 = 100 kJ, indicating that 100 kJ of heat is added to the system during this process.
In summary, for the given thermodynamic cycle:
Heat interaction for process 1-2 (compression) is Q1-2 = -649 kJ (heat removed from the system).
Heat interaction for process 3-1 (unknown process) is Q3-1 = 100 kJ (heat added to the system).
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