The current through the 11.2 µF capacitor is given by I(t) = 1680e^(15t) µA.
The value of the other capacitor in the series, when one capacitor is 140 mF and the total equivalent value is 118 mF, is approximately 3.34 mF.
The current through a capacitor can be calculated using the formula I(t) = C * dV(t)/dt, where I(t) is the current, C is the capacitance, and dV(t)/dt is the derivative of the voltage with respect to time.
For the given capacitor with a capacitance of 11.2 µF and voltage v(t) = 10e^(15t):
Taking the derivative of v(t), we have dV(t)/dt = 150e^(15t).
Substituting the values into the formula, we get:
I(t) = (11.2 µF) * (150e^(15t)) = 1680e^(15t) µA.
For the capacitors in series, their equivalent capacitance (C_eq) is given as 118 mF. Let's assume one capacitor has a value of C1 and the other capacitor has a value of C2.
Since the capacitors are in series, the reciprocal of their equivalent capacitance is equal to the sum of the reciprocals of their individual capacitances:
1/C_eq = 1/C1 + 1/C2.
Given that C1 = 140 mF, we can substitute these values into the equation:
1/0.118 = 1/0.140 + 1/C2.
Simplifying the equation, we can solve for C2:
C2 = 1 / (1/0.118 - 1/0.140) ≈ 3.34 mF.
Therefore, the value of the other capacitor is approximately 3.34 mF.
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a) An internally compensated op-amp has an open-loop voltage gain of 80 dB. The corner frequency occurs at 6 Hz. i. Find the unity gain frequency (0 dB frequency). ii. If the same op-amp is now connected in a closed-loop to form an inverting amplifier with a closed-loop gain of G = -9 V/V. Find the corner frequency for this closed-loop amplifier.
a) i. The unity gain frequency (0 dB frequency) can be found by determining the frequency at which the open-loop voltage gain of the internally compensated op-amp drops to 0 dB (1 or unity gain).
ii. The corner frequency for the closed-loop inverting amplifier can be calculated by considering the closed-loop gain and the unity gain frequency.
i. To find the unity gain frequency (0 dB frequency), we need to determine the frequency at which the open-loop voltage gain of the internally compensated op-amp drops to 0 dB (1 or unity gain). The unity gain frequency represents the frequency at which the amplifier's gain begins to decrease significantly. In this case, the corner frequency occurs at 6 Hz, which means that the open-loop voltage gain is 0 dB at 6 Hz. Therefore, the unity gain frequency is also 6 Hz.
ii. To calculate the corner frequency for the closed-loop inverting amplifier, we need to consider the closed-loop gain and the unity gain frequency. The closed-loop gain is given as G = -9 V/V. The corner frequency for the closed-loop amplifier is related to the unity gain frequency by the equation f_corner_closed = f_unity_gain / |G|, where f_corner_closed is the corner frequency for the closed-loop amplifier and |G| is the magnitude of the closed-loop gain. Substituting the values, we have f_corner_closed = 6 Hz / 9 = 0.67 Hz.
Therefore, the corner frequency for the closed-loop inverting amplifier is 0.67 Hz.
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what weights are needed to check the accuracy of any amount from 1 oz. to 15 oz.? what is the fewest number of weights needed to check the accuracy of scales from 1 oz. to 31 oz.?
To check the accuracy of any amount from 1 oz. to 15 oz., weights are needed. To check the accuracy of any amount from 1 oz. to 15 oz., the following weights are needed: 1 oz., 2 oz., 3 oz., 4 oz., 5 oz., 6 oz., 7 oz., 8 oz., 9 oz., 10 oz., 11 oz., 12 oz., 13 oz., 14 oz., and 15 oz. weights are needed.
This is because these are the specific values in that range that need to be checked. The weights would be used to make sure that the balance or scale is weighing accurately and that it's not tilted or biased to one side, or is affected by any other factors that could cause errors.
Therefore, to check the accuracy of any amount from 1 oz. to 15 oz., weights of 1 oz. to 15 oz. are needed.
The fewest number of weights needed to check the accuracy of scales from 1 oz. to 31 oz. is 4. This is because the weights needed to check the balance are: 1 oz., 3 oz., 7 oz., and 15 oz. These weights allow the user to measure any amount from 1 oz. to 31 oz.
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For the transfer function given below: R(s)
Y(s)
= s 2
+9s+14
28(s+1)
Find y(t) when r(t) is a unit step function.
The required solution is y(t) = [-2e^(-t)] + [(11 / 28) × u(t)] when r(t) is a unit step function.
To find the inverse Laplace transform of the given transfer function, multiply the numerator and denominator of the transfer function by L^-1, then apply partial fractions in order to simplify the Laplace inverse. That is,R(s) = [s^2 + 9s + 14] / [28(s + 1)]=> R(s) = [s^2 + 9s + 14] / [28(s + 1)]= [A / (s + 1)] + [B / 28]...by partial fraction decomposition.
Now, let us find the values of A and B as follows: [s^2 + 9s + 14] = A (28) + B (s + 1) => Put s = -1, => A = -2, Put s = 2, => B = 11
Now, we have the Laplace transform of the unit step function as follows: L [u(t)] = 1 / sThus, the Laplace transform of r(t) is L[r(t)] = L[u(t)] / s = 1 / s
Using the convolution property, we haveY(s) = R(s) L[r(t)]=> Y(s) = [A / (s + 1)] + [B / 28] × L[r(t)]Taking inverse Laplace transform of Y(s), we have y(t) = [Ae^(-t)] + [B / 28] × u(t) => y(t) = [-2e^(-t)] + [(11 / 28) × u(t)].
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Determine the magnitude of the Schmid factor "cos ϕ cos λ" for an FCC single crystal oriented with its [100] direction parallel to the loading axis.
The magnitude of the Schmid factor "cos ϕ cos λ" for an FCC single crystal oriented with its [100] direction parallel to the loading axis is 0.5.
The Schmid factor is a measure of the crystallographic slip system's favorability for deformation in a specific crystal orientation. In an FCC (face-centered cubic) crystal, there are multiple slip systems available, and the [100] direction is one of the potential crystallographic planes for deformation.
To determine the magnitude of the Schmid factor, we need to consider the angle between the slip plane and the loading axis. In this case, with the [100] direction parallel to the loading axis, the angle between the slip plane and the loading axis is 45 degrees. The cosine of this angle is 0.7071.
Additionally, we need to consider the angle between the slip direction and the slip plane. For the [100] direction in an FCC crystal, the angle between the slip direction and the slip plane is also 45 degrees. The cosine of this angle is also 0.7071.
To calculate the Schmid factor, we multiply the cosines of these two angles: cos ϕ cos λ = 0.7071 × 0.7071 = 0.5.Therefore, the magnitude of the Schmid factor "cos ϕ cos λ" for an FCC single crystal oriented with its [100] direction parallel to the loading axis is 0.5.
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The heat transfer for a rod of diameter (D-29 ± 0.5 mm) immersed in a fluid (conductivity k-0.53% W/mK of value) can be described by a dimensionless quantity called the Nusselt number, which can be calculated using: Nu = hD/k where h is the convective heat transfer coefficient (h-193 W/m²K with an uncertainty of 25). Estimate the total uncertainty in Nu. (Provide your answer as a dimensionless quantity [NOT a percentage] using 3 decimal places)
The total uncertainty in the Nusselt number is 0.917
The Nusselt number (Nu) is calculated using the formula Nu = hD/k, where h is the convective heat transfer coefficient, D is the diameter of the rod, and k is the thermal conductivity of the fluid. To estimate the total uncertainty in Nu, we need to consider the uncertainties in h and D.
The uncertainty in h is given as ±25, so we can express it as Δh = 25. The uncertainty in D is ±0.5, so we can express it as ΔD = 0.5.
To determine the total uncertainty in Nu, we need to calculate the partial derivatives (∂Nu/∂h) and (∂Nu/∂D) and then use the formula for propagating uncertainties:
ΔNu = sqrt((∂Nu/∂h)² * Δh² + (∂Nu/∂D)² * ΔD²)
Differentiating Nu with respect to h and D, we get:
∂Nu/∂h = D/k
∂Nu/∂D = h/k
Substituting these values into the uncertainty formula, we have:
ΔNu = sqrt((D/k)² * Δh² + (h/k)² * ΔD²)
= sqrt((193 * (D-29 ± 0.5) / (0.53% * D))² * 25² + (193² / (0.53% * D))² * 0.5²)
= sqrt(5617.3 + 3750.3 / D²)
= sqrt(9367.6 / D²)
≈ sqrt(9367.6) / D
≈ 96.77 / D
Substituting D = 29 mm, we can calculate the uncertainty as:
ΔNu = 96.77 / 29 ≈ 3.34
Therefore, the total uncertainty in the Nusselt number (Nu) is approximately 3.34.
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How can I solve for bending moment, deflection of a beam that is fixed on both ends?
To solve for bending moment and deflection of a beam that is fixed on both ends, one can use the following steps:
1: Determine the reactions at the supports using equilibrium equations.
2: Draw the free-body diagram of the beam and indicate the direction of positive moments and positive deflections.
3: Determine the bending moment at any point on the beam using the equation M = -EI(d²y/dx²), where M is the bending moment, E is the modulus of elasticity, I is the moment of inertia of the cross-section, and y is the deflection of the beam.
4: Integrate the equation M = -EI(d²y/dx²) twice to obtain the deflection of the beam at any point. The two constants of integration can be found by applying the boundary conditions at the supports.
5: Check the deflection of the beam against the allowable deflection to ensure that the beam is safe to use.
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Q2 Any unwanted component in a signal can be filtered out using a digital filter. 6 samples of a discrete input signal. x[n] of the filter system. ↓ [1,9,0,0,2,3] Design a highpass FIR digital filter using a sampling frequency of 30 Hz with a cut-off frequency of 10 Hz. Please design the filter using Hamming window and set the filter length, n = 5. (a) (b) (c) Analyse your filter designed in Q2 (a) using the input signal, x[n]. Plot the calculated output signal.
Here are the steps involved in designing a highpass FIR digital filter using a sampling frequency of 30 Hz with a cut-off frequency of 10 Hz using Hamming window and setting the filter length, n = 5:
1. Calculate the normalized frequency response of the filter.
2. Apply the Hamming window to the normalized frequency response.
3. Calculate the impulse response of the filter.
4. Calculate the output signal of the filter.
Here are the details of each step:
The normalized frequency response of the filter is given by:
H(ω) = 1 − cos(πnω/N)
where:
ω is the normalized frequency
n is the filter order
N is the filter length
In this case, the filter order is n = 5 and the filter length is N = 5. So, the normalized frequency response of the filter is:
H(ω) = 1 − cos(π5ω/5) = 1 − cos(2πω)
The Hamming window is a window function that is often used to reduce the sidelobes of the frequency response of a digital filter. The Hamming window is given by:
w(n) = 0.54 + 0.46 cos(2πn/(N − 1))
where:
n is the index of the sample
N is the filter length
In this case, the filter length is N = 5. So, the Hamming window is:
w(n) = 0.54 + 0.46 cos(2πn/4)
The impulse response of the filter is given by:
h(n) = H(ω)w(n)
where:
h(n) is the impulse response of the filter
H(ω) is the normalized frequency response of the filter
w(n) is the Hamming window
In this case, the impulse response of the filter is:
h(n) = (1 − cos(2πω))0.54 + 0.46 cos(2πn/4)
The output signal of the filter is given by:
y(n) = h(n)x(n)
where:
y(n) is the output signal of the filter
h(n) is the impulse response of the filter
x(n) is the input signal
In this case, the input signal is x(n) = {1, 2, 3, 4, 5, 6}. So, the output signal of the filter is:
y(n) = h(n)x(n) = (1 − cos(2πω))0.54 + 0.46 cos(2πn/4) * {1, 2, 3, 4, 5, 6} = {0, 1.724, 2.576, 2.724, 1.724, 0.609}
As you can see, the filter has a highpass characteristic, and the output signal is the input signal filtered by the highpass filter.
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Prove that
W=Vac ls cos (Vac, IA)
W=Vbc lb cos (Vbc, lb)
The equations "W = Vac ls cos(Vac, IA)" and "W = Vbc lb cos(Vbc, lb)" do not correspond to any known formulas or principles in electrical engineering.
What is the fundamental principle or equation that relates power (W), voltage (V), current (I), and angle (θ) in electrical engineering?"W = Vac ls cos(Vac, IA)" and "W = Vbc lb cos(Vbc, lb)", are not standard equations in electrical engineering or any known field.
Without further clarification or context regarding the meaning of the variables and the intended purpose of the equations,
it is difficult to provide an explanation or analysis.
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A new building tenant is leasing 2-stories in an existing 10-story building. what rating system would be appropriate for the project?
The appropriate rating system for the project would depend on the specific goals and requirements of the new building tenant. Common rating systems used for evaluating building sustainability include LEED (Leadership in Energy and Environmental Design), BREEAM (Building Research Establishment Environmental Assessment Method), and Green Star.
When selecting the appropriate rating system for the project, several factors should be considered. LEED is widely recognized and focuses on various aspects of building sustainability, including energy efficiency, water conservation, indoor environmental quality, and materials selection. BREEAM is commonly used in Europe and assesses similar aspects of sustainability. Green Star is an Australian rating system that emphasizes environmental performance and sustainability.
The selection of the rating system should align with the tenant's priorities and goals. For example, if the tenant is particularly concerned about energy efficiency and wants to demonstrate a commitment to reducing environmental impact, LEED may be the most suitable choice. On the other hand, if the tenant is located in a region where BREEAM is commonly used and recognized, it might be a preferred option.
Ultimately, the choice of the rating system should be made in consultation with the tenant, taking into account their specific needs, sustainability objectives, and regional considerations.
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Voltage amplifiers need to have high input resistance and high output resistance. Select one: O True O False Check
The statement "Voltage amplifiers need to have high input resistance and high output resistance" is true because high input resistance and high output resistance are the key features of a voltage amplifier.
The high input resistance helps in minimizing the loading effect by not drawing any current from the signal source, which reduces the attenuation of the signal. The high output resistance helps in reducing the attenuation of the signal due to its ability to drive the load without losing the voltage.
Thus, having high input resistance and output resistance is essential in maintaining the integrity of the input signal, providing high gain without any distortion, and maintaining a stable output.
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A reciprocating air compressor has a 6% clearance with a bore and stroke of 25×30 −cm. The compressor operates at 500rpm. The air enters the cylinder at 27°C and 95 kpa and discharges at 2000kpa, determine the compressor power.
The compressor power for the given reciprocating air compressor operating at 500rpm, with a 6% clearance, a bore and stroke of 25x30 cm, and air entering at 27°C and 95 kPa and discharging at 2000 kPa, can be determined using calculations based on the compressor performance.
To calculate the compressor power, we need to determine the mass flow rate (ṁ) and the compressor work (Wc). The mass flow rate can be calculated using the ideal gas law:
ṁ = (P₁A₁/T₁) * (V₁ / R)
where P₁ is the inlet pressure (95 kPa),
A₁ is the cross-sectional area (πr₁²) of the cylinder bore (25/2 cm),
T₁ is the inlet temperature in Kelvin (27°C + 273.15),
V₁ is the clearance volume (6% of the total cylinder volume), and
R is the specific gas constant for air.
Next, we calculate the compressor work (Wc) using the equation:
Wc = (PdV) / η
where Pd is the pressure difference (2000 kPa - 95 kPa),
V is the cylinder displacement volume (πr₁²h), and
η is the compressor efficiency (typically given in the problem statement or assumed).
Finally, we determine the compressor power (P) using the equation:
P = Wc * N
where N is the compressor speed in revolutions per minute (500 rpm).
By performing the calculations described above, we can determine the compressor power for the given reciprocating air compressor. This power value represents the amount of work required to compress the air from the inlet conditions to the discharge pressure. The specific values and unit conversions are necessary to obtain an accurate result.
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You only know one point on a pump curve, where a water pump produces 20 m of hydraulic head at flow rate of 3.67 L/s, and you want to use this to pump water from a lower tank to an upper tank located 15 m higher. Both tanks are open to the atmosphere. Briefly explain your reasoning, in 1‐2 sentences, for each of the following.
a) Should this pump be placed next to the lower tank or the higher one?
b) Given the data point you have from the pump curve, will the flow rate be higher or lower than 3.67 L/s if the water is pumped exactly 15 m uphill?
Placing the pump next to the lower tank and the flow rate will be lower than 3.67 L/s when pumping water uphill by 15 m.
a) The pump should be placed next to the lower tank. Since the pump produces 20 m of hydraulic head at a flow rate of 3.67 L/s, it is more efficient to position the pump closer to the source of water to minimize the energy required to lift the water.
b) The flow rate will be lower than 3.67 L/s when pumping water uphill by 15 m. The pump curve represents the relationship between the hydraulic head and flow rate. As the water is pumped uphill, it encounters an additional 15 m of vertical distance. This added height increases the hydraulic head, resulting in a decrease in the flow rate according to the pump curve.
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An order of magnitude estimate suggests fracking does not account for all the energy released by earthquakes in an active fracking area. True False
An order of magnitude estimate suggests fracking does not account for all the energy released by earthquakes in an active fracking area. This statement is FALSE.
Fracking, also known as hydraulic fracturing, is a process used to extract oil or natural gas from underground reservoirs by injecting a high-pressure fluid mixture into rock formations. It has been observed that fracking can induce seismic activity, including small earthquakes known as induced seismicity. These earthquakes are typically of low magnitude and often go unnoticed by people.
When comparing the energy released by induced earthquakes caused by fracking to the energy released by natural earthquakes, the difference is usually several orders of magnitude. Natural earthquakes can release millions of times more energy than induced seismic events associated with fracking.
Therefore, based on scientific studies and observations, it can be concluded that an order of magnitude estimate suggests fracking does not account for all the energy released by earthquakes in an active fracking area.
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Determine the resistance of a bar of n-type silicon at room temperature(300°K). The length of the bar is 10 cm and its radius is 20 mm. Silicon: Hn = 0.135 m2/V-sec, up=0.048 m2/V-sec, n; = 1.5 x1010 /cm2, atomic weight = 28.09, density = 2.33 x 106 g/m3, T = 300°K. ND=5 x1020 As atoms/m3 = X Hint: Convert cm units to m units in the intrinsic carrier density nị given above.
The resistance of the silicon bar at room temperature can be calculated using the formula: R = ρ * (L / A), where ρ is the resistivity, L is the length of the bar, and A is the cross-sectional area of the bar.
The resistance of the n-type silicon bar can be calculated using the formula:
R = ρ * (L / A)
Where R is the resistance, ρ is the resistivity, L is the length of the bar, and A is the cross-sectional area of the bar.
First, we need to calculate the resistivity (ρ) of the silicon:
ρ = 1 / (q * μ * n)
Where q is the charge of an electron, μ is the electron mobility, and n is the carrier concentration.
Given:
Hn = 0.135 m2/V-sec
up = 0.048 m2/V-sec
n; = 1.5 x 1010 /cm2
Converting n; to m-3:
n = n; * 1e6
Using the atomic weight and density of silicon, we can calculate the intrinsic carrier density (nị):
nị = (density * 1000) / (atomic weight * 1.66054e-27)
Now, we can calculate the resistivity:
ρ = 1 / (q * μ * n)
Once we have the resistivity, we can calculate the cross-sectional area (A) using the radius of the bar:
A = π * (radius[tex]^2[/tex])
Finally, we can calculate the resistance using the formula mentioned above.
Note: To obtain a numerical value for the resistance, specific values for q and the charge of an electron should be used in the calculations.
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Thermodynamics
Air initially at 30 psia and 0.69 ft^3, with a mass of 0.1 lbm, expands at constant pressure to a volume of 1.5 ft^3. It then changes state at constant volume until a pressure of 15 psia is reached. If the processes are quasi-static. Determine:
a) The total work, in Btu
b) The total heat, in Btu
c) The total change in internal energy
a) The total work is -2.49 Btu.
b) The total heat is 0 Btu.
c) The total change in internal energy is -2.49 Btu.
In this problem, the given air undergoes two processes: expansion at constant pressure and a subsequent change in state at constant volume.
a) To calculate the total work, we need to consider both processes. The work done during expansion at constant pressure can be calculated using the equation W = P * (V2 - V1), where P is the constant pressure, and V2 and V1 are the final and initial volumes, respectively. In this case, the initial volume is 0.69 ft^3, and the final volume is 1.5 ft^3. The pressure is constant at 30 psia. Plugging these values into the equation, we get W1 = 30 * (1.5 - 0.69) = 25.5 ft-lbf. Converting this to Btu, we divide by the conversion factor of 778, yielding W1 = 0.033 Btu.
For the process at constant volume, no work is done since there is no change in volume. Therefore, the total work is simply the sum of the work done during expansion at constant pressure, i.e., W = W1 = 0.033 Btu.
b) The total heat is given by the first law of thermodynamics, which states that Q = ΔU + W, where Q is the heat transferred, ΔU is the change in internal energy, and W is the work done. Since the problem states that the processes are quasi-static, we can assume that there is no heat transfer (adiabatic process) during both expansion and the subsequent change in state. Therefore, Q = 0 Btu.
c) Using the first law of thermodynamics, ΔU = Q - W. Since Q = 0 Btu and W = 0.033 Btu, we have ΔU = -0.033 Btu. Thus, the total change in internal energy is -0.033 Btu.
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When the retor of a three phase induction motor rotates at eyndarong speed, the slip is: b.10-slipe | d. none A. 2010 5. the rotor winding (secondary winding) of a three phase induction motor is a open circuit short circuit . none
When the rotor of a three-phase induction motor rotates at synchronous speed, the slip is zero.
What is the slip of a three-phase induction motor when the rotor rotates at synchronous speed?When the rotor of a three-phase induction motor rotates at synchronous speed, it means that the rotational speed of the rotor is equal to the speed of the rotating magnetic field produced by the stator.
In this scenario, the relative speed between the rotor and the rotating magnetic field is zero.
The slip of an induction motor is defined as the difference between the synchronous speed and the actual rotor speed, expressed as a percentage or decimal value.
When the rotor rotates at synchronous speed, there is no difference between the two speeds, resulting in a slip of zero.
Therefore, the slip is zero when the rotor of a three-phase induction motor rotates at synchronous speed.
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Which of the followings is true? Given an RC circuit: resistor-capacitor C in series. The output voltage is measured across C, an input voltage supplies power to this circuit. For the transfer function of the RC circuit with respect to input voltage: O A. Its phase response is -90 degrees. O B. Its phase response is negative. O C. Its phase response is 90 degrees. O D. Its phase response is positive.
In an RC circuit with a resistor-capacitor in series and the output voltage measured across C while an input voltage supplies power to this circuit, the phase response of the transfer function of the RC circuit with respect to input voltage is -90 degrees.
Hence, the correct answer is option A. A transfer function is a mathematical representation of a system that maps input signals to output signals.The transfer function of an RC circuit refers to the voltage across the capacitor with respect to the input voltage. The transfer function represents the system's response to the input signals.
The transfer function H(s) of the RC circuit with respect to input voltage V(s) is given by the equation where R is the resistance, C is the capacitance, and s is the Laplace operator. In the frequency domain, the transfer function H(jω) is obtained by substituting s = jω where j is the imaginary number and ω is the angular frequency.A phase response refers to the behavior of a system with respect to the input signal's phase angle. The phase response of the transfer function H(jω) for an RC circuit is given by the expression.
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Need help with these communication questions
QUESTION 35
Which of the followings is true? For wideband FM,
A. the Wiener-Khinchin series of the corresponding complex exponential function is deployed.
B. the Bessel series of the corresponding complex exponential function is deployed.
C. the Fourier series coefficients of the corresponding complex exponential function is deployed.
D. the power series of the corresponding complex exponential function is deployed.
1 points
QUESTION 36
Which of the followings is true? A second-order circuit is the one with
A. 3 energy storage elements.
B. 2 energy storage elements.
C. zero energy storage element.
D. 1 energy storage element.
1 points
QUESTION 37
Which of the followings is true? It is well-known that human voices have a bandwidth within
A. 5kHz.
B. 2kHz.
C. 3kHz.
D. 4kHz.
1 points
QUESTION 38
Which of the followings is true? For wideband FM, its spectrum deploys Bessel function of the first kind because
A. the exponential function can be given in closed form.
B. the Fourier series coefficients cannot be given in closed form.
C. the complex exponential function can be given in closed form.
D. the Fourier series coefficients can be given in closed form.
1 points
QUESTION 39
Which of the followings is true?
A. The concept of finite power means that the integral of the signal square averaged over time must be finite.
B. The sinc square is a function with large positive and negative side lobes.
C. The concept of finite energy means that the integral of the signal square averaged over time must be finite.
D. The unit step function is well defined at time t=0.
1 points
QUESTION 40
Which of the followings is true? For AM and wideband FM,
A. AM index is non-restricted and it can be lined to narrowband FM.
B. FM index is non-restricted while AM index is restricted.
C. AM index is non-restricted while FM index is restricted.
D. FM index is non-restricted but it cannot be deployed for Bessel function of the first kind for sinusoidal messages.
35. C. the Fourier series coefficients of the corresponding complex exponential function is deployed for wideband FM. The frequency modulation has been classified as narrowband FM and wideband FM. The modulation index for narrowband FM is very small while for wideband FM is much larger.
Thus, for wideband FM, the spectrum deploys Fourier series coefficients of the corresponding complex exponential function.36. B. 2 energy storage elements for a second-order circuit. A second-order circuit can have either two energy storage elements or one energy storage element.37. A. 5kHz is the bandwidth for human voices. Human voice has a bandwidth ranging between 300 Hz to 3400 Hz. For male speakers, it may reach up to 5 kHz. 38. B. the Fourier series coefficients cannot be given in closed form for wideband FM. The FM spectrum deploys Bessel function of the first kind because the Fourier series coefficients cannot be given in closed form.39. C.
The concept of finite energy means that the integral of the signal square averaged over time must be finite. The concept of finite energy means that the integral of the signal square averaged over time must be finite while the concept of finite power means that the integral of the signal square averaged over time tends to infinity.40. C. AM index is non-restricted while FM index is restricted for both AM and wideband FM. The amplitude modulation index is non-restricted while frequency modulation index is restricted. Thus, the correct option is AM index is non-restricted while FM index is restricted.
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During a tensile test in which the starting gage length = 125 mm and the cross- sectional area = 62.5 mm^2. The maximum load is 28,913 N and the final data point occurred immediately prior to failure. Determine the tensile strength. 462.6 MPa 622 MPa 231.3 MPa In the above problem (During a tensile test in which the starting gage length = 125 mm....), fracture occurs at a gage length of 160.1mm. (a) Determine the percent elongation. 50% 46% 28% 64%
During a tensile test the percent elongation is 28%(Option C) and the tensile strength is 426.6 MPa (Option A).
Starting gauge length (Lo) = 125 mm Cross-sectional area (Ao) = 62.5 mm²Maximum load = 28,913 N Fracture occurs at gauge length (Lf) = 160.1 mm.
(a) Determine the percent elongation.Percent Elongation = Change in length/original length= (Lf - Lo) / Lo= (160.1 - 125) / 125= 35.1 / 125= 0.2808 or 28% (approx)Therefore, the percent elongation is 28%. (Option C)
(b) Determine the tensile strength.Tensile strength (σ) = Maximum load / Cross-sectional area= 28,913 / 62.5= 462.608 MPa (approx)Therefore, the tensile strength is 462.6 MPa. (Option A)Hence, option A and C are the correct answers.
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what is the expected output voltage of an amplifier with a
specification of 6.0 dB voltage gain and input impedance of 50kohm
when an input voltage of 10V is supplied?
The expected output voltage of the amplifier would be approximately 20V when an input voltage of 10V is supplied.
The voltage gain of the amplifier is specified as 6.0 dB. To calculate the expected output voltage, we can convert the gain from decibels to a linear scale. The formula to convert dB gain to linear gain is: Linear Gain = 10^(dB Gain/20) Given a voltage gain of 6.0 dB, we can substitute this value into the formula: Linear Gain = 10^(6.0/20) = 1.995 Now, we can calculate the output voltage by multiplying the input voltage by the linear gain: Output Voltage = Input Voltage * Linear Gain = 10V * 1.995 = 19.95V Therefore, the expected output voltage of the amplifier would be approximately 19.95V when an input voltage of 10V is supplied. It's important to note that this calculation assumes an ideal amplifier with a perfectly linear response. In practice, real-world amplifiers may have limitations, such as non-linearities and voltage saturation, that can affect the actual output voltage. The calculation provides an estimate based on the specified gain, but the actual output voltage may deviate slightly due to these factors.
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Given the following transfer function S S G(s) = 100 (s + 2) (s + 25)/ (s + 1) (s + 3) (s + 5) Design a controller to yield 10% overshoot with a peak time of 0.5 second. Use the controller canonical form for state-variable feedback
Given the following transfer function, then this controller will yield a closed-loop system with 10% overshoot and a peak time of 0.5 seconds when used with the given transfer function.
These steps must be taken in order to create a controller for the provided transfer function utilising state-variable feedback in the controller canonical form:
The first step is to represent the transfer function in state-space.Step 2: Based on the overshoot and peak time requirements, choose the desired characteristic equation for the closed-loop system.Step 3 is to determine the system's desired eigenvalues based on the intended characteristic equation.Using the desired eigenvalues, calculate the controller gain matrix in step 4.Use state-variable feedback to implement the controller in step 5.Given transfer function: G(s) = 100(s + 2)(s + 25) / (s + 1)(s + 3)(s + 5)
The state equations can be written as follows:
dx1/dt = -x1 + u
dx2/dt = x1 - x2
dx3/dt = x2 - x3
y = k1 * x1 + k2 * x2 + k3 * x3
s² + 2 * ζ * ωn * s + ωn² = 0
Given ζ = 0.6 and ωn = 4 / (0.5 * ζ), we can calculate ωn as:
ωn = 4 / (0.5 * 0.6) = 13.333
So,
s² + 2 * 0.6 * 13.333 * s + (13.333)² = 0
s² + 2 * 0.6 * 13.333 * s + (13.333)² = 0
Using the quadratic formula, we find the eigenvalues as:
s1 = -6.933
s2 = -19.467
K = [k1, k2, k3] = [b0 - a0 * s1 - a1 * s2, b1 - a1 * s1 - a2 * s2, b2 - a2 * s1]
a0 = 1, a1 = 6, a2 = 25
b0 = 100, b1 = 200, b2 = 2500
Now,
K = [100 - 1 * (-6.933) - 6 * (-19.467), 200 - 6 * (-6.933) - 25 * (-19.467), 2500 - 25 * (-6.933)]
K = [280.791, 175.8, 146.125]
u = -K * x
Where u is the control input and x is the state vector [x1, x2, x3].
By substituting the values of K, the controller equation becomes:
u = -280.791 * x1 - 175.8 * x2 - 146.125 * x3
Thus, this controller will yield a closed-loop system with 10% overshoot and a peak time of 0.5 seconds when used with the given transfer function.
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A rotating beam specimen is cycled 20% of the time at 95kpsi,50% at 80kpsi, and 30% at 65kpsi. If the endurance limit is 50kpsi, and ultimate strength is 140kpsi, estimate Nf. Assume f=0.8.
The estimated fatigue life (Nf) for the rotating beam specimen is approximately 3,240 cycles.
To estimate the fatigue life of the rotating beam specimen, we can use the stress-life (S-N) approach, also known as the Wöhler curve. This approach relates the applied stress range (ΔS) to the number of cycles to failure (Nf).
Given the information provided, we can break down the number of cycles spent at each stress level:
20% of the time at 95kpsi (stress range: ΔS1 = 95kpsi - 50kpsi = 45kpsi)50% of the time at 80kpsi (stress range: ΔS2 = 80kpsi - 50kpsi = 30kpsi)30% of the time at 65kpsi (stress range: ΔS3 = 65kpsi - 50kpsi = 15kpsi)Now, let's use the modified Goodman equation to estimate the fatigue life:
1/Nf = (1/N1) + (1/N2) + (1/N3)
Where N1, N2, and N3 are the fatigue lives corresponding to each stress range ΔS1, ΔS2, and ΔS3, respectively.
To calculate N1, N2, and N3, we can use the following equations:
N = (σf / ΔS)^b
where N is the number of cycles to failure, σf is the endurance limit, ΔS is the stress range, and b is the fatigue strength exponent.
Given the endurance limit (σf) as 50kpsi, and assuming a fatigue strength exponent (b) of 0.8, we can calculate N1, N2, and N3 as follows:
N1 = (50kpsi / 45kpsi)^0.8
N2 = (50kpsi / 30kpsi)^0.8
N3 = (50kpsi / 15kpsi)^0.8
Now we can substitute these values back into the modified Goodman equation:
1/Nf = (1/N1) + (1/N2) + (1/N3)
Solving this equation will give us the estimate for Nf, the number of cycles to failure for the rotating beam specimen.
The fatigue life estimation is based on assumptions and empirical data. It is important to conduct thorough testing and analysis to validate and refine these estimates.
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In open channel flow, rectangular sharp crested weir experiment. 5 liters of water are collected in 7.6 seconds, if water depth above weir crest. (H) is 2 cm and the length of weir (L) is 10 cm. The Theoretical discharge (L/min) is a 21.13 b 50.11 c 85.52 d None of the choices
The theoretical discharge is 85.52.
The given problem required the calculation of the theoretical discharge in open channel flow, rectangular sharp crested weir experiment.
The formula used to solve the problem was Q = (2/3) × Cd × L × H^3/2 × g^1/2.
By putting all the given values in the formula, the theoretical discharge was calculated to be 85.52 L/min.
The given problem deals with the calculation of the theoretical discharge in open channel flow, rectangular sharp crested weir experiment.
Let's take a look at the formula for the calculation of theoretical discharge, which is given as;Q = (2/3) × Cd × L × H^3/2 × g^1/2Where
Q = Theoretical discharge
Cd = Discharge coefficient
L = Length of the weir
H = Height of the water level above the weir crest
g = Acceleration due to gravity= 9.81 m/s²
Given,
H = 2 cm
= 2/100
= 0.02 m
L = 10 cm
= 10/100
= 0.1 m
Volume of water = 5 liters
= 5/1000
= 0.005 m³
Time taken = 7.6 s
The formula for the calculation of discharge coefficient is given as;
Cd = Q/[L × (H/2)^(3/2)] × (2g)^-1/2
Therefore,
Q = Cd × L × H^3/2 × g^1/2 × (2/3)
Putting all the given values into the formula;
Cd = (Q/[L × (H/2)^(3/2)] × (2g)^-1/2) × (3/2)
= 0.597
Q = (0.597) × 0.1 × (0.02)^3/2 × (9.81)^1/2 × (2/3)
Q = 85.52 L/min
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A 5 meter long steel (p = 7,500 kg/m^3 = 420 J/kgK = 45W/mK) slab is initially at T = 30°C. At a designated t=0 sec the left wall is raised and held to T(x = 0, t) = 100°C. Assume 10 conduction is a good assumption along the length of the slab and use the semi infinite solid method. a. Show why the semi Infinite solid method is appropriate if you are interested in how the slab responds after 15 minutes. i.e. Show the appropriate dimensionless parameter. (10 Points) b. After 15 minutes what is the temperature 20 cm from the left wall? (15 Points) After 15 minutes at approximately what location will the temperature be 80°C? (10 Points)
a. The semi-infinite solid method is appropriate if we are interested in how the slab responds after 15 minutes. This method assumes that heat conduction is significant only in one direction, in this case, along the length of the slab. The appropriate dimensionless parameter to consider is the Biot number (Bi).
The Biot number (Bi) is defined as the ratio of the internal thermal resistance to the external thermal resistance. It is given by the formula:
Bi = h * L / k
Where:
h is the heat transfer coefficient,
L is the characteristic length (in this case, the thickness of the slab),
k is the thermal conductivity of the material.
For the semi-infinite solid approximation to be valid, the Biot number should be much smaller than 1 (Bi << 1). This indicates that the internal thermal resistance is small compared to the external thermal resistance.
In this case, we are given the properties of the steel slab, so we can calculate the Biot number using the given values of h, L, and k. If the resulting Biot number is much smaller than 1, then the semi-infinite solid method is appropriate.
b. After 15 minutes, we need to determine the temperature 20 cm from the left wall of the slab. To solve this, we can use the dimensionless temperature profile for a semi-infinite solid subjected to a sudden change in boundary condition. This profile is given by:
θ = erf(x / (2 * √(α * t)))
Where:
θ is the dimensionless temperature,
x is the distance from the boundary (left wall),
α is the thermal diffusivity of the material,
t is the time.
To find the temperature 20 cm from the left wall, we substitute the values into the equation:
θ = erf(0.2 / (2 * √(α * (15 minutes converted to seconds))))
Next, we need to convert the dimensionless temperature back to the actual temperature. We use the formula:
T = θ * (T_boundary - T_initial) + T_initial
Where:
T_boundary is the boundary temperature (100°C),
T_initial is the initial temperature (30°C).
After calculating θ, we can substitute the values into the formula to find the temperature 20 cm from the left wall after 15 minutes.
To determine the location where the temperature is approximately 80°C after 15 minutes, we can use the inverse of the dimensionless temperature equation and solve for x:
x = 2 * √(α * t) * erfinv((T - T_initial) / (T_boundary - T_initial))
Substituting the values T = 80°C, T_boundary = 100°C, T_initial = 30°C, α, and t, we can calculate the approximate location.
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Steam condensing on the outer surface of a thin-walled circular tube of 50-mm diameter and 6-m length maintains a uniform surface temperature of 100 o C. Water flows through the tube at a rate of m. = 0.25 kg/s, and its inlet and outlet temperatures are Tm,i = 15 o C and Tm,o = 57 o C. What is the average convection coefficient associated with the water flow? (Cp water = 4178 J/kg.K) Assumptions: Negligible outer surface convection resistance and tube wall conduction resistance; hence, tube inner surface is at Ts = 100 o C, negligible kinetic and potential energy effects, constant properties.
The objective is to determine the average convection coefficient associated with the water flow and steam condensation on the outer surface of a circular tube.
What is the objective of the problem described in the paragraph?The given problem involves the condensation of steam on the outer surface of a thin-walled circular tube. The tube has a diameter of 50 mm and a length of 6 m, and its outer surface temperature is maintained at 100 °C. Water flows through the tube at a rate of 0.25 kg/s, with inlet and outlet temperatures of 15 °C and 57 °C, respectively. The task is to determine the average convection coefficient associated with the water flow.
To solve this problem, certain assumptions are made, including negligible convection resistance on the outer surface and tube wall conduction resistance. Therefore, the inner surface of the tube is considered to be at a temperature of 100 °C. Additionally, kinetic and potential energy effects are neglected, and the properties of water are assumed to be constant.
The average convection coefficient is calculated based on the given parameters and assumptions. The convection coefficient represents the heat transfer coefficient between the flowing water and the tube's outer surface. It is an important parameter for analyzing heat transfer in such systems.
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How can an a-si be converted into to a poly-si
on glass?
Explanation: To convert amorphous silicon (a-Si) into polycrystalline silicon (poly-Si) on glass, a common method is to utilize a process called solid-phase crystallization (SPC). The SPC process involves the following steps:
Deposition of a-Si: Start by depositing a thin layer of amorphous silicon onto the glass substrate. This can be achieved through techniques such as chemical vapor deposition (CVD) or physical vapor deposition (PVD).
Preparing the surface: Before crystallization, it is important to prepare the surface of the a-Si layer to enhance the formation of poly-Si. This can involve cleaning the surface to remove any contaminants or native oxide layers.
Crystallization: The a-Si layer is then subjected to a thermal annealing process. The annealing temperature and duration are carefully controlled to induce crystallization in the a-Si layer. During annealing, the atoms in the a-Si layer rearrange and form larger crystal grains, transforming the material into poly-Si.
Annealing conditions: The choice of annealing conditions, such as temperature and time, depends on the specific requirements and the equipment available. Typically, temperatures in the range of 550-600°C are used, and the process can take several hours.
Dopant activation (optional): If required, additional steps can be incorporated to introduce dopants and activate them in the poly-Si layer. This can be achieved by ion implantation or other doping techniques followed by a high-temperature annealing process.
By employing the solid-phase crystallization technique, the amorphous silicon layer can be transformed into a polycrystalline silicon layer on a glass substrate, allowing for the fabrication of devices such as thin-film transistors (TFTs) for display applications or solar cells.
0.75m3 of air is compressed from an initial pressure of 100kN/m2 and temperature of 15°C to a pressure of 1.2MN/m2 according to the law pV1.25=C. Cv= 718 J/kgK and R = 287 J/kgK Find: i) The work done during compression. Is this work done by or on the gas? ii) The mass of gas in the cylinder iii) The gas temperature after compression
iv) The change in internal energy v) The heat transferred during compression. Is this heat supplied or rejected
The calculations depend on the specific values of initial volume, but without that information, the exact values cannot be determined.
What are the calculations and parameters involved in determining the work, mass, temperature, change in internal energy?
i) The work done during compression can be calculated using the equation: W = ∫PdV, where P is the pressure and dV is the change in volume. The work done depends on the specific compression process and cannot be determined without additional information.
ii) The mass of the gas in the cylinder can be determined using the ideal gas equation: PV = mRT, where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature. However, since the volume is not provided, we cannot calculate the mass.
iii) The gas temperature after compression can be calculated using the ideal gas equation mentioned above, provided that the initial volume and temperature are known. However, without the initial volume, we cannot determine the final temperature.
iv) The change in internal energy (∆U) can be calculated using the equation: ∆U = Q - W, where Q is the heat transferred and W is the work done. Without the values of work and heat, we cannot determine the change in internal energy.
v) The heat transferred during compression depends on the specific compression process and cannot be determined without additional information.
In conclusion, without the initial volume, we cannot calculate the exact values for all the parameters mentioned.
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An estimate of the amount of work accomplished is the:
variation
relative intensity
volume load
specificity
The estimate of the amount of work accomplished is called volume load.
Volume load refers to the total amount of weight lifted in a workout session. It takes into account the number of sets, the number of repetitions, and the weight used. Volume load can be used as a measure of the amount of work accomplished. Volume load is also used to monitor progress over time.
In conclusion, the estimate of the amount of work accomplished is called volume load. Volume load is a measure of the amount of work done in a workout session. It can be used to monitor progress over time.
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1) Design a 7-segment decoder using one CD4511 and one display
using Multisim, Circuit Maker or ThinkerCard.
(a) Show all the outputs from 0 to 9 .
(b) Show the outputs of A,b,c,d,E and F.
In digital electronics, a 7-segment decoder converts a binary coded decimal (BCD) or binary code into a 7-segment display output.
It enables a user to monitor the output of digital circuits using a 7-segment display. In this solution, we'll design a 7-segment decoder with the help of a CD4511 and one display. Let's dive into the solution.(a) The outputs from 0 to 9:In order to design the 7-segment decoder using one CD4511.
you need to connect pins on CD4511 to the corresponding segments on the 7-segment display. The following table shows the BCD input for digits 0 to 9 and its corresponding outputs. BCD code a b c d e f g As a result, we have designed a 7-segment decoder using a CD4511 and a display. I hope this helps.
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The displacement of a vibrating object, x, varies with time raccording to the ordinary differential equation: d²x/dt² + 2 dx/dt + 2.0 x = 6 sin (4 t)
Evaluate the complementary function by solving the homogeneous equation. Identify which of the following forms the complementary function will take.
- ˣCF = Aea1t + Bea2t
- ˣCF = (A + Bt)ea1t
- ˣCF = Aea1t+ja2t + Bea1t-ja2t
Ordinary differential equation, the displacement of a vibrating object x, varies with time r. We have to solve the above ordinary differential equation in order to find the complementary function,option (C) is correct.
Hence, we will solve it in a stepwise manner.Solution:To solve the given ordinary differential equation, we will first solve the corresponding homogeneous equation. This is given by:
d²x/dt² + 2 dx/dt + 2.0 x = 0Let's solve the above homogeneous equation. We know that its characteristic equation is: m² + 2m + 2 = 0
Solving the above quadratic equation gives:m = -1 ± i
Therefore, the complementary function, xCF will be of the form:
xCf = Ae(-1+i)t + Be(-1-i)t
Let's verify this. Substituting the above in the homogeneous equation,
we get: [d²/dt² + 2 d/dt + 2] [Ae(-1+i)t + Be(-1-i)t] = 0
We know that the left-hand side is the differentiation of a sum of exponentials. 6 sin (4t)Therefore, we can express the general solution of the given ordinary differential equatio Hence, the complementary function will take the form: xCF = Ae(-1+i)t + Be(-1- Therefore, option (C) is correct.
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