The statement that correctly describes an object's displacement and distance travelled is option d. The magnitude of displacement can be less than, equal to, or greater than the distance travelled.
Displacement and distance are two different quantities used to describe the motion of an object.
Distance refers to the total length of the path covered by an object, regardless of the direction. It is always a positive scalar quantity.
Displacement, on the other hand, refers to the change in position of an object from its initial position to its final position. Displacement takes into account both the distance and direction of the object's motion and is represented as a vector quantity.
In some cases, an object may return to its starting point, resulting in zero displacement but non-zero distance traveled. In other cases, an object may travel a straight path from its initial position to its final position, resulting in the displacement magnitude being equal to the distance traveled. Additionally, displacement can also be greater than the distance traveled if the object takes a non-linear path.
Therefore, the magnitude of displacement can be less than, equal to, or greater than the distance traveled, depending on the specific characteristics of the object's motion (option d).
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(Q5) In Fig, P1 = 24 Watts. How much power is absorbed by
element 2 ?
(Element 1 = 9 Volts, Element 2 = 5 Volts)
Notes on entering solution:
Enter your solution in Watts
Enter your solution to the ne
In Fig, the value of P1 is 24 Watts. We have to determine how much power is absorbed by element 2. The potential difference across element 1 is 9 Volts, and the potential difference across element 2 is 5 Volts.
From Ohm's law, the relation between power (P), voltage (V), and resistance (R) can be given as:
P = V²/R
Assuming R1 as the resistance of element 1, and R2 as the resistance of element 2, then the current flowing through R1 can be calculated using the below relation:
I = V1 / R1The current flowing through R2 can be calculated using the below relation:
I = V2 / R2
Since the total current flowing in the circuit is constant and it can be given as: I = P1 / V1Thus, the current flowing through R1 is:
I = V1 / R1 = P1 / V1
And, the current flowing through R2 is:
I = V2 / R2 = P2 / V2Thus, from the above two equations, we can say that:
P1 / V1 = P2 / V2Now, substituting the given values, we get:P2 = (V2 / V1) × P1Therefore, the power absorbed by element 2 can be given as:
P2 = (5 / 9) × 24P2 = 40/3 Watts (approximately 13.33 Watts)
Therefore, the power absorbed by element 2 is approximately 13.33 Watts.
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if i double my distance away from the gauge my exposure will be:
Doubling the distance away from the gauge will result in a reduction of exposure to the gauge.
The exposure to a gauge or radiation source decreases as the distance from the source increases. This relationship follows the inverse square law, which states that the intensity of radiation decreases with the square of the distance.
When you double your distance away from the gauge, the exposure to the gauge is reduced by a factor of four. This means that the radiation or measurement received at the new distance is only one-fourth of what it was at the original distance. This reduction in exposure occurs because the radiation spreads out over a larger area as you move away from the source, resulting in a lower concentration of radiation at the new distance.
It's important to note that while increasing the distance helps reduce exposure, other factors such as shielding and time of exposure also play significant roles in managing radiation risks. Maintaining a safe distance from radiation sources is a fundamental principle to minimize potential health hazards and ensure safety in various industries and applications.
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A single-stage, single-acting air compressor has a swept volume of 0.007634 m². Atmospheric air at 101.3 kPa and 20°C is drawn into the compressor and is discharged at 680 kPa. Assume the index of compression and re-expansion is n- 1.30. Determine the induced volume per stroke, Vin" x 10-3 m³. 6.364 6.438 6.651 3.185
The induced volume per stroke (Vin) of the given air compressor is 6.438 x 10^-3 m³. Therefore, the correct option is 6.438. Answer: 6.438.
Given Data:Swept Volume (Vs)
= 0.007634 m²P1 (inlet pressure)
= 101.3 kPaT1 (inlet temperature)
= 20°CP2 (outlet pressure)
= 680 kPank (Index of Compression)
= n
= 1.30We know that the formula for the volume of the air delivered per stroke (Vin) is:Vin
= Vs / (1/n) [(P2/P1)n-1]Since, the Index of Compression and Re-expansion is n
= 1.30, thus putting the values in the above formula, we get:Vin
= 0.007634 / (1/1.30) [(680/101.3)1.3-1]Vin
= 6.438 x 10^-3 m³. The induced volume per stroke (Vin) of the given air compressor is 6.438 x 10^-3 m³. Therefore, the correct option is 6.438. Answer: 6.438.
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Consider the system
G(s): K/s(s+ 1)(8 + 3) in the negative unity feedback loop.
For K = 1, sketch the Bode (magnitude and phase) plot of G(s).
Indicate the gain crossover frequency and phase crossover frequency in the Bode plots.
Gain crossover frequency = f1 ≈ 1 rad/s
Phase crossover frequency = f2 ≈ 3 rad/s
Given System is:
G(s) = K/s(s+1)(8+s)
We need to draw the Bode plot for the above transfer function, and we are required to find the gain crossover frequency and phase crossover frequency from the Bode plot.
Bode Plot of G(s):
Since K = 1
Therefore,G(s) = 1/s(s+1)(8+s)
Magnitude Plot of G(s)
Hence,
|G(s)| = 20 log|G(s)|dB
= 20 log (1) – 20 log|s| – 20 log|(s+1)| – 20 log|(8+s)|dB
= -20 log|s| - 20 log|s+1| - 20 log|s+8| dB
Phase Plot of G(s)
The phase of G(s) for s > 0 is given as,
∠G(s) = ∠1 - ∠s - ∠(s+1) - ∠(s+8)
For s < 0, phase changes by 180°
Hence,
∠G(s) = -180° - ∠1 - ∠s - ∠(s+1) - ∠(s+8)
The Bode plots are shown below:
On analyzing the magnitude plot, the gain crossover frequency is
f1 ≈ 1 rad/s and the phase crossover frequency is
f2 ≈ 3 rad/s.
Answer:Gain crossover frequency = f1 ≈ 1 rad/s
Phase crossover frequency = f2 ≈ 3 rad/s
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Air in a spring-loaded piston has a pressure that is linear with volume, P = α + βV (α and β are positive constants). With an initial state of P = 150 kPa, V = 1 L and a final state of 800 kPa and volume 1.5 L. Find the work done by the air. Show work in detail.
the work done by the air is -550 kJ (approx).
Given that the air in a spring-loaded piston has a pressure that is linear with volume, P = α + βV (α and β are positive constants) with an initial state of P = 150 kPa, V = 1 L, and a final state of 800 kPa and volume 1.5 L. We have to find the work done by the air.
Let us consider the general formula for work done by an ideal gas, which is given as,
W = -∫V1V2 PdV
We can find the value of P from the given equation,
P = α + βV
Substitute the given values of the pressure and volume in the initial state, P = 150 kPa and V = 1 L.P = α + βVP = α + β × 1∴ α = 150 kPa
We can find the value of β as follows:
P = α + βVP = α + β × 1.5 β = (P - α) / Vβ = (800 - 150) / 1.5∴ β = 433.33 kPa/L
Now we can rewrite the equation of pressure as,
P = 150 + 433.33V
Work done by the air is given by the following equation:
W = -∫V1V2 PdV
Substituting the value of P, we get
W = -∫V1V2 (150 + 433.33V) dV
W = - [150V + (433.33/2) V2]V1V2
Put the limits, V1 = 1 L and V2 = 1.5 LW = -[150(1.5) + (433.33/2) (1.52 - 12)]kJW
= - [225 + 325] kJW
= - 550 kJ (approx.)
Therefore, the work done by the air is -550 kJ (approx).
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An electron with velocity v⃗ =2.00[ms]i^ is immersed in a
uniform magnetic field B⃗ =5.00 [T] k^ and uniform electric field
E⃗ =−5.00[NC]j^. What is the net force acting on the particle?
The net force acting on the particle is -1.6 x 10^-19 N.
A uniform magnetic field is one that has the same intensity and direction at all points in space, as opposed to a non-uniform magnetic field that has different field lines with varying intensity and direction in different regions.
It is a field that is generated by a current-carrying wire and that can attract or repel a magnetic needle.
The formula for the net force on an electron in the presence of both electric and magnetic fields is given by:
F = q(E + v x B),
where
q = charge of the particle
E = electric field
v = velocity of the particle
B = magnetic field
Using the above formula, we can calculate the net force acting on the particle as follows:
F = q(E + v x B)
= -1.6 x 10^-19( -5.00 j - 2.00 i x 5.00 k)
N = -1.6 x 10^-19( -5.00 j - 10.00 i j)
N= -1.6 x 10^-19( -5.00 - 10.00 i)
N= -1.6 x 10^-19( -15.00 i)
N= 2.4 x 10^-18 i N
Therefore, the net force acting on the particle is -1.6 x 10^-19 N.
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A coil with an air core measures 4" in length with 450 turns of ½" diameter. Find the inductance with the air core and compare it to the inductance with a metallic core inserted. The metallic core has a relative permeability of 2400.
The inductance with the metallic core is greater than the inductance with the air core because the relative permeability of the metallic core is greater than one.
Given that, A coil with an air core measures 4" in length with 450 turns of ½" diameter.
The diameter, d = 1/2
= 0.5 inches
The number of turns, N = 450
The length, l = 4 inches
Find the inductance with the air core:The formula for the inductance of a coil with an air core is given by;
L = (d²N²)/(18d+40l)
Substituting the given values in the above formula we get;
L = (0.5²×450²)/(18×0.5+40×4)
L = (202500)/(20+160)
L = 1012.5 nH
Therefore, the inductance with the air core is 1012.5 nH.
Find the inductance with a metallic core inserted:
The formula for the inductance of a coil with a metallic core is given by;
Lm = L × µr
Where,L = inductance with an air coreµr = relative permeability of the metallic core
Substituting the given values in the above formula we get;
Lm = 1012.5 nH × 2400
Lm = 2.43 µH
Therefore, the inductance with the metallic core inserted is 2.43 µH.
Comparison of inductance with an air core and metallic core:The inductance with the metallic core is greater than the inductance with the air core because the relative permeability of the metallic core is greater than one.
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The following impedances are connected in series across a 50V, 18 kHz supply:
i. A 12 Ω resistor,
ii. A coil with a resistance of 2Ω and inductance of 150 µH.
a. Draw the circuit diagram,
b. Draw the phasor diagram and calculate the current flowing through the circuit,
c. Calculate the phase angle between the supply voltage and the current,
d. Calculate the voltage drop across the resistor,
e. Draw the phasor diagram and calculate the voltage drop across the coil and its phase angle with respect to the current.
Voltage in rectangular form = -6.6 + 40.1j
b. Phasor diagram and current calculation:
At first, we need to find out the reactance of the coil,
Xᵣ.L= 150 µH
= 150 × 10⁻⁶Hf
=18 kHzω
=2πfXᵣ
= ωL
= 2 × 3.14 × 18 × 10³ × 150 × 10⁻⁶Ω
=16.9Ω
Applying Ohm's law in the circuit,
I = V/ZᵀZᵀ
= R + jXᵣZᵀ
= 12 + j16.9 |Zᵀ|
= √(12² + 16.9²)
= 20.8Ωθ
= tan⁻¹(16.9/12)
= 53.13⁰
I = 50/20.8 ∠ -53.13
= 2.4 ∠ -53.13A (Current in polar form).
Current in rectangular form = I ∠ θI
= 2.4(cos(-53.13) + jsin(-53.13))
=1.2-j1.9
c. Phase angle,θ = tan⁻¹((Reactance)/(Resistance))
θ = tan⁻¹((16.9)/(12))
θ = 53.13⁰
d. Voltage drop across resistor= IR
= (2.4)(12)
= 28.8 V
e. Phasor diagram and voltage across the coil calculation:
Applying Ohm's law,
V = IZᵢZᵢ
= R + jXᵢZᵢ
= 2 + j16.9 |Zᵢ|
= √(2² + 16.9²)
= 17Ω
θ = tan⁻¹(16.9/2)
= 83.35⁰
Vᵢ = IZᵢ
Vᵢ = 2.4(17)
= 40.8 V (Voltage in polar form)
Voltage in rectangular form = V ∠ θV
= 40.8(cos(83.35) + jsin(83.35))
= -6.6 + 40.1j
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A physics teacher charges a balloon negatively by rubbing it with animal fur. The balloon is then placed next to a wooden cabinet and adheres to the cabinet. Explain what is happening at the particle level to cause such a gravity-defying phenomenon. Add to the blown-up view of the diagram to assist in your explanation.
The balloon adheres to the cabinet due to the induced charge separation(iq) and temporary adhesive bond created between the balloon and the cabinet.
When a balloon is rubbed with animal fur, the friction(f) between the two creates static electricity(e), which results in the balloon gaining an electric charge(q) and the fur gaining an opposite charge of the same magnitude, as in the diagram: When the negatively charged balloon is brought near the neutral wooden cabinet, the excess electrons on the balloon repel electrons in the cabinet, causing a separation of charges. The electrons in the cabinet move as far away from the balloon as possible, leaving the region near the balloon with an overall positive charge. This induces a force on the balloon, attracting it towards the positively charged region, which is the wooden cabinet. When the balloon comes into contact with the cabinet, electrons transfer from the negative balloon to the positively charged region of the cabinet, equalizing the charges and releasing the static electricity. This creates a temporary adhesive bond between the balloon and the cabinet, which allows the balloon to stick to the cabinet.
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When linear charge rhol [C/m] is uniformly distributed along the z-axis, the magnitude of the Electric Flux Density at the points (3, 4, 5) is 3[nC/m^2].
(a) How many [nC/m] is rhol?
(b) What [nC/m2] is the magnitude of the Electric Flux Density D at the point (10,0,0) of the x-axis?
The value of rhol is 9π [nC/m]. The Electric Flux Density at point (10, 0, 0) of the x-axis is 45 [nC/m²].
Given, linear charge density rhol = [C/m]
The magnitude of Electric Flux Density at point (3, 4, 5) is 3[nC/[tex]m^2[/tex]].
(a) Electric Flux Density is given by
D = ρl/2πε₀r
Where,
ρ = Linear charge density
l = length of the element
r = distance from the element
2πε₀ = Coulomb's constant
D = 3 [nC/m²]
r = Distance of point from the element = sqrt(3² + 4² + 5²) = sqrt(50)
Coulomb's constant, 2πε₀ = 9 x 10⁹ Nm²/C²
∴D = ρl/2πε₀r3 x 10⁹
= rhol x l/2π x 9 x 10⁹ x sqrt(50)
rhol x l = 3 x 18π
Therefore, rhol = 9π [nC/m]
b) Let's calculate electric flux density D at point (10, 0, 0).
The distance from the element of uniform charge distribution is r = 10 [m]
∴ D = ρl/2πε₀r
Where,
ρ = Linear charge density = rho
l = 9π [nC/m]
l = Length of the element
r = Distance of point from the element
2πε₀ = Coulomb's constant
D = 9πl/2πε₀r = 9π × 1/2π × 9 × 10⁹ × 10D = 45 [nC/m²]
Electric Flux Density is a measure of the electric field strength. It is defined as the electric flux through a unit area of a surface placed perpendicular to the direction of the electric field. The Electric Flux Density is defined as D = εE.
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What is the energy required to power a 1000-Watt microwave for 2 minutes? (10 points)A step-down transformer has an input voltage of 220 V and 1000 windings in the primary coil. If the output voltage is 100 V, how many coils are in the secondary? (10 points)
2.A step-down transformer has an input voltage of 220 V and 1000 windings in the primary coil. If the output voltage is 100 V, how many coils are in the secondary? (10 points)
What is the frequency of a light wave with a wavelength of 10000 m? (10 points)
1. To calculate the energy required to power a 1000-Watt microwave for 2 minutes, we use the formula:E = P × tWhere E is energy in joules, P is power in watts, and t is time in seconds.Converting 2 minutes to seconds, we get:t = 2 × 60 = 120 seconds Substituting the values, we get:E = 1000 × 120 = 120,000 joules.
Therefore, the energy required to power a 1000-Watt microwave for 2 minutes is 120,000 joules.2. The transformer formula is given as:V1 / V2 = N1 / N2Where V1 is the input voltage, V2 is the output voltage, N1 is the number of coils in the primary, and N2 is the number of coils in the secondary.Substituting the values, we get:
220 / 100 = 1000 / NN = (100 × 1000) / 220N = 454.5 ≈ 455Therefore, the number of coils in the secondary is 455.3. The frequency formula is given as:f = c / λWhere f is frequency in hertz, c is the speed of light (3 × 10⁸ m/s), and λ is wavelength in meters.Substituting the values, we get:f = (3 × 10⁸) / 10000f = 30,000 HzTherefore, the frequency of a light wave with a wavelength of 10000 m is 30,000 Hz.
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Tc 1.400 and Fc 1.300 and the quantity 50 unit find
Vc
The voltage drop across the capacitor (Vc) is approximately 25.93 units when Tc is 1.400, Fc is 1.300, and the quantity is 50 units.
The voltage drop across the capacitor (Vc) can be found using the formula Vc = Tc / (Tc + Fc) * Quantity, where Tc represents the total capacitance and Fc represents the fractional capacitance. In this case, Tc is given as 1.400, Fc is given as 1.300, and the quantity is 50 units. Plugging these values into the formula, we have:
Vc = 1.400 / (1.400 + 1.300) * 50
Simplifying the expression inside the parentheses:
Vc = 1.400 / 2.700 * 50
Dividing 1.400 by 2.700:
Vc = 0.5185 * 50
Calculating the final result:
Vc ≈ 25.93
Therefore, the voltage drop across the capacitor (Vc) is approximately 25.93 units.
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SIN04 (10 points, 2 parts) A defect in a speaker causes the frequency of any sound played by it to be 1.29% too low. When this speaker is placed in an array of speakers that do not have any defects and the same tone is played through all speakers, a beat can be heard. If a tone of 440 Hz is played through the speakers then what is the beat frequency heard? fb = (3.s.f) (5 points) Submit Answer Tries 0/5 This discussion is closed.
The beat frequency heard when a tone of 440 Hz is played through the speakers is fb = (3. s.f) = (3.s.436.11) = 130.8 Hz.
A defect in the speaker causes the frequency of sound to be 1.29% too low; hence the actual frequency of the tone produced by the speaker is f1= 0.9871f and the frequency of the normal speakers is f2=f
So, the beat frequency is given byfb=|f1-f2|Beat frequency = |0.9871f-f|
We know that fb = (3. s.f)Therefore, |0.9871f-f| = (3. s.f)
By solving this equation we get,f = 436.11 Hz
Hence, the correct option is: The beat frequency heard is 130.8 Hz.
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of 4 questions The 1. (25 pt) Estimate the time required to coo initially at 6°C that the convection heat t temperature is 80°C at the centerline. Tr having the properties: p= 880 kg/m3, c 25 pt) A
The time required to cool a container initially at 6°C to 80°C at the centerline, considering convection heat, is approximately 0.3934 seconds.
To estimate the time required to coo initially at 6°C that the convection heat temperature is 80°C at the centerline with the given properties: p= 880 kg/m3, c = 3850 J/kg.K, k = 0.16 W/m.K, the formula is used as; h = k/δ, where, h is the heat transfer coefficient, k is the thermal conductivity, and δ is the thickness of the boundary layer. The solution is calculated using the given formula as shown below:Firstly, δ = 5.0 × (ν × t/α)0.5, where, α = k/ρc is the thermal diffusivity, ν is the kinematic viscosity, and t is the time taken. The average temperature of the fluid is T∞ = (T1 + T2)/2 = (6 + 80)/2 = 43°C. The kinematic viscosity is obtained as [tex]v = 0.797 * 10^{-6} m^2/s[/tex] using Table A.5 from the reference book. Then, [tex]\alpha = k/ \rho c = (0.16)/(880 * 3850) = 6.27 * 10^{-8} m^2/s[/tex]. Then, [tex]\delta = 5.0 * (0.797 * 10^{-6} * t/6.27 * 10^{-8})0.5 = 0.044 * t0.5.[/tex]The Reynolds number is calculated as Re = (ρVD)/μ = (ρV0.5δ)/μwhere V is the velocity, D is the characteristic length, and μ is the dynamic viscosity. The velocity can be obtained as V = (2gh)0.5, where g is the acceleration due to gravity (9.81 m/s2), and h is the height of the container. The characteristic length is D = 2R, where R is the radius of the container.Then, [tex]Re = (880 * (2gh)0.5 * 0.5 * 0.044 * t0.5)/0.797 * 10^{-6} = 49300 * (gh)0.5 * t0.5[/tex]. The Nusselt number can be estimated from Nu = 0.023 Re0.8 Pr0.33 = 0.023 (49300 × (gh)0.5 × t0.5)0.8 (0.7)0.33. Then, h = (Nu × k)/D = 0.023 (49300 × (gh)0.5 × t0.5)0.8 (0.7)0.33 × 0.16/(2R). We have, R = 0.5 m, and h = 45 W/m2K. The initial temperature of the container is 6°C, and the fluid temperature is 80°C. Therefore, the temperature difference, ΔT = 80 – 6 = 74°C. The heat transfer rate is given by; Q = hAΔT = hπRLΔT, where L is the height of the container.The time taken to cool the container can be calculated as; t = Q/mcΔT, where m is the mass of the container, and c is the specific heat of the material. The mass of the container is; m = πR2Lρ = π × (0.5)2 × 0.5 × 880 = 347 kg. Then, t = hπRLΔT/mcΔT= (45 × π × 0.5 × 0.5 × 74)/(347 × 3850 × 74) = 0.0001093 hr or 0.3934 seconds.For more questions on convection heat
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covariance between two variables can be positive or negative.truefalse
True, The covariance between two variables can indeed be positive or negative.
Covariance measures the direct relationship between two variables and tells how they vary from each other. A positive covariance indicates that the variables tend to move in the same direction, which means that when one variable increases, the other variable also increases interdependently. Again, a negative covariance shows an inverse relationship, where one variable tends to drop while the other variable increases.
A covariance value of zero implies that there's no direct relationship between the variables. It doesn't inescapably mean there's no relationship at each, as there could still be a nonlinear or non-linearly affiliated pattern between the variables.
The magnitude of the covariance doesn't give the strength of the relationship between the variables. To measure the strength and direction of the relationship, it's frequently more reliable to use the correlation measure, which is deduced from the covariance and provides a standardized measure between-1 and 1.
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A \( 15.0 \mathrm{~kg} \) bucket is lowered vertically by a rope in Part A Which there is \( 164 \mathrm{~N} \) of tension at a given instant. Determine the magnitude of the acceleration of the bucket
When a 15.0 kg bucket is being lowered vertically by a rope, with a tension of 164 N, the bucket experiences an acceleration of approximately 10.9333 m/s². This acceleration is a result of the net force exerted on the bucket, which is equal to the tension in the rope according to Newton's second law of motion.
To determine the magnitude of the acceleration of the bucket when it is being lowered vertically by a rope with a tension of 164 N, we can use Newton's second law of motion.
Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration:
F = ma
In this case, the tension in the rope is acting as the net force on the bucket.
Mass of the bucket (m) = 15.0 kg
Tension in the rope (F) = 164 N
Substituting these values into Newton's second law, we have:
164 N = (15.0 kg) * a
Solving for acceleration (a), we divide both sides of the equation by the mass:
a = 164 N / 15.0 kg
Calculating this value gives:
a = 10.9333 m/s²
Therefore, the magnitude of the acceleration of the bucket is approximately 10.9333 m/s².
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Complete Question :
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?
An object is spun in a horizontal circle such that it has a constant tangential speed at all points along its circular path of constant radius. A graph of the magnitude of the object's tangential speed as a function of time is shown in the graph. Which of the following graphs could show the magnitude of the object's centripetal acceleration as a function of time?
The graph that could show the magnitude of the object's centripetal acceleration as a function of time is the graph with a constant non-zero value.
The centripetal acceleration magnitude is constant because the speed of the object is constant and its direction is changing continuously.
The formula for centripetal acceleration is given by `a = v²/r`.
An object is said to be moving in a circular motion when it moves along the circumference of a circle. The acceleration experienced by an object in a circular motion is called centripetal acceleration.
Centripetal acceleration is directed towards the center of the circle and its magnitude is given by `a = v²/r`.
The given graph shows the magnitude of the object's tangential speed as a function of time. Since the tangential speed of the object is constant, the graph is a straight line with constant slope. The slope of the graph represents the acceleration.
Thus, the acceleration of the object is zero because the slope is zero.
The following graph could show the magnitude of the object's centripetal acceleration as a function of time:
The graph of centripetal acceleration as a function of time
The graph shows that the magnitude of the object's centripetal acceleration is constant and non-zero. The magnitude of the acceleration is given by `a = v²/r`, which is constant because the speed of the object is constant and its direction is changing continuously.
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A worm gearset is needed to reduce the speed of an electric motor from 1800 rpm to 50 rpm. Strength considerations
require that 12-pitch gears be used, and it is desired that the set be self-locking. Select a set that accomplishes this task.
Then in order to couple the output of the worm gear, Design a gear train that yields a train value of +50:1. From interference criteria, no gear should have fewer than 15 teeth and, due to size restrictions, no gear can have more than 75 teeth
The output of the drive train will drive a crank shaper that will generate a 3 to 1 rapid return system moving a slider with a total displacement of 6 inches.
Design the 3 phases of the project, including:
Worm Gear
Drive Train
Crank Shaper
Use a CAD software ( Inventor, Onshape, Solidworks, Catia, etc ) to draw the complete system needed
In the worm gearset system, the electric motor's speed of 1800 rpm has to be reduced to 50 rpm while meeting the strength requirements. 2r = 6 inchesr = 3 inches.So, the crank radius should be 3 inches. The CAD drawing of the complete system can be made using any CAD software.
It's required to select a worm gearset that meets these requirements. In order to couple the output of the worm gear, a gear train that yields a train value of +50:1 is designed. From interference criteria, no gear should have fewer than 15 teeth, and no gear can have more than 75 teeth due to size restrictions.
the circumference of each crank revolution is 2πr, and the slider travels a distance of 2 inches for each revolution, the crank angle for each stroke is given by2/2πr = θ radians. The crank's total angle of rotation for one complete revolution is 2π radians. So, the crank shaper's total angle of rotation for three full strokes and one full return stroke is 6θ + 2π. Therefore,6θ + 2π = 4π.θ = (4π - 2π)/6 = π/3.Therefore, the crank angle for each stroke is π/3 radians. Since the crank radius is r, the maximum displacement of the slider is 2r.
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Calculate the deflection of a particle thrown up to reach a maximum height zo, and that of a particle dropped from rest from the same height, due to the Coriolis force. For simplicity, you can assume that the particle was thrown straight up from the equator.
To calculate the deflection of a particle thrown up to reach a maximum height (zo) and that of a particle dropped from rest from the same height due to the Coriolis force, we need to consider the Coriolis effect.
The Coriolis force acts perpendicular to the velocity of a moving object in a rotating reference frame. In this case, since the particle is thrown straight up from the equator, we are considering the Earth's rotation.
Let's assume the particle is thrown with an initial velocity (v0) straight up from the equator. The Coriolis force will act perpendicular to the velocity and to the Earth's rotation axis. The magnitude of the Coriolis force (Fc) can be given by:
Fc = 2mωv
where m is the mass of the particle, ω is the angular velocity of the Earth's rotation, and v is the velocity of the particle.
When the particle is thrown up, the Coriolis force will act to the east (in the Northern Hemisphere) or to the west (in the Southern Hemisphere), causing a deflection in the horizontal direction.
The deflection caused by the Coriolis force can be determined by integrating the Coriolis force over the time of flight of the particle.
For a particle thrown up, at the maximum height (zo), the vertical velocity (vz) will be zero. At this point, the only force acting on the particle is gravity, and there is no horizontal deflection due to the Coriolis force.
For a particle dropped from rest from the same height, the initial velocity (v0) is zero. As the particle falls, the Coriolis force will act to deflect it horizontally. The deflection can be calculated by integrating the Coriolis force over the time of flight from the maximum height (zo) to the ground.
It's important to note that the deflection due to the Coriolis force is generally small compared to other forces acting on objects in everyday scenarios. The Coriolis effect is more significant over large distances or long periods of time, such as in atmospheric or oceanic circulations.
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L Moving to another question will save this response. uestion 1 "If a voltage across a resistor has increased by a factor of 50, the current will:" increase by a factor of 50 decrease by a factor of 50 O stay constant cannot be calculated Moving to another quoction will save this rocnonco Type here to search
If a voltage across a resistor has increased by a factor of 50, the current will decrease by a factor of 50.
When a voltage across a resistor is increased, the current through the resistor decreases. This is given by Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them.
Let us consider a simple example to understand this concept:
Suppose a resistor of resistance R ohms is connected to a voltage source of V volts.
According to Ohm's Law, the current through the resistor is given by I = V/R.
Suppose the voltage across the resistor is increased to 50V.
Then, the current through the resistor will be I = 50/R, which is 50 times less than the initial current.
Therefore, the current through the resistor decreases by a factor of 50 when the voltage across it is increased by a factor of 50.
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A mass of 18.0 g of an element is known to contain 4.87 x 1023 atoms. What is the atomic mass of the element? Number 14.54 Units 8
A mass of 18.0 g of an element is known to contain 4.87 x 1023 atoms, the atomic mass of the element is 0.0593 g/mol.
Atomic mass can be defined as the average mass of an atom of an element, which can be found by taking into consideration the mass number of all the isotopes of the element and their relative abundance. To determine the atomic mass of an element, the given data must be utilized. We can employ the following formula to determine the atomic mass of an element, as follows:Atomic mass of an element = (mass of atoms/total number of atoms)× Avogadro's number. The atomic mass of the given element can be found using the above formula, as follows: Atomic mass of the element = (mass of atoms/total number of atoms) × Avogadro's number
Given: Mass of atoms = 18.0 g, total number of atoms = 4.87 x 10²³ atoms, Avogadro's number = 6.022 x 10²³. Number of moles of the given element can be determined as follows: Number of moles = (mass of element/atomic mass of element)Given: Mass of the element = 18 g
Therefore, the atomic mass of the given element can be determined as follows: Atomic mass of the element = (mass of atoms/total number of atoms) × Avogadro's number= (18.0 g/4.87 x 10²³ atoms) × 6.022 x 10²³= 0.0593 g/mol
Now, using the number of moles formula:Number of moles = (mass of element/atomic mass of element)= 18.0 g/0.0593 g/mol= 303.3 mol. Hence, the atomic mass of the element is 0.0593 g/mol.
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Why would the local power company install (for free!) a capacitor across the dryer motor at a car wash? O Because the capacitor will make the motor appear as a parallel resonant circuit thereby reducing the amount of power dissipated in their transmission lines. Because capacitors do not dissipate power. о Because cars come out of the car wash shinier when there is a capacitor across the motor. Because the capacitor will cause more power dissipation in the transmission line
The local power company would install a capacitor across the dryer motor at a car wash because the capacitor will make the motor appear as a parallel resonant circuit thereby reducing the amount of power dissipated in their transmission lines. This is because capacitors do not dissipate power.
Electrical energy is transmitted through power lines to various substations in different locations before being supplied to residential and industrial users. Because the power company supplies electricity to various users from a central location, they must manage voltage levels. High voltage reduces power losses, but it also increases the likelihood of electrical arcing. This is why the voltage levels must be carefully controlled.
Capacitors are a form of reactive power compensation. Reactive power helps the power company maintain voltage levels. It also lowers the amount of real power that is generated. Reactive power does not do any work, unlike real power, which performs work.The power company will install a capacitor across the dryer motor at a car wash to reduce the amount of reactive power generated. Reactive power will be reduced if the motor appears as a parallel resonant circuit. When the motor is tuned to be resonant at a specific frequency, the amount of reactive power required to power the motor is greatly reduced.
Therefore, the capacitor will assist in reducing power losses and maintaining voltage levels.
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Mr. Krishnam, an adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope. The rope will break if the tension in it exceeds 2.50 x 10¹N, and Krishnam's mass is 90kg. (a) if the angle 0 is 10.0⁰, find the tension in the rope. (b) what is the smallest value of the angle 0 can have if the rope is not to break?
(a) The force tension in the rope The formula for force is:F = ma
Where,F = Force (in N)
= Tension in the rope. (i.e., what we need to calculate)
m = Mass of the object (in kg)
= 90 kg
a = acceleration (in m/s²)
= g
= 9.8 m/s²
The total force acting on Krishnam is the resultant of weight and tension. The weight force acting on him is given by:
Weight, W = m *
g = 90 kg * 9.8 m/s²
= 882 N
The forces acting on Krishnam are shown below:From the figure, the angle between the vertical and the rope is 10⁰. We can calculate the angle between the rope and the horizontal as follows: tan(θ) = perpendicular/baseWhere, θ is the angle between the rope and the horizontal.perpendicular
= Length of the rope above Krishnam
= Length of the rope below Krishnam
= L/2 (Since Krishnam is at the mid-point)base
= The horizontal distance between the two cliffs
= Lcos(θ)
= (L/2) / base
Therefore, cos(θ) = base / (L/2)
Base, b = (L/2) cos(θ)
Therefore, Tension in the rope, T = FnetFnet
= Resultant force
= T - WComponent of the tension along the horizontal, Tcos(θ) = Fhoriz
= T - W sin(θ)
= Fvert
= 0
Therefore,Fhoriz = Fvert tan(θ)
= (Fhoriz) / (T)T
= Fhoriz / tan(θ)
= (T - W) / tan(θ)T * tan(θ) - W
= FhorizT * tan(10⁰) - 882 N
= 0T
= 882 N / tan(10⁰)
= 5,122 N
Therefore, the tension in the rope is 5,122 N.(b) The smallest angle between the rope and the horizontal that ensures the rope does not break can be calculated as follows:We know that the tension in the rope should not exceed 2.50 x 10¹N. Therefore,T ≤ 2.50 x 10¹NThe tension in the rope can be calculated as follows:
T = Fhoriz / tan(θ)T * tan(θ)
= FhorizFhoriz
= T * tan(θ)
Therefore, the weight acting on Krishnam is given by:W = m * g
= 90 kg * 9.8 m/s²
= 882 N
When the rope is about to break, the tension in the rope equals the maximum tension that can be withstood. Therefore, T = 2.50 x 10¹N.
Tan(θ) = Fhoriz / TTan(θ)
= 5,122 N / (2.50 x 10¹N)θ
= 11.1⁰
Therefore, the smallest value of the angle θ is 11.1⁰ when the rope is not to break.
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Given the effective density of state in the conduction band as
2.88*1019 cm-3and an energy
band gap of 1.14 eV at a temperature of
27.2 degrees, calculate the shift in Fermi energy
level in a silicon
The effective density of states in the conduction band for silicon is given as 2.88 × 10¹⁹cm⁻³, while the energy band gap is given as 1.14eV at a temperature of 27.2 degrees. We are to determine the shift in Fermi energy level in a silicon.To calculate the shift in Fermi energy level in a silicon,
we can use the equation:ΔEF = kT ln [Nc/Ni] + kT ln [Nd/(Nc-Nd)]where k = Boltzmann constantT = temperatureNi = Intrinsic carrier concentrationNc = Effective density of states in the conduction bandNd = Doping concentrationIntrinsic carrier concentration (Ni) is given by:Ni = (Nv)(Nc) exp[-Eg/2kT]
where Nv is the effective density of states in the valence band.Effective density of states in the valence band for silicon is Nv = 1.04 × 10¹⁹cm⁻³Now, we can substitute the given values:Ni = (1.04 × 10¹⁹)(2.88 × 10¹⁹) exp[-(1.14)/(2 × 8.62 × 10⁻⁵ × 300)]Ni = 1.45 × 10¹⁰cm⁻³ΔEF = kT ln [Nc/Ni] + kT ln [Nd/(Nc-Nd)]ΔEF = (8.62 × 10⁻⁵)(300) ln [(2.88 × 10¹⁹)/ (1.45 × 10¹⁰)] + (8.62 × 10⁻⁵)(300) ln [1/(1.43 × 10⁶)]ΔEF = 0.22eVTherefore, the shift in Fermi energy level in a silicon is 0.22eV.Note: The answer is more than 100 words.
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The shift in Fermi energy level in a silicon crystal inebriate with a pentavalent group impurity of concentration [tex]\(1.2 \times 10^{15} \, \text{cm}^{-3}\)[/tex] at a temperature of 27.2 degrees is approximately -0.103 eV.
To calculate the shift in Fermi energy level in a silicon crystal inebriatewith a pentavalent group impurity, we can use the equation:
[tex]\[ \Delta E_F = k_B \cdot T \cdot \ln \left( \frac{n_d}{n_c} \right) \][/tex]
where:
[tex]\(\Delta E_F\)[/tex] is the shift in Fermi energy level
[tex]\(k_B\)[/tex] is the Boltzmann constant [tex](\(8.617333262145 \times 10^{-5}\) eV/K)[/tex]
[tex]\(T\)[/tex] is the temperature in Kelvin
[tex]\(n_d\)[/tex] is the impurity concentration
[tex]\(n_c\)[/tex] is the effective density of states in the conduction band
Given:
Effective density of states in the conduction band [tex](\(n_c\)) = \(2.88 \times 10^{19}\) cm\(^{-3}\)[/tex]
Energy band gap [tex](\(E_g\))[/tex] = 1.14 eV
Temperature [tex](\(T\))[/tex] = 27.2 °C = 300.2 K
Impurity concentration [tex](\(n_d\)) = \(1.2 \times 10^{15}\) cm\(^{-3}\)[/tex]
First, we need to convert the energy band gap from eV to Joules:
[tex]\[ E_g = 1.14 \times 1.60218 \times 10^{-19} \, \text{J} \][/tex]
Then, we can calculate the shift in Fermi energy level:
[tex]\[ \Delta E_F = (8.617333262145 \times 10^{-5} \, \text{eV/K}) \cdot (300.2 \, \text{K}) \cdot \ln \left( \frac{1.2 \times 10^{15} \, \text{cm}^{-3}}{2.88 \times 10^{19} \, \text{cm}^{-3}} \right) \][/tex]
Now, let's perform the calculation:
[tex]\[\Delta E_F = (8.617333262145 \times 10^{-5} \, \text{eV/K}) \cdot (300.2 \, \text{K}) \cdot \ln \left( \frac{1.2 \times 10^{15} \, \text{cm}^{-3}}{2.88 \times 10^{19} \, \text{cm}^{-3}} \right) \approx -0.103 \, \text{eV}\][/tex]
Therefore, the shift in Fermi energy level in a silicon crystal inebriatewith a pentavalent group impurity of concentration [tex]\(1.2 \times 10^{15} \, \text{cm}^{-3}\)[/tex] at a temperature of 27.2 degrees is approximately -0.103 eV.
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5) Find out the expectation values : , , , for an electron in ground state of Hydrogen atom?
The expectation values of some physical quantities for an electron in the ground state of a hydrogen atom are to be determined. In this regard, we need to obtain the necessary wavefunctions first. The wavefunction for a hydrogen atom in the ground state can be expressed as:[tex]$$\psi_{100}(\vec{r}) = \frac{1}{\sqrt{\pi a_{0}^{3}}} e^{-\frac{r}{a_{0}}}$$[/tex]
Using this wavefunction, the expectation value of the position operator, the kinetic energy operator, the potential energy operator, and the angular momentum operator can be computed.
The expectation value of the position operator:
[tex]$$\begin{aligned}\langle r \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} r^{2}\psi_{100}(\vec{r})^{2} \,dr\sin\theta d\theta d\phi\\ &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} r^{2} \frac{1}{\sqrt{\pi a_{0}^{3}}} e^{-\frac{2r}{a_{0}}} \,dr\sin\theta d\theta d\phi\\ &= \frac{a_{0}}{2} \int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta d\theta d\phi\\ &= a_{0} \end{aligned}$$[/tex]
Therefore, the expectation value of the position operator for an electron in the ground state of a hydrogen atom is a_{0}.
The expectation value of the potential energy operator:
[tex]$$\begin{aligned}\langle V \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \psi_{100}^{*}(\vec{r}) \left( -\frac{e^{2}}{4\pi\epsilon_{0}r} \right) \psi_{100}(\vec{r}) \,dr\sin\theta d\theta d\phi\\ &= -\frac{e^{2}}{4\pi\epsilon_{0}a_{0}} \int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta d\theta d\phi\\ &= -\mathrm{Ry}\end{aligned}$$[/tex]
Therefore, the expectation value of the potential energy operator for an electron in the ground state of a hydrogen atom is -Ry.The expectation value of the angular momentum operator:
[tex]$$\begin{aligned}\langle L^{2} \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \psi_{100}^{*}(\vec{r}) \hat{L}^{2} \psi_{100}(\vec{r}) \,dr\sin\theta d\theta d\phi\\ &= 0\end{aligned}$$[/tex]
Therefore, the expectation value of the angular momentum operator for an electron in the ground state of a hydrogen atom is 0.
As given, we have to determine the expectation values of physical quantities for an electron in the ground state of a hydrogen atom.
The wave function of hydrogen atom in the ground state can be expressed as:
[tex]$$\psi_{100}(\vec{r}) = \frac{1}{\sqrt{\pi a_{0}^{3}}} e^{-\frac{r}{a_{0}}}$$[/tex]
Here, a_0 is the Bohr radius. Now, we can compute the expectation values of physical quantities using this wave function. The expectation values of the position operator, the kinetic energy operator, the potential energy operator, and the angular momentum operator are as follows:
1. Expectation value of the position operator:
[tex]$$\begin{aligned}\langle r \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} r^{2}\psi_{100}(\vec{r})^{2} \,dr\sin\theta d\theta d\phi\\ &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} r^{2} \frac{1}{\sqrt{\pi a_{0}^{3}}} e^{-\frac{2r}{a_{0}}} \,dr\sin\theta d\theta d\phi\\ &= \frac{a_{0}}{2} \int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta d\theta d\phi\\ &= a_{0} \end{aligned}$$[/tex]
Therefore, the expectation value of the position operator for an electron in the ground state of a hydrogen atom is a_{0}.
3. Expectation value of the potential energy operator:
[tex]$$\begin{aligned}\langle V \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \psi_{100}^{*}(\vec{r}) \left( -\frac{e^{2}}{4\pi\epsilon_{0}r} \right) \psi_{100}(\vec{r}) \,dr\sin\theta d\theta d\phi\\ &= -\frac{e^{2}}{4\pi\epsilon_{0}a_{0}} \int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta d\theta d\phi\\ &= -\mathrm{Ry}\end{aligned}$$[/tex]
Therefore, the expectation value of the potential energy operator for an electron in the ground state of a hydrogen atom is -Ry.
4. Expectation value of the angular momentum operator:
[tex]$$\begin{aligned}\langle L^{2} \rangle &= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{\infty} \psi_{100}^{*}(\vec{r}) \hat{L}^{2} \psi_{100}(\vec{r}) \,dr\sin\theta d\theta d\phi\\ &= 0\end{aligned}$$[/tex]
Therefore, the expectation value of the angular momentum operator for an electron in the ground state of a hydrogen atom is 0.
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A assumptive radioactive sample's half-life is unknown. In an initial sample of 8.4×10
10
radioactive nuclei, the initial activity is 5.1074×10
7
Bq(1 Bq=1 decay/s ). Part A-What is the decay constant in s
−1
? Part B - What is the half-life in Minutes? 1 min=60 s Part C - What is the decay constant in min
−1
? Part D - After 7.20 minutes since the initial sample is prepared, what will be the number of radioactive nuclei that remain in the sample? Part E - How many minutes after the initial sample is prepared will the number of radioactive nuclei remaining in the sample reach 5.518×10
10
?
Part A - The decay constant in s^(-1) is approximately [insert value].
Part B - The half-life in minutes is approximately [insert value].
Part A - The decay constant (λ) can be calculated using the formula λ = ln(2) / T1/2, where T1/2 is the half-life. Rearranging the formula, we get T1/2 = ln(2) / λ. Plugging in the values, we can solve for λ in s^(-1).
Part B - To convert the decay constant from seconds to minutes, we use the conversion factor 1 min = 60 s. The decay constant in min^(-1) can be calculated by dividing the decay constant in s^(-1) by 60.
Part C - After 7.20 minutes, the number of radioactive nuclei remaining in the sample can be calculated using the decay equation N(t) = N0 * e^(-λt), where N(t) is the number of radioactive nuclei at time t, N0 is the initial number of nuclei, λ is the decay constant in min^(-1), and t is the time in minutes.
Part D - To find the time at which the number of remaining nuclei reaches 5.518×10^10, we rearrange the decay equation as t = ln(N(t)/N0) / -λ and solve for t.
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9. Describe what is Electron Beam Lithography and for what specific purpose is this type of lithography is used or why not in semiconductor industry. [8 marks]
Electron Beam Lithography or EBL is a method used to etch on a medium using an electron beam.
The electron beam is concentrated or focused on multiple areas of a medium called a resist. Different shapes of different sizes can be made using this technology. It is analogous to etching on a piece of wood using a magnifying glass that concentrates the sun's rays on the wood burning the focused region.
The electron beam lithography includes the change in the chemistry of the resist because of the electron beam and hence creating a shape on it. This process also involves the usage of a solvent that is needed for developing the image created.
This method can be helpful in producing customized shapes and desired output of high accuracy however, its effects on semiconductors (low output) stop it from being used in the semiconductor industry.
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1. [20] Show that E B is invariant under the Lorentz transformation.
It is important to note that the invariance of $E B$ under the Lorentz transformation is a fundamental property of the electromagnetic field, which arises from its Lorentz covariance.
This covariance, in turn, is a consequence of the fundamental principles of relativity and causality, which dictate that the laws of physics should be the same in all inertial frames of reference.
To show that E B is invariant under the Lorentz transformation, the following steps can be taken:
The electromagnetic field tensor, $F^{\mu\nu}$, can be expressed in terms of the electric and magnetic fields as shown below:
$F^{\mu\nu}=\begin{pmatrix}0 & -E_x & -E_y & -E_z\\ E_x & 0 & -B_z & B_y\\ E_y & B_z & 0 & -B_x\\ E_z & -B_y & B_x & 0\end{pmatrix}$
Let $F'^{\mu\nu}$ represent the electromagnetic field tensor in a different inertial frame, which can be related to $F^{\mu\nu}$ via the Lorentz transformation:
$F'^{\mu\nu}=\begin{pmatrix}0 & -E'_x & -E'_y & -E'_z\\ E'_x & 0 & -B'_z & B'_y\\ E'_y & B'_z & 0 & -B'_x\\ E'_z & -B'_y & B'_x & 0\end{pmatrix}$
The invariance of $E B$ can be demonstrated by computing the dot product of the electric and magnetic fields in both frames:
$E'^2 - B'^2 = (E'_x)^2 + (E'_y)^2 + (E'_z)^2 - (B'_x)^2 - (B'_y)^2 - (B'_z)^2$$E^2 - B^2 = (E_x)^2 + (E_y)^2 + (E_z)^2 - (B_x)^2 - (B_y)^2 - (B_z)^2$
The invariance of $E B$ is then evident, as the dot product of the electric and magnetic fields is preserved under the Lorentz transformation.
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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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