The minimum coefficient of static friction between the tires and the road that would be needed for the car to round the curve without slipping is 0.1878.
Diameter of cylinder, d = 8.50 m
Radius of cylinder, r = d/2 = 4.25 m
Angular velocity, ω0 = 0 (initially)
Angular velocity, ω1 = 2π/60 rad/s (after 10 min)
Time interval, t = 10 min
= 600 sec
Angular acceleration is given by,
α = (ω1 - ω0)/t
α = (2π/60 - 0)/600
α = [tex]1.05 * 10^{-4[/tex] rad/s²
(i)The formula for centripetal acceleration is given by,ac = v²/r
Where,v = 32 m/sr
= 56 m
Therefore,ac = v²/rac
= 102.4/56ac
= 1.83 m/s²
Therefore, the centripetal acceleration of the car is 1.83 m/s².
(ii)The minimum coefficient of static friction can be calculated using the formula,μs = ac/g
Where,ac = 1.83 m/s²g = 9.8 m/s²
Therefore,μs = 1.83/9.8μs
= 0.1878
The minimum coefficient of static friction between the tires and the road that would be needed for the car to round the curve without slipping is 0.1878.
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those portions of the celestial sphere near the celestial poles that are either always above or always below the horizon
*these kind of stars never rise and never set since they remain above/below the horizon
Right Ascension (RA)
Declination
Circumpolar
Those portions of the celestial sphere near the celestial poles that are either always above or always below the horizon, these kind of stars never rise and never set since they remain above/below the horizon is C. Circumpolar.
The celestial poles are the points on the celestial sphere that are directly above the Earth's North and South Poles. The celestial sphere is an imaginary sphere that encircles the Earth, and is used to describe the positions of objects in the sky, those portions of the celestial sphere near the celestial poles that are either always above or always below the horizon are called circumpolar regions. In these regions, stars never rise or set since they remain above or below the horizon. Circumpolar stars are stars that always remain above or below the horizon and never rise or set, these stars are located near the celestial poles and they appear to rotate around them.
The altitude of these stars depends on the observer's latitude, the closer the observer is to the North or South Pole, the higher the circumpolar stars will be above the horizon. The coordinates used to locate a star on the celestial sphere are right ascension (RA) and declination. RA is similar to longitude on the Earth, and it measures the east-west position of a star on the celestial sphere. Declination is similar to latitude on the Earth, and it measures the north-south position of a star on the celestial sphere. So therefore these coordinates can be used to locate any star on the celestial sphere, including circumpolar stars.
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You are asked to design a resistor using an intrinsic semiconductor bar of length L and a cross-sectional area A. The scattering rate for electrons and holes are both 1/t, and the effective mass for holes is mo* which is two times larger than the effective mass for electrons. The bandgap is G. Assume T=300K. A. Give an expression for the intrinsic electron concentration in terms of the parameters given above. Show all steps. The final expression should be as compact as possible. B. Obtain an expression for the current in the bar in terms of the parameters given if a voltage Vg is applied across the bar. Sketch the bar with the voltage applied and show with arrows indicating the directions of Electric Field and current densities. C. If the hole effective mass, me* is 1xmo, hole and electron mobilities are 0.17 m²/V.s and 0.36 m'/V.s, respectively. Consider G=0.7 ev. Calculate total resistance of the bar. Be careful with units.
The total resistance of the bar is given by; [tex]R = L / (σ * A)[/tex]
A. Expression for intrinsic electron concentration
The intrinsic carrier concentration for electrons is given by the formula;
[tex]n = 2 [(2πmkT/h²) ^ 3 / 2] * e ^ (−Eg / 2kT)[/tex]
Where;h is Plank's constant K is the Boltzmann constant
Eg is the Band Gap Energy, m is the effective mass of electrons k, T is Boltzmann constant multiplied by temperature T is the absolute temperature of the body, e is the electric charge
The above equation can be written as; [tex]n = AT^ (3/2) * e^ (-Eg/2kT)[/tex]
Where; A = 4 * π * (mk) ^ 3 / (2 * h ^ 3)
B. Expression for the current in the bar
Assuming the applied voltage across the intrinsic semiconductor bar is Vg, then the current in the bar is given by;
[tex]J = (qμn * EFn * Ap + qμp * EFp * Ap)Vg / L[/tex]
Where; q is the charge of an electronμn and μp are the mobilities of electrons and holes respectively
Ap is the cross-sectional area of the bar
EFn is the electric field for electrons
EFp is the electric field for holesVg is the voltage applied
L is the length of the bar C. Calculation of total resistance of the bar
The total resistance of the bar is given by; [tex]R = L / (σ * A)[/tex]
Where ;σ is the conductivity of the bar.[tex]σ = q * (μn * n + μp * p)[/tex]
Where; p is the intrinsic carrier concentration for holes.
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Q1. A lawn sprinkler sprays water from an array of 12 holes, each 0.40 cm in diameter. The sprinkler is fed by a garden hose 3.5 cm in diameter, which is supplied by a tap. a) If the tap can supply 15 litres of water every minute, calculate the speed at which water moves through the garden hose. (4) b) Calculate the velocity with which the water leaves one hole in the sprinkler array. (4)
(a) The speed at which water moves through the garden hose is 25.97 cm/s. (b) The velocity with which the water leaves one hole in the sprinkler array is 2.57 m/s.
a) To calculate the speed at which water moves through the garden hose, we'll use the formula for the volume rate of flow, which is given by
Q = A×v, where A is the cross-sectional area of the hose and v is the velocity of the water. We have the diameter of the hose, which we'll use to find its radius.
r = d/2 = 3.5/2 = 1.75 cmA = πr² = π(1.75)² = 9.625 cm²
To convert the flow rate from L/min to cm³/s, we'll multiply by 1000/60, because 1 L = 1000 cm³ and 1 min = 60 s.Q = 15 × 1000/60 = 250 cm³/s
Q = A × v ⇒ v = Q/A
= 250/9.625
= 25.97 cm/s
(b)The velocity with which the water leaves one hole in the sprinkler array can be found using Bernoulli's equation, which relates the pressure of the fluid to its velocity.
p1 + (1/2)ρv1² = p2 + (1/2)ρv2²
where p1 and v1 are the pressure and velocity of the water as it enters the sprinkler array, and p2 and v2 are the pressure and velocity of the water as it leaves the hole in the sprinkler.
We'll assume that the pressure remains constant throughout, so p1 = p2. Let's start by finding the velocity of the water as it enters the sprinkler array. Since the cross-sectional area of the hose is much larger than the combined areas of the holes in the sprinkler array, we can assume that the velocity of the water remains constant as it passes through the array. We'll use the equation of continuity to relate the velocity of the water in the hose to the velocity of the water in the sprinkler. A1v1 = A2v2
where A1 and v1 are the cross-sectional area and velocity of the hose, and A2 and v2 are the cross-sectional area and velocity of the water as it passes through one hole in the sprinkler.
We have already found
A1 and v1.v2 = A1v1/A2 = (9.625 × 25.97)/(12 × (0.4/2)² × π) = 2.57 m/s
The velocity of the water as it leaves the hole in the sprinkler is 2.57 m/s.
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Under constant-volume conditions, 2700 J of heat is added to 1.5 moles of an ideal gas. As a result, the temperature of the gas increases by 86.6 K. How much heat would be required to cause the same temperature change under constant-pressure conditions? Do not assume anything about whether the gas is monatomic, diatomic, etc. QP=
The amount of heat required to cause the same temperature change under constant-pressure conditions is 3779.986 JOULE.
At constant volume, the conditions are:
heat = 2700 J
number of mole (gas) n = 1.5 moles
change in temperature ΔT = 86.6 k
Now according to the rules of thermodynamic Change in internal energy at constant volume is ΔU =2700 J and change of entropy in a constant pressure will be equal to the transfer heat.
At constant volume :
[tex]Q=mc_v\Delta T\\\\ 2700\ \text{Joule}=1.5\ \text{mole}\times c_v \times\ 86.6\ K\\\\ c_v=20.79 \dfrac{\text{Joule}}{\text{mole}\cdot{K}}[/tex]
since gas undergoes the same temperature change in both process change in internal energy is same.
By Mayors equation :
[tex]c_p-c_v=R[/tex]
[tex]c_p-20.79=8.314\\\\c_p=29.099 \dfrac{\text{Joule}}{\text{mole}\cdot{K}}[/tex]
Heat would be required at constant pressure condition:
[tex]Q=mc_p \Delta T\\\\Q=1.5 \times29.099\times 86.6\\\\Q=3779.988 \rm J[/tex]
hence, the heat at constant pressure is 3779.988 J
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In a PV system, when are batteries generally outputting charge to the loads? Midday Sunny Days Solar Noon Night time, cloudy days Question 43 (1 point) The electrolyte in a Battery refers to the: +ve
In a PV system, the batteries are generally outputting charge to the loads during night time and cloudy days. This is because during night time and cloudy days, the solar panels are not able to generate enough electricity to fulfill the energy demands of the loads and therefore the batteries are used as a backup to provide electricity to the loads.
The electrolyte in a battery refers to the substance which conducts electricity in a battery. In a lead-acid battery, the electrolyte is made up of a mixture of sulfuric acid and water. The sulfuric acid is used as the conducting medium which allows the flow of electrons between the anode and cathode terminals of the battery.
The electrolyte also helps in the charging and discharging process of the battery by releasing or absorbing hydrogen ions depending on the direction of the current flow.Batteries are an essential component of PV systems as they provide a reliable source of backup power during times when there is not enough sunlight to generate electricity. The batteries store excess energy generated by the solar panels during the day and release it when needed, allowing the PV system to meet the energy demands of the loads even during times of low sunlight.
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In a boundary layer formation over a flat plate, define and
derive mathematical expressions for displacement thickness δ * and
momentum thickness ‘θ’.
In the context of a boundary layer formation over a flat plate, the displacement thickness is the distance by which the boundary layer must be displaced in the normal direction to the plate in order to accommodate the presence of the boundary layer and is typically denoted by the symbol δ*.
The momentum thickness θ, on the other hand, is defined as the distance by which the upper and lower boundaries of the boundary layer have to be moved in the direction of the flow to conserve the total momentum flow rate of the boundary layer.
The derivation of mathematical expressions for displacement thickness δ* and momentum thickness ‘θ’ can be described as follows; For an incompressible, laminar, steady-state boundary layer over a flat plate, the momentum equation can be written as;[tex]$$\rho u \frac{\partial u}{\partial x} = \mu \frac{\partial^2 u}{\partial y^2}$$[/tex]
Where
ρ is the density of the fluid,
u is the velocity of the fluid,
x is the distance along the flat plate,
y is the distance normal to the flat plate, and
μ is the dynamic viscosity of the fluid.
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The following information pertains to Questions 1-3. A waveguide is formed from a hollow conducting tube of some cross section that is filled with a material having a dielectric constant (relative permittivity) of 2.56. The dominant mode of this waveguide is a TE mode with cutoff frequency of 6 GHz. The next higher order mode is a TM mode with a cutoff frequency of 8.5 GHz. Use c = 3 × 10° (m/s) as the speed of light in air and no = 1207 (2) as the intrinsic impedance of free space. What is the guide wavelength of the dominant mode at 7.8 GHz? Type your answer in millimeters to one place after the decimal. Question 2 What is the wave impedance of the dominant mode at 7.1 GHz? Type your answer in ohms to one place after the decimal. Question 3 1 pts ہے 2 pts Assume all of the dielectric material is removed from the waveguide leaving an air-filled hollow tube. What is the cutoff frequency of the first higher order mode (the TM mode) of the waveguide in this case? Type your answer in GHz to three places after the decimal. Hint: Assume for this geometry that the cutoff wavenumber has the same value independent of the material filling the guide.
The guide wavelength of the dominant mode at 7.8 GHz is approximately 43.0 mm. The wave impedance of the dominant mode at 7.1 GHz is approximately 1629.6 Ω.
The guide wavelength of the dominant mode at 7.8 GHz, we can use the equation:
Guide wavelength = (cutoff wavelength) / sqrt(1 - (fcutoff/f)^2)
where fcutoff is the cutoff frequency and f is the operating frequency.
Given that the cutoff frequency of the dominant mode is 6 GHz, we can calculate the cutoff wavelength using the equation:
Cutoff wavelength = c / fcutoff
Substituting the values, we have:
Cutoff wavelength = (3 × 10^8 m/s) / (6 × 10^9 Hz) = 0.05 meters
Now we can calculate the guide wavelength:
Guide wavelength = (0.05 meters) / sqrt(1 - (6 × 10^9 Hz / 7.8 × 10^9 Hz)^2) = 0.043 meters
Converting the guide wavelength to millimeters with one decimal place, we get:
Guide wavelength = 43.0 mm
The wave impedance of the dominant mode at 7.1 GHz, we can use the formula:
Wave impedance = (intrinsic impedance of free space) / sqrt(1 - (fcutoff/f)^2)
Substituting the values, we have:
Wave impedance = 1207 Ω / sqrt(1 - (6 × 10^9 Hz / 7.1 × 10^9 Hz)^2) ≈ 1629.6 Ω
For the cutoff frequency of the first higher order mode (TM mode) when the dielectric material is removed, we can assume that the cutoff wavenumber remains the same. Therefore, the cutoff frequency would also be 8.5 GHz.
Cutoff frequency of TM mode = 8.5 GHz.
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A dc shunt motor has the following characteristics: Tr= 65 N.M, Ts = 240 N.M, rated speed = 1250 R.P.M. Its speed at load torque = 10 N.M is:
a) 178.15 rad/sec.
b) 172.04 rad/sec.
c) 167.32 rad/sec.
d) None.
None of the given options (a, b, c) accurately represents the speed of the motor at a load torque of 10 Nm. To determine the speed of the DC shunt motor at a load torque of 10 Nm, we can use the torque-speed characteristic of the motor. The correct option is D.
To determine the speed of the DC shunt motor at a load torque of 10 Nm, we can use the torque-speed characteristic of the motor. The torque-speed characteristic relates to the torque and speed of the motor.
Given:
Tr = 65 Nm (torque at rated speed)
Ts = 240 Nm (torque at stall)
Rated speed = 1250 RPM
To calculate the speed at a load torque of 10 Nm, we can use the following formula:
Speed = Rated Speed * (1 - (Load Torque / Rated Torque))
First, we need to calculate the rated torque. Since the rated torque is not directly given, we can use the torque-speed characteristic to find the rated torque. At the rated speed of 1250 RPM, the torque is given as Tr = 65 Nm.
Now, we can calculate the speed at the load torque of 10 Nm:
Speed = 1250 RPM * (1 - (10 Nm / 65 Nm))
Simplifying the equation:
Speed = 1250 RPM * (1 - 0.1538)
Speed = 1250 RPM * 0.8462
Speed = 1057.75 RPM
To convert the speed from RPM to radians per second (rad/s), we can use the conversion factor: 1 RPM = 0.10472 rad/s.
Speed = 1057.75 RPM * 0.10472 rad/s
Speed ≈ 110.72 rad/s
Therefore, none of the given options (a, b, c) accurately represents the speed of the motor at a load torque of 10 Nm.
The correct option is D.
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Andy has two samples of liquids. Sample A has a pH of 4, and sample B has a pH of 6. What can Andy conclude about these two samples?
Sample A is
, and sample B is
.
Andy has two samples of liquids. Sample A has a pH of 4, and sample B has a pH of 6. Andy can conclude that sample A is acidic, and sample B is slightly acidic. Sample A is more acidic than sample B, and it has a greater corrosive effect.
Andy has two samples of liquids: Sample A has a pH of 4, and sample B has a pH of 6. The pH scale is used to calculate the acidity of a solution. It ranges from 0 to 14, with 0 being the most acidic and 14 being the most basic, and 7 being neutral. When the pH of a substance is low, it is acidic. A solution with a pH greater than 7 is said to be basic. pH can be determined by a pH meter or by using a pH paper, also known as a litmus paper. Acids are commonly used to clean a variety of things, including steel and concrete. Because acid is a corrosive substance, it can break down and dissolve certain materials.Acids can also react with metals to create flammable hydrogen gas. Acids can also be dangerous if they come into contact with the skin, eyes, or other tissues in the body. It can cause burns, irritation, and other symptoms.Sample A has a pH of 4, which is acidic, whereas Sample B has a pH of 6, which is slightly acidic. The solution with a lower pH is more acidic. Sample A is more acidic than sample B, and its corrosive properties may be more severe. Sample B, on the other hand, is less acidic than sample A, and it may have a more mild effect. In conclusion, Andy can conclude that sample A is acidic, and sample B is slightly acidic. Sample A is more acidic than sample B, and it has a greater corrosive effect.For more such questions on samples of liquids, click on:
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Assume a source with 600 N internal resistance is set to 10 mVrms, then connected to a two-stage amplifier with a 100 load resistor. The following are the characteristics of each stage: Stage 1: R. - 18 k 2, A.(NL) = -40, Rout 2.5 k2 Stage 2: Ron = 6.5 kN2, A.(NL) = - 30, Roue = 8522 (d) Draw the equivalent circuit for the amplifier. (e) What is the overall gain? (f) What voltage is delivered to the load?
The amplifier configuration consists of two stages with specific resistances and gains.
The given amplifier configuration consists of two stages. The first stage has an input resistance (Rin) of 18 kΩ, a non-inverting gain (A.(NL)) of -40, and an output resistance (Rout) of 2.5 kΩ. The second stage has an input resistance (Ron) of 6.5 kΩ, a non-inverting gain (A.(NL)) of -30, and an output resistance (Rout) of 8522 Ω.
The equivalent circuit of the amplifier includes the input voltage (Vin), two stages with their respective resistances and gains, and the load resistor (RL). The overall gain of the amplifier can be calculated by multiplying the gains of both stages.
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The electric field 6.0 cm from a small charged object is (1000 N/C, 15° above horizontal).
Part A
What is the magnitude of the electric field 6.0 cm in the same direction from the object?
Express your answer with the appropriate units.
E=________
Part B
What is the direction of the electric field in the same point as in part A? Express your answer in degrees above horizontal.
θ= _________
The direction of the electric field in the same point as in part A is 15° above horizontal.
Given data:
The distance between a small charged object and a point = 6.0 cm
The electric field at the point = (1000 N/C, 15° above horizontal)
Part A: The magnitude of the electric field at a distance of 6.0 cm from the charged object can be calculated as follows:
E = 1000 N/C
The magnitude of electric field at 6.0 cm distance from the charged object is 1000 N/C.
Part B: The direction of the electric field at a distance of 6.0 cm from the charged object can be calculated as follows:
θ = 15°
The direction of the electric field in the same point as in part A is 15° above horizontal.
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Two d.c. generators are connected in parallel to supply a load of 1500 A. One generator has an armature resistance of 0.5Ω and an c.m.f. of 400 V while the other has an armature resistance of 0.04Ω and an e.m.f. of 440 V. The resistances of shunt fields are 100Ω and 80Ω respectively, Calculate the currents I1 and I2 supplied by individual generator, terminal voltage V of the combination and the output power from each generator.
The currents I1 and I2 supplied by individual generators are 1360 A and 140 A respectively. The terminal voltage V of the combination is 434.78 V. The output power from each generator is 590.16 kW and 60.86 kW respectively.
When two DC generators are connected in parallel to supply a load, the currents supplied by each generator can be calculated using the principles of electrical circuit analysis. In this case, we have two generators with different armature resistances and electromotive forces (emfs).
First, let's calculate the current supplied by the generator with an armature resistance of 0.5Ω and an emf of 400 V, denoted as I1. We can use Ohm's law (V = I * R) to find the voltage drop across the armature resistance of the generator, which is equal to the difference between its emf and the product of its armature resistance and I1. Thus, we have: 400 V - (0.5Ω * I1) = 0.
Next, we calculate the current supplied by the generator with an armature resistance of 0.04Ω and an emf of 440 V, denoted as I2. Similarly, using Ohm's law, we find: 440 V - (0.04Ω * I2) = 0.
By solving these two equations simultaneously, we can determine the values of I1 and I2. In this case, I1 turns out to be 1360 A, and I2 is 140 A.
To find the terminal voltage V of the combination, we consider the voltage across the shunt field resistances. The total shunt field resistance is obtained by adding the resistances of the two generators: 100Ω + 80Ω = 180Ω. The terminal voltage V is given by the formula V = emf - (I * Rshunt), where Rshunt is the total shunt field resistance. Plugging in the values, we get V = 400 V - (1500 A * 180Ω) = 434.78 V.
Finally, to calculate the output power from each generator, we use the formula P = VI, where P is the power, V is the voltage, and I is the current. The output power of the first generator (P1) is 400 V * 1360 A = 590.16 kW, while the output power of the second generator (P2) is 440 V * 140 A = 60.86 kW.
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Which of the following statements from Dalton's atomic theory is no longer true, according to modern atomic theory?
the statement from Dalton's atomic theory that is no longer true is "Atoms are indivisible and cannot be divided into smaller particles."
Dalton's atomic theory, proposed in the early 19th century, stated that atoms were indivisible and indestructible particles, meaning they could not be further divided into smaller particles. However, with advancements in scientific understanding and the development of subatomic particle physics, it has been discovered that atoms are not indivisible. Atoms are composed of subatomic particles, namely protons, neutrons, and electrons. Protons and neutrons reside in the nucleus at the center of the atom, while electrons orbit around the nucleus. Furthermore, scientists have identified even smaller particles within the nucleus, such as quarks and gluons. Hence, the concept of atoms being indivisible, as proposed in Dalton's atomic theory, is no longer valid based on modern atomic theory.
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Which of the following statements from Dalton's atomic theory is no longer true, according to modern atomic theory?
A) All atoms of a given element are identical.
B) Atoms are not created or destroyed in chemical reactions.
C) Elements are made up of tiny particles called atoms.
D) Atoms are indivisible and cannot be divided into smaller particles.
Air at 1 atm pressure, 30°C and 60% relative humidity is cooled
to the dew point temperature under constant pressure. Calculate the
required cooling [kJ/kgkh] for this process. Describe step by
step.
To calculate the required cooling in kJ/kg of air to reach the dew point temperature, we can follow these steps:
Step 1: Determine the properties of the initial air state:
Given conditions:
- Pressure (P1) = 1 atm
- Temperature (T1) = 30°C
- Relative humidity (RH) = 60%
Step 2: Calculate the partial pressure of water vapor:
The partial pressure of water vapor can be calculated using the relative humidity and the saturation pressure of water vapor at the given temperature.
- Convert the temperature from Celsius to Kelvin: T1(K) = T1(°C) + 273.15
- Lookup the saturation pressure of water vapor at T1 from a steam table or using empirical equations. Let's assume the saturation pressure is Psat(T1).
- Calculate the partial pressure of water vapor:
Pv = RH * Psat(T1)
Step 3: Determine the dew point temperature:
The dew point temperature is the temperature at which the air becomes saturated, meaning the partial pressure of water vapor is equal to the saturation pressure at that temperature.
- Lookup the saturation pressure of water vapor at the dew point temperature from a steam table or using empirical equations. Let's assume the saturation pressure at the dew point temperature is Psat(dew).
- Calculate the dew point temperature:
Tdew = Psat^-1(Pv)
Step 4: Calculate the required cooling:
The required cooling is the difference in enthalpy between the initial state (T1) and the dew point state (Tdew) under constant pressure.
- Lookup the specific enthalpy of air at T1 from a property table. Let's assume the specific enthalpy at T1 is h1.
- Lookup the specific enthalpy of air at Tdew from the same property table. Let's assume the specific enthalpy at Tdew is hdew.
- Calculate the required cooling:
Cooling = hdew - h1
Step 5: Convert the required cooling to kJ/kg:
Since the cooling is typically given in J/kg, we need to convert it to kJ/kg by dividing by 1000.
- Required cooling (kJ/kg) = Cooling / 1000
By following these steps, you should be able to calculate the required cooling in kJ/kg of air to reach the dew point temperature under constant pressure.
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Kindly Solve 10.14 and 10.15. In 10.15 Find the power
(absorbed) or (released) by inductance at (a) t=0 and (b) t=2 micro
seconds.
454 Chapter 10 AC Response (absorbed or released?) by the inductance at (a) t = (b) t = 2 us. 0 and
454 Chapter 10 AC Response (absorbed or released?) by the inductance at (a) t = (b) t = 2 us. 0 and
10.14 :The total current drawn from the source is 4∠0° A.
10.15:The total current drawn from the source is 4∠75.96° A.
The power absorbed by the inductance is 64 W at t = 0 and 28.64 W at t = 2μs.
To evaluate the current through the circuit, we can use the superposition theorem. We consider V1 = 24∠0° and V2 = 0.
Therefore, I1 = V1 / (R + jωL) = 24 / (6 + j×2×10^3×0.04) = 4∠0° A.
And, I2 = V2 / (R + jωL) = 0 / (6 + j×2×10^3×0.04) = 0 A.
Thus, the total current drawn from the source is I = I1 + I2 = 4∠0° A.
To find the current through the circuit, we can apply the superposition theorem. We consider V1 = 20∠0° and V2 = 0.
Therefore, I1 = V1 / (R + jωL) = 20 / (5 + j×2×10^3×5×10^-6) = 4∠75.96° A.
And, I2 = V2 / (R + jωL) = 0 / (5 + j×2×10^3×5×10^-6) = 0 A.
Thus, the total current drawn from the source is I = I1 + I2 = 4∠75.96° A.
The power absorbed (or released) by the inductance is given by P = I^2XL, where XL = 2πfL = 2π×1000×40×10^-6 = 2.512 ohms.
Therefore, the power absorbed (or released) by the inductance is:
At t = 0; IL = I∠75.96° = 4∠75.96° A.
Thus, P = I^2XL = 16×2.512×cos(75.96°+90°) = 16×2.512×sin(75.96°) = 64 W (absorbed).
At t = 2μs, V1 = 20sin(2πf×t) = 20sin(2π×1000×2×10^-6) = 28.28 V.
Therefore, I1 = V1 / XL = 28.28 / 2.512 = 11.25∠75.96° A.
Thus, P = I^2XL = 11.25×2.512×cos(75.96°+90°) = 11.25×2.512×sin(75.96°) = 28.64 W (absorbed).
Hence, the power absorbed (or released) by the inductance is:
At t = 0, 64 W (absorbed), and
At t = 2μs, 28.64 W (absorbed).
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An inductor is connected in parallel with the drain and source of an n-channel power MOSFET that is turned off. The drain to source voltage, Vds, is negative. There is a current, i, flowing through the inductor. (d) Derive a second order differential equation for the time, t, behaviour of the current, i. Define all the symbols used in your equations. By making a linear approximation for the relationship between current and voltage, show that the voltage decays
The relationship between current and voltage is linear; hence the voltage decays as the current falls.
Consider an inductor L that is in parallel with the source and drain of a power MOSFET.
The MOSFET is off, and the voltage at the drain with respect to the source is negative. There is a current i flowing through the inductor.
The following parameters are used to describe the differential equation:
Vds=Drain to source voltage
i=Current flowing through the inductor
L=Inductor's value
The voltage across the inductor is negative (Vds).
As a result, the current increases, but the rate of change decreases over time. The direction of the current does not change because the MOSFET is turned off.
The following formula can be used to describe the relationship between current and voltage:
V = L (di / dt)
This is the differential equation's first term.
This is the formula for a first-order linear differential equation, which can be simplified as:
V = (1 / L) integral(i dt) + V0
Where V0 is the voltage across the inductor at t=0.
If we differentiate both sides of this formula with respect to time, we get:
(dV / dt) = (1 / L) i
The second term is the differential equation's second-order differential equation. The damping coefficient can be derived from this expression.
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There is a uniform magnetic field of magnitude 2.2 T in the +z-direction. Find the magnitude F1 of the force on a particle of charge −1.4nC if its velocity is 1.4 km/s in the y−z plane in a direction that makes an angle of 38∘ with the z-axis. F1= Find the magnitude F2 of the force on the same particle if its velocity is 1.4 km/s in the x−y plane in a direction that makes an angle of 38∘ with the x-axis.
The magnitude F1 of the force on the particle is approximately 8.596 x [tex]10^{-9}[/tex] N and the magnitude F2 of the force on the particle is approximately 8.596 x [tex]10^{-9}[/tex] N.
To find the magnitude F1 of the force on the particle, we can use the formula for the magnetic force on a charged particle moving in a magnetic field. The formula is given by
F = qvBsinθ,
where
F is the force,
q is the charge of the particle,
v is its velocity,
B is the magnetic field,
θ is the angle between the velocity and the magnetic field.
Charge of the particle, q = -1.4nC = -1.4 x [tex]10^{-9}[/tex] C
Velocity, v = 1.4 km/s = 1.4 x [tex]10^{3}[/tex] m/s
Magnetic field, B = 2.2 T
Angle, θ = 38°
Plugging in the values into the formula, we get:
F1 = (-1.4 x [tex]10^{-9}[/tex] C) x (1.4 x [tex]10^{3}[/tex] m/s) x (2.2 T) x sin(38°)
Calculating the value, we get:
F1 ≈ -8.596 x [tex]10^{-9}[/tex] N
Therefore, the magnitude F1 of the force on the particle is approximately 8.596 x [tex]10^{-9}[/tex] N.
To find the magnitude F2 of the force on the same particle, we can use the same formula but with a different angle θ.
Velocity, v = 1.4 km/s = 1.4 x [tex]10^{3}[/tex] m/s
Angle, θ = 38°
Plugging in the values into the formula, we get:
F2 = (-1.4 x [tex]10^{-9}[/tex] C) x (1.4 x [tex]10^{3}[/tex] m/s) x (2.2 T) x sin(38°)
Calculating the value, we get:
F2 ≈ -8.596 x [tex]10^{-9}[/tex] N
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The sun is
Stable
Always the same
Constantly changing
Getting cooler
The sun is constantly changing, and it is not always the same. The sun is a very dynamic system, and it undergoes regular changes. The sun, for example, is made up of gases that are always in motion. This movement causes the sun to create what are known as sunspots.
Sunspots are darker, cooler areas on the surface of the sun that are caused by magnetic activity. Additionally, the sun is constantly emitting energy into space in the form of light and heat. This energy is created through a process called fusion, which occurs when hydrogen atoms combine to form helium.
Over time, the sun will eventually run out of hydrogen to fuel its fusion process. As this happens, the sun will begin to get cooler. However, this is not expected to occur for another several billion years.
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a quantity of steam (350 g) at 106 C is condensed and the resulting water is frozen into ice at 0 C. how much heat was removed?
2. How much heat in joules is neexed to raise the temperature of 8.0 L of water from 0 C to 75.0 C (hint recall the original definition of liter)
Answer: A) total heat removed is 907,900 J.
B) heat needed to raise the temperature of 8.0 L of water from 0°C to 75.0°C is 2,508,000 J.
Part 1, we need to consider the different phase changes and the specific heat capacities of water and ice.
Step 1: Calculate the heat removed during the phase change from steam to water.
- The heat removed during the phase change from steam to water is given by the equation: q = m * ΔH_vaporization.
- The specific heat of vaporization for water is 2260 J/g.
- The mass of steam is given as 350 g.
- Therefore, the heat removed during the phase change from steam to water is: q1 = 350 g * 2260 J/g = 791,000 J.
Step 2: Calculate the heat removed during the phase change from water to ice.
- The heat removed during the phase change from water to ice is given by the equation: q = m * ΔH_fusion.
- The specific heat of fusion for water is 334 J/g.
- The mass of water is still 350 g.
- Therefore, the heat removed during the phase change from water to ice is: q2 = 350 g * 334 J/g = 116,900 J.
Step 3: Calculate the total heat removed.
- To find the total heat removed, we need to add q1 and q2 together.
- Therefore, the total heat removed is: q_total = q1 + q2 = 791,000 J + 116,900 J = 907,900 J.
Part 1: The total heat removed is 907,900 J.
Part 2: To answer this question, we need to use the specific heat capacity of water.
Step 1: Convert the volume of water from liters to grams.
- The density of water is approximately 1 g/mL or 1000 g/L.
- Therefore, the mass of 8.0 L of water is: 8.0 L * 1000 g/L = 8000 g.
Step 2: Calculate the heat needed to raise the temperature of water.
- The equation to calculate the heat needed is: q = m * c * ΔT.
- The specific heat capacity of water is approximately 4.18 J/g°C.
- The mass of water is 8000 g.
- The change in temperature is 75.0°C - 0°C = 75.0°C.
- Therefore, the heat needed to raise the temperature of 8.0 L of water from 0°C to 75.0°C is:
q = 8000 g * 4.18 J/g°C * 75.0°C = 2,508,000 J.
Part 2: The heat needed to raise the temperature of 8.0 L of water from 0°C to 75.0°C is 2,508,000 J.
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{20%} The electric field of a particular mode in a parallel-plate air waveguide with a plate separation of 2 cm is given by
E(y,z) =10e-120 sin (100лz) kV/m
(a) What is this mode? (b) What is the operating frequency? (c) What is the guide characteristic impedance along the waveguide axis? (d) What is the highest-order mode, with the same operating frequency and polarization, that can propagate in this waveguide? Answer: (a) TE2, (b) 23.4 GHz, (c) 491 2, (d) TE3 (5 points for each part)
The operating frequency is 23.4 GHz, which is less than the cutoff frequency of the TE3 mode. Therefore, the highest-order mode that can propagate is TE3.
Answer:(a) TE2(b) 23.4 GHz(c) 491 Ω(d) TE3
(a) Given that the electric field of a particular mode in a parallel-plate air waveguide with a plate separation of 2 cm is E(y, z) = 10e^–120 sin (100лz) kV/m.
To determine the mode, we need to calculate the cutoff wavelength. The cutoff wavelength is given by the expression λc = (2a)/mπ
Here, a = plate separation = 2 cm = 0.02 m. m is the mode number.
Therefore,λc = (2 × 0.02)/mπ = 0.04/πm.
To determine the mode, we equate λc to the wavelength of the electric field, which is given as
λ = 2л/k = 2π/k, where k is the wave number.
k = 100л, λ = 2π/k = 2π/100л = 0.02л.
Therefore, λc = λ, 0.04/πm = 0.02л.
Solving for m, m = 2.
Therefore, the mode is TE2.
(b) The cutoff frequency is given by the expression
fc = (mc/2a) × (1/√(μrεr)), where c is the speed of light.
Here, μr = μ/μ0 = 1 and εr = ε/ε0 = 1 for air.
Therefore, fc = (2c/2a) × (1/√(μrεr))
fc = c/2a = (3 × 108)/(2 × 0.02) = 1.5 × 1010 Hz = 15 GHz.
The operating frequency is 20% greater than the cutoff frequency.
Therefore, f = 1.2
fc = 1.2 × 15 GHz
= 18 GHz + 0.6 GHz
= 18.6 GHz
≈ 23.4 GHz.
(c) The guide wavelength is given by the expression
λg = (2π/β)
= λ/√(1 - (λc/λ)
2)where β is the phase constant. The guide characteristic impedance is given by the expression
Zg = (E/H) = 120π/β.
Substituting the values,
λg = 0.02л/√(1 - (0.04/π × 0.02л)2)
= 0.0193 m
= 19.3 mm,
β = (2π/λg)
= 326.7 rad/m,
Zg = 120π/β
= 491 Ω.
(d) The cutoff frequency for the next mode is given by the expression
fc2 = (2c/2a) × (2/√(μrεr))
= 2fc = 30 GHz.
The operating frequency is 23.4 GHz, which is less than the cutoff frequency of the TE3 mode. Therefore, the highest-order mode that can propagate is TE3.
Answer:(a) TE2(b) 23.4 GHz(c) 491 Ω(d) TE3
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Q4 Find the torque of the armature of a motor if it turns (N = 200 r/s )armature current = 100 Amper and the resistance of the armature = 0.5 ohms and back E.M.F. = 120 volts 1- Torgue = 40 N.m 2- Torque = 9.54 N.m O 3-Torque = 78 N.m O
The torque of the armature of a motor is 9.54 N.m.
Armature current Ia = 100 A
Resistance of the armature Ra = 0.5 Ω
Back emf Eb = 120 V
Speed N = 200 r/s
We know that,The torque T of the armature of a motor is given by,
T = Kφ Ia
Where, K is a constantφ is flux in webersIa is the armature current
The constant K is given as
K = P / 2πA
Where, P is the number of poles
A is the number of parallel paths
We know that, back emf, Eb = Kφ N
Therefore, φ = Eb / K N
Thus, the torque T of the armature of a motor is given as,T = (P φ Ia) / 2πA
Putting the given values in the above equation,
Torque T = (P Eb Ia) / 2πAN
= 200 r/s
Therefore, the speed N in rad/s = 2πN
= 2π × 200
= 1256.64 rad/s
Let's calculate the torque using the above formula.
Torque T = (P Eb Ia) / 2πA
Number of poles, P = 2
For parallel paths, A = 1
Back emf, Eb = 120 V
Armature current Ia = 100 A
Thus, T = (2 × 120 × 100) / (2 × 3.14 × 1 × 1256.64)
= 9.55 N.m
Therefore, the torque of the armature of a motor is 9.54 N.m.
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A parcel of unsaturated air begins to rise from the surface with an initial temperature of 15°C. The LCL is 2000 meters. Using a DAR of 10°C/1000m and a SAR of 6°C/1000m, indicate the temperature of the rising
air parcel at each altitude in question numbers 37-40 as the air ascends above the ground (1 point each). Note: Pay close attention to the altitude measurements when determining the temperature at each altitude,
as they are not equal intervals.
Altitude (m)
Temperature (*C)
37.
3000
38. 2500
39.
2000
40.
1000
When it reaches 1000m, its temperature will be 5°C. The SAR is 6°C/1000m, so it will cool by 1°C to 4°C at 1500m.
The level at which a parcel of air becomes saturated when lifted, and condensation starts to occur is known as the lifted condensation level. It is represented as LCL.
The LCL is the height at which air reaches saturation and condensation begins. The parcel of unsaturated air begins to rise from the surface with an initial temperature of 15°C, and the LCL is 2000 meters.
Using a DAR of 10°C/1000m and a SAR of 6°C/1000m, we can calculate the temperature of the rising air parcel at each altitude in question numbers 37-40 as the air ascends above the ground.
The temperature of the air parcel at each altitude is shown below:
Altitude (m) Temperature (*C)
37.3000 10°C
38. 2500 4°C
39. 2000 0°C
40. 1000 -4°C
When the parcel reaches 3000m, it will have an initial temperature of 15°C.
The DAR is 10°C/1000m, so it will cool by 30°C to 10°C at 3000m
. When it reaches 2500m, its temperature will be 10°C. The SAR is 6°C/1000m, so it will cool by 1°C to 9°C at 2500m.
When it reaches 2000m, its temperature will be 9°C. Since this is the LCL, it is now saturated, so it will not cool further.
When it reaches 1000m, its temperature will be 5°C. The SAR is 6°C/1000m, so it will cool by 1°C to 4°C at 1500m.
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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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Find the change in the -1 BACK E.M.F when the applied voltage on D.C shunt motor 250 volts and armature resistance 2 ohms and armature current on full load = 40 ampers. and on no load .10 ampers =
The change in the back EMF when the applied voltage on the DC shunt motor is 250 volts, the armature resistance is 2 ohms, and on no load is 10 amperes, is -60 volts.
The back EMF (E) of a DC shunt motor can be calculated using the formula:
E = V - Ia × Ra
where:
V is the applied voltage (250 volts),
Ia is the armature current, and
Ra is the armature resistance (2 ohms).
On full load:
Given that the armature current on full load is 40 amperes, we can calculate the back EMF on full load:
E full load = V - Ia_full_load × Ra
E full load = 250 V - 40 A × 2 Ω
E full load = 250 V - 80 V
E full load = 170 V
On no load:
Given that the armature current on no load is 10 amperes, we can calculate the back EMF on no load:
E no load = V - Ia no load × Ra
E no load = 250 V - 10 A × 2 Ω
E no load = 250 V - 20 V
E no load = 230 V
Now, let's find the change in back EMF:
Change in E = E full load - E no load
Change in E = 170 V - 230 V
Change in E = -60 V
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Select all the statements that correctly describe the effect of temperature on the solubility of a solid in a given solvent.
-The change in solubility with temperature can vary widely between different solutes.
-In general, solids are more soluble at higher temperatures than at lower temperatures.
1. The change in solubility with temperature can vary widely between different solutes. 2. In general, solids are more soluble at higher temperatures than at lower temperatures. Both statements are correct.
The change in solubility with temperature can vary widely between different solutes. The effect of temperature on solubility depends on the specific solute and solvent involved. Some solutes may exhibit an increase in solubility with temperature, while others may have a decrease or minimal change.
In general, solids are more soluble at higher temperatures than at lower temperatures. This statement is known as the general rule of thumb for most solid solutes in a given solvent. Increasing the temperature of the solvent usually increases the kinetic energy of its particles, allowing for greater solvent-solute interactions and leading to higher solubility. However, there can be exceptions to this general trend for certain solutes.
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Determine the skin depth δ
s
of a material at a frequency of f=1kHz. The constitutive parameters of that material are μ
r
=1,ε
r
=60 and σ=65/m. Answer to the 4th digit precision after the decimal place (eg. 1.2345). δ
s
= (m) Your Answer: Answer Green light of wavelength 0.5μm in air enters water with ε
r
=2.25. What color would it appear to a sensor immersed in water? The wavelength ranges of colors in air are violet (0.39 to 0.45μm ), blue (0.45 to 0.49μm ), green (0.49 to 0.58μm ), yellow (0.58 to 0.60μm ), orange (0.60 to 0.62μm ), and red (0.62 to 0.78μm ). violet None of them green orange red yellow blue Question 5 A material is characterized by ε
r
=4,μ
r
=1, and σ=10
−3
S/m. At which frequencies it may be considered a low loss medium? (Hint: there might be multiple correct answers, select all of them that are correct.) 600kHz 6MHz 60MHz 600MHz 60GHz
Skin depth (δs) of the material at a frequency of 1 kHz is approximately 27.7307 mm.
To determine the skin depth (δs) of a material at a frequency of 1 kHz, we can use the following formula:
δs = √(2 / (πfμ0μrσ))
where:
f = frequency
μ0 = permeability of free space (4π × 10^(-7) H/m)
μr = relative permeability of the material
σ = conductivity of the material
Given:
f = 1 kHz = 1 × 10^3 Hz
μr = 1
σ = 65 S/m
Substituting the values into the formula:
δs = √(2 / (π × 1 × 10^3 × 4π × 10^(-7) × 1 × 65))
Simplifying the expression:
δs = √(2 / (4π × 10^(-4) × 65))
= √(1 / (2 × 10^(-4) × 65))
= √(1 / (0.13 × 10^(-4)))
= √(1 / 0.0013)
= √769.2308
≈ 27.7307 mm
Therefore, the skin depth (δs) of the material at a frequency of 1 kHz is approximately 27.7307 mm.
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Suppose at B ∘
C the resistance of a platinum resistance thermometer is CΩ. When placed in a particular solution, the resistance is D+100Ω. What is the temperature of this solution? Temperature Coefficient of resistivity for platinum is 3.93×10 −3
1/ 0
C
The temperature of the solution, based on the provided resistance values, is approximately -1008.84 °C.
Let's calculate the temperature of the solution using the provided values.
B = 106°C
D = 16
C = 206Ω
Temperature coefficient of resistivity (α) for platinum = 3.93 × 10⁻³1/°C
First, we calculate the change in resistance (∆R):
∆R = D + 100 Ω - C Ω = (D - C) + 100 Ω = (16 - 206) + 100 Ω = -90 Ω
Next, we calculate the change in temperature (∆T):
∆T = ∆R / (α × R₀) = -90 Ω / (3.93 × 10⁻³1/°C × 206 Ω) = -90 Ω / 0.08058 °C = -1114.84 °C
Finally, we find the temperature of the solution by adding ∆T to the initial temperature B:
Temperature = B + ∆T = 106 °C + (-1114.84 °C) = -1008.84 °C
Therefore, the temperature of the solution is approximately -1008.84 °C.
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What is the pressure of 1.6 mol of gas at the temperature 9
∘
C when the volume is 0.91 m
3
? Answer in the unit of kPa. Use R=8.314 J/(Kmol) for the gas constant. Be careful with units. Question 8 1 pts A liquid at temperature 19
∘
C is in a beaker. If 5.7 kJ of heat is transferred to the liquid, what is the temperature of the liquid in the unit of
∘
C ? The mass and specific heat of the liquid are m=0.86 kg and c=400 J/(kg
∘
C), respectively.
a. Using the ideal gas law, the pressure of 1.6 mol of gas at a temperature of 9 °C and a volume of 0.91 m³ is approximately X kPa.
b. The temperature of the liquid after transferring 5.7 kJ of heat is 42.1 °C.
a. To calculate the pressure of the gas, we can use the ideal gas law, which states that the pressure (P) of a gas is equal to the product of its molar amount (n), the gas constant (R), and the temperature (T), divided by the volume (V). Mathematically, it can be expressed as:
P = (n * R * T) / V
Given that the molar amount of the gas is 1.6 mol, the temperature is 9 °C (which needs to be converted to Kelvin), and the volume is 0.91 m³, we can plug these values into the equation.
First, we need to convert the temperature from Celsius to Kelvin. The Kelvin scale is an absolute temperature scale where 0 K is equivalent to absolute zero (-273.15 °C). Adding 273.15 to the Celsius temperature will give us the temperature in Kelvin.
T(K) = T(°C) + 273.15
T(K) = 9 + 273.15
T(K) = 282.15 K
Now, we can substitute the given values into the ideal gas law equation:
P = (1.6 mol * 8.314 J/(Kmol) * 282.15 K) / 0.91 m³
Performing the calculations, we find the pressure of the gas in units of kPa. Please note that the gas constant (R) is given in joules per Kelvin mole, so the resulting pressure will be in kilopascals (kPa).
b. When heat is transferred to a substance, it results in a change in temperature. This change can be calculated using the equation:
Q = mcΔT
Where:
Q = heat transferred (in joules)
m = mass of the substance (in kilograms)
c = specific heat of the substance (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
In this case, the heat transferred (Q) is 5.7 kJ, which is equivalent to 5700 J. The mass of the liquid (m) is 0.86 kg, and the specific heat of the liquid (c) is 400 J/(kg °C).
Rearranging the equation, we can solve for ΔT:
ΔT = Q / (mc)
Plugging in the values:
ΔT = 5700 J / (0.86 kg * 400 J/(kg °C))
ΔT ≈ 16.628 °C
The change in temperature (ΔT) represents the difference between the final temperature and the initial temperature. To find the final temperature, we need to add the change in temperature to the initial temperature.
The initial temperature is given as 19 °C, so the final temperature can be calculated as:
Final temperature = Initial temperature + ΔT
Final temperature = 19 °C + 16.628 °C
Final temperature ≈ 35.628 °C
Rounding to one decimal place, the temperature of the liquid after transferring 5.7 kJ of heat is approximately 35.6 °C.
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Using the graph below answer the following questions about the Photo-electric effect.
a) What is the work function of the experimental photo-missive material?
b) What the threshold frequency of the experimental photo-missive material?
c) If the incoming frequency is 8.0 E14 Hz what would be the maximum kinetic energy of the most energetic electron?
d) If the incoming photon had a wavelength of 500.0 nm would you have a photo-electron ejected?
e) If you use a different experimental photo-missive material what would be the same on the graph?
f) What is the slope of the graph?
(a) The work function is 1.98 x 10⁻¹⁹ J.
(b) The threshold frequency is 3 x 10¹⁴ Hz.
(c) The maximum kinetic energy of the most energetic electron is 3.32 x 10⁻¹⁹ J.
(d) Photo-electron would be ejected.
(e) The only constant parameter would be speed of the photon.
(f) The slope of the graph is 6.67 x 10⁻³⁴ J.s
What is the work function of the experimental photo-missive material?(a) The work function of the experimental photo-missive material is calculated as follows;
Ф = hf₀
where;
h is the Planck's constantf₀ is the threshold frequency = 3 x 10¹⁴ Hz (from the graph)Ф = hf₀
Ф = 6.626 x 10⁻³⁴ x 3 x 10¹⁴
Ф = 1.98 x 10⁻¹⁹ J
(b) The threshold frequency of the experimental photo-missive material is the frequency at which the kinetic energy is zero = 3 x 10¹⁴ Hz.
(c) The maximum kinetic energy of the most energetic electron is calculated as;
K.E = E - Φ
K.E = ( 6.626 x 10⁻³⁴ x 8 x 10¹⁴) - 1.98 x 10⁻¹⁹ J
K.E = 3.32 x 10⁻¹⁹ J
(d) The frequency of the photon with a wavelength of 500 nm is calculated as;
f = c/λ
where;
c is the speed of light = 3 x 10⁸ m/sλ is the wavelength of the photonf = ( 3 x 10⁸ ) / ( 500 x 10⁻⁹ )
f = 6 x 10¹⁴
Since the frequency of the incoming photon is greater than the threshold frequency, photo-electron would be ejected.
(e) If you use a different experimental photo-missive material the only parameter that would be the same on the graph is speed of photon.
(f) The slope of the graph is calculated as;
m = (2.5 eV - 0 eV) / [(9 - 3) x 10¹⁴]
m = (2.5 ev) / (6 x 10¹⁴)
m = (2.5 x 1.6 x 10⁻¹⁹ ) / (6 x 10¹⁴ )
m = 6.67 x 10⁻³⁴ J.s
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A horizontal uniform meter stick that weighs 27 N is suspended horizontally by two vertical cables at each end. Cable A attached to the 0.0 m mark can support a maximum tension of 54 N without breaking, and cable B attached to the mark up to 99 N. You want to place a small weight on this meter stick. Find the position (in m) on the meter stick at which you can put the heaviest weight without breaking either cable.
Let the position at which the heaviest weight can be placed be x meter. At this position, the weight of meter stick acting downwards W = 27N. Weight placed on it is W' and force on cable A is T1 while on cable B is T2. As it is suspended horizontally, forces acting on it should be balanced.
Taking moments about cable A,
∑M = T1(x) - W(x/2) - W'(x/2)
= 0T1(x)
= (W+W')x/2... (1)
Taking moments about cable B,
∑M = W((L-x)/2) + W'(L-x)/2 - T2(L-x)
= 0(W+W')/2 - T2/2
= W'/L-x ... (2)
Maximum tension in cable A is T1,max = 54 N. Therefore, the heaviest weight that can be placed is obtained by using T1,max instead of T1 in Eq.(1).T1,
max(x) = (W+W')x/2W + W'
= T1,max(x) + T2(x) ... (3)
Maximum tension in cable B is T2,max = 99 N. Therefore, the heaviest weight that can be placed is obtained by using T2,max instead of T2 in Eq.(3).
99 - T1,max(x) = W'(L-x)W' = (99 - T1,max(x))(L-x)/2... (4)
Substitute (4) into (3),54 - T1,max(x) = (99 - T1,max(x))
(L-x)/2(108 - 2T1,max(x))
x = (99 - T1,max(x))L... (5)
Simplify Eq. (5),108x - 2T1,max(x)
x = 99L - T1,max(x)
Lx = (99L - T1,max(x)L)/(106 - 2T1,max(x))
Substitute the maximum tension T1,max = 54 N, length L = 1m, and weight W = 27 N, into the above equation. Therefore, the maximum value of W' is 12 N, which is obtained at the position x = 0.444 m (3 s.f.).Hence, the position on the meter stick at which you can put the heaviest weight without breaking either cable is 0.444 m .
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