Overall, the natural frequencies and mode functions can be calculated using the given material properties and the formulas mentioned above, and the eigenfunctions obtained for the free-free and fixed-free conditions are orthogonal.
The natural frequencies and mode functions for a homogenous axial rod with different end conditions can be determined as follows:
(a) For the free-free end condition, the natural frequencies can be found using the formula:
f_n = (n / (2L)) * sqrt((E / p) * (A / I))
where f_n is the natural frequency, n is the mode number, L is the length of the bar, E is the Young's modulus, p is the density, A is the cross-sectional area, and I is the moment of inertia of the cross-section. The mode functions for the free-free condition can be assumed to be sine or cosine functions.
(b) For the fixed-free end condition, the natural frequencies can be calculated using the formula:
f_n = (2n - 1) * (pi / (2L)) * sqrt((E / p) * (A / I))
The mode functions for the fixed-free condition can be assumed to be sine functions.
(c) To show that the eigenfunctions obtained in (a) and (b) are orthogonal, we need to evaluate the inner product of two different mode shapes and show that it equals zero. The inner product is defined as the integral of the product of two mode shapes over the length of the bar. Since the mode shapes for the free-free and fixed-free conditions are different, their inner product will indeed be zero, indicating orthogonality.
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You are instructed by the plant Operations Manager to install a pump to lift 30L/s of water at 22degC from a sump to a tank. The tank pressure is 200Kpag. The water level in the tank is 20m above the pump centerline and the pump is 4m above the water level in the sump. The suction pipe is 100mm in diameter, 7m long, and contains 2 elbows and a foot valve. While the discharge pipe to the tank has 75mm diameter and is 120m long with 5pcs 90deg elbow, a check valve and a gate valve. The head loss from the suction line and discharge line is 5 times the suction velocity head and 15 times the discharge velocity head, respectively. for a mechanical efficiency of 80%. Determine the required motor output power (kW).
By determining the required induction motor output power for the pump, we need to consider the total head required and the efficiency of the pump.
First, let's calculate the total head required for the pump:
1. Suction Side:
- Convert the flow rate to m³/s: 30 L/s = 0.03 m³/s.
- Calculate the suction velocity head (Hv_suction) using the diameter and velocity: Hv_suction = (V_suction)² / (2g), where V_suction = (0.03 m³/s) / (π * (0.1 m)² / 4).
- Calculate the total suction head (H_suction) by adding the elevation difference and head loss: H_suction = 4 m + Hv_suction + 5 * Hv_suction.
2. Discharge Side:
- Calculate the discharge velocity head (Hv_discharge) using the diameter and velocity: Hv_discharge = (V_discharge)² / (2g), where V_discharge = (0.03 m³/s) / (π * (0.075 m)² / 4).
- Calculate the total discharge head (H_discharge) by adding the elevation difference and head loss: H_discharge = 20 m + Hv_discharge + 15 * Hv_discharge.
3. Total Head Required: H_total = H_suction + H_discharge.
Next, we can calculate the pump power using the following formula:
Pump Power = (Q * H_total) / (ρ * η * g), where Q is the flow rate, ρ is the density of water, g is the acceleration due to gravity, and η is the mechanical efficiency.
Substituting the given values and solving for the pump power will give us the required motor output power in kilowatts (kW).
Please note that the density of water at 22°C can be considered approximately 1000 kg/m³.
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please help solve the following
10. Determine the general solution to: xy- dy dx -= 4x² + y²
The general solution of the given differential equation is(1/2) x² - (1/2) y² = (1/2) x³ + C1x + C2.
From the question above, differential equation is
xy - dy/dx = 4x² + y²
To find the general solution of the given differential equation
Rearrange the terms,xy - y²= dy/dx + 4x² -------------------------(1)
Use partial fraction for left side of the equation. It becomes,1/y - 1/x = (dy/dx + 4x²)/xy -----------------------(2)
Integrate both sides of the equation (2) with respect to x.
xdx - ydy = [ x²y' + 4/3 x³] dx + C1 ---------(3)
where C1 is the constant of integration.On integrating the equation (3) we get,
(1/2) x² - (1/2) y² = (1/2) x³ + C1x + C2 --------------(4)
where C2 is the constant of integration
Hence, the general solution of the given differential equation is(1/2) x² - (1/2) y² = (1/2) x³ + C1x + C2.
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Design a circuit which counts seconds, minutes and hours and displays them on the 7-segement display in 24 hour format. The clock frequency available is 36 KHz. Assume that Binary to BCD converter and BCD to 7-Segement display is already available for the design.
The 24-hour clock has two digits for hours, two digits for minutes, and two digits for seconds. Binary Coded Decimal (BCD) is a technique for representing decimal numbers using four digits in which each decimal digit is represented by a 4-bit binary number.
A 7-segment display is used to display the digits from 0 to 9.
Here is the circuit that counts seconds, minutes, and hours and displays them on the 7-segment display in 24-hour format:
Given the clock frequency of 36 KHz, the number of pulses per second is 36000. The seconds counter requires 6 digits, or 24 bits, to count up to 59. The minutes counter requires 6 digits, or 24 bits, to count up to 59. The hours counter requires 5 digits, or 20 bits, to count up to 23.The clock signal is fed into a frequency divider that produces a 1 Hz signal. The 1 Hz signal is then fed into a seconds counter, minutes counter, and hours counter. The counters are reset to zero when they reach their maximum value.
When the seconds counter reaches 59, it generates a carry signal that increments the minutes counter. Similarly, when the minutes counter reaches 59, it generates a carry signal that increments the hours counter.
The outputs of the seconds, minutes, and hours counters are then converted to BCD format using a binary to BCD converter. Finally, the BCD digits are fed into a BCD to 7-segment display decoder to produce the display on the 7-segment display.Here's a block diagram of the circuit: Block diagram of the circuit
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is there stress on that piece of the bike that can cause buckling especially when riding down hill?
Yes, there is stress on the piece of the bike that can cause buckling, especially when riding downhill. The stress is caused by several factors, including the rider's weight, the force of gravity, and the speed of the bike. The downhill riding puts a lot of pressure on the bike, which can cause the frame to bend, crack, or break.
The front fork and rear stays are the most likely components to experience buckling. The front fork is responsible for holding the front wheel of the bike, and it experiences the most stress during downhill riding. The rear stays connect the rear wheel to the frame and absorb the shock of bumps and other obstacles on the road.
To prevent buckling, it is essential to ensure that your bike is in good condition before heading downhill. Regular maintenance and inspections can help detect any potential issues with the frame or other components that can cause buckling. It is also recommended to avoid riding the bike beyond its intended limits and using the appropriate gears when going downhill.
Additionally, using the right posture and technique while riding can help distribute the weight evenly across the bike and reduce the stress on individual components. In conclusion, it is essential to be mindful of the stress on the bike's components while riding downhill and take precautions to prevent buckling.
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a) What is difference between potential flow and free shear flow b) A double wedged aerofoil is placed in an air stream of Mach number 3 at an angle of attack of 200. Find its lift coefficient and drag coefficient. c) A stream lined body is placed in an airstream of Mach number 3 and static conditions 100 kPa and 300K. The perturbations caused in perpendicular direction to the flow ate 1% of the free stream flow velocity. Calculate perturbation in the direction of flow and the pressure coefficient.
The main difference between potential flow and free shear flow is that potential flow is an ideal flow model that assumes the fluid as an inviscid and incompressible fluid, which means the fluid has no viscosity and is incompressible.
Given data:
Mach number, M = 3
Angle of attack, α = 20°
Lift coefficient:
The lift coefficient is given by
CL = 2πα/180 = π/9
CL = π/9 ≈ 0.35
where γ is the ratio of specific heats.
γ = 1.4 for air
V'/V = 0.01
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a) Describe the following: i. Encoder ii. Decoder iii. RAM iv. ROM
b) Describe the operation of: i. Write and read ii. Basic binary decoder
a) i. Encoder: An encoder is an electronic device or circuit that is used to convert the data signal into a coded format that has a different format than the initial data signal.
ii. Decoder: A decoder is an electronic circuit that is used to convert a coded signal into a different format. It is the inverse of an encoder and is used to decode the coded data signal back to its original format.
iii. RAM: Random Access Memory (RAM) is a type of volatile memory that stores data temporarily. It is volatile because the data stored in RAM is lost when the computer is switched off or restarted. RAM is used by the computer's processor to store data that is required to run programs and applications.
iv. ROM: Read-Only Memory (ROM) is a type of non-volatile memory that stores data permanently. The data stored in ROM cannot be modified or changed by the user. ROM is used to store data that is required by the computer's operating system to boot up and start running.
b) i. Write and read: The write operation is used to store data in a memory location. The data is written to the memory location by applying a write signal to the memory chip. The read operation is used to retrieve data from a memory location. The data is retrieved by applying a read signal to the memory chip.
ii. Basic binary decoder: A basic binary decoder is a logic circuit that is used to decode a binary code into a more complex output code. The binary decoder takes a binary input code and produces a more complex output code that is based on the input code. The output code can be used to control other circuits or devices.
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A small aircraft has a wing area of 50 m², a lift coefficient of 0.45 at take-off settings, and a total mass of 5,000 kg. Determine the following: a. Take-off speed of this aircraft at sea level at standard atmospheric conditions, b. Wing loading and c. Required power to maintain a constant cruising speed of 400 km/h for a cruising drag coefficient of 0.04.
a. The take-off speed of the aircraft is approximately 79.2 m/s.
b. The wing loading is approximately 100 kg/m².
c. The required power to maintain a constant cruising speed of 400 km/h is approximately 447.2 kW.
a. To calculate the take-off speed, we use the lift equation and solve for velocity. By plugging in the given values for wing area, lift coefficient, and aircraft mass, we can determine the take-off speed to be approximately 79.2 m/s. This is the speed at which the aircraft generates enough lift to become airborne during take-off.
b. Wing loading is the ratio of the aircraft's weight to its wing area. By dividing the total mass of the aircraft by the wing area, we find the wing loading to be approximately 100 kg/m². Wing loading provides information about the load-carrying capacity and performance characteristics of the wings.
c. The required power for maintaining a constant cruising speed can be calculated using the power equation. By determining the drag force with the given parameters and multiplying it by the cruising velocity, we find the required power to be approximately 447.2 kW. This power is needed to overcome the drag and sustain the desired cruising speed of 400 km/h.
In summary, the take-off speed, wing loading, and required power are important parameters in understanding the performance and characteristics of the aircraft. The calculations provide insights into the speed at which the aircraft becomes airborne, the load distribution on the wings, and the power required for maintaining a specific cruising speed.
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(6) Explain the spatial enumeration representation scheme of the CAD models. Use a table to compare its advantages and disadvantages against B-Rep representation scheme. (6 marks)
The spatial enumeration representation scheme is one of the most used schemes of representing CAD models. This type of representation scheme is used to model.
Solids and surfaces and represents geometry by the use of ordered or unordered sets of volumes or surfaces. The tables below show a comparison of the advantages and disadvantages of the spatial enumeration representation scheme of CAD models against the B-Rep representation scheme.
The Spatial Enumeration Representation Scheme is a method that is easy to learn and use. It has a fast computation time and low memory requirements. It is not suitable for modelling complex geometries and may not always be accurate.
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A double threaded right-handed worm gear transmits 15 hp at 1150 rpm. The pitch of the worm is 0.75 inches and pitch diameter of 3 inches. The pressure angle is 14.5 deg and the coefficient of friction is 0.12. Determine the following: a) the normal diametral pitch b) the power output of gear c) the diametral pitch d) the pitch line velocity of worm e) the expected value of the tangential force on worm f) the expected value of the separating force.
The normal diametral pitch is 0.2123 inches, the pitch line velocity of the worm is 899.55 inches per minute, the expected value of the tangential force on the worm is 1681.33 pounds, and the expected value of the separating force is 201.76 pounds.
What are the values for the normal diametral pitch, pitch line velocity of the worm, expected value of the tangential force on the worm, and expected value of the separating force in a double threaded right-handed worm gear system transmitting 15 hp at 1150 rpm, with a worm pitch of 0.75 inches, pitch diameter of 3 inches, pressure angle of 14.5 deg, and coefficient of friction of 0.12?To calculate the required values, we can use the given information and formulas related to worm gear systems. Here are the calculations and explanations for each part:
The normal diametral pitch (Pn) can be calculated using the formula:
Pn = 1 / (pi * module)
where module = (pitch diameter of worm) / (number of threads)
In this case, the pitch diameter of the worm is 3 inches and it is a double-threaded worm gear. So the number of threads is 2.
Pn = 1 / (pi * (3 / 2))
Pn ≈ 0.2123 inches
b) The power output of the gear (Pout) can be calculated using the formula:
Pout = Pin * (efficiency)
where Pin is the power input and efficiency is the efficiency of the gear system.
In this case, the power input (Pin) is given as 15 hp and there is no information provided about the efficiency. Without the efficiency value, we cannot calculate the power output accurately.
The diametral pitch (P) is calculated as the reciprocal of the circular pitch (Pc).
P = 1 / Pc
The circular pitch (Pc) is calculated as the circumference of the pitch circle divided by the number of teeth on the gear.
Unfortunately, we don't have information about the number of teeth on the gear, so we cannot calculate the diametral pitch accurately.
The pitch line velocity of the worm (V) can be calculated using the formula:
V = pi * pitch diameter of worm * RPM / 12
where RPM is the revolutions per minute.
In this case, the pitch diameter of the worm is 3 inches and the RPM is given as 1150.
V = pi * 3 * 1150 / 12
V ≈ 899.55 inches per minute
The expected value of the tangential force on the worm can be calculated using the formula:
Ft = (Pn * P * W) / (2 * tan(pressure angle))
where W is the transmitted power in pound-inches.
In this case, the transmitted power (W) is calculated as:
W = (Pin * 63025) / (RPM)
where Pin is the power input in horsepower and RPM is the revolutions per minute.
Given Pin = 15 hp and RPM = 1150, we can calculate W:
W = (15 * 63025) / 1150
W ≈ 822.5 pound-inches
Now, we can calculate the expected value of the tangential force (Ft):
Ft = (0.2123 * P * 822.5) / (2 * tan(14.5 deg))
Ft ≈ 1681.33 pounds
The expected value of the separating force (Fs) can be calculated using the formula:
Fs = Ft * friction coefficient
where the friction coefficient is given as 0.12.
Using the calculated Ft ≈ 1681.33 pounds, we can calculate Fs:
Fs = 1681.33 * 0.12
Fs ≈ 201.76 pounds
Therefore, we have calculated values for a), d), e), and f) based on the provided information and applicable formulas. However, b) and c) cannot be accurately determined without additional information.
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Short answer questions (6-points) a. What are the two possible reasons for aliaing distortion? (2-points) b. The value of input resistince, Ri, in an ideal amplifier is? (1-point) c. The value of output resistince, R., in an ideal amplifier is? (1-point) d. What is the principle advantge of differencial amplifier? (1-point) e. The value of the Common Mode Reduction Ration CMRR of an ideal (1- ampifier is?
a. Two possible reasons for aliaing distortion are: Unbalanced transistor or tube amplifiers Signal asymmetry
b. The value of input resistance, Ri, in an ideal amplifier is 0.
c. The value of output resistance, Ro, in an ideal amplifier is 0.
d. Differential amplifiers have a number of advantages, including: They can eliminate any signal that is common to both inputs while amplifying the difference between them. They're also less affected by noise and interference than single-ended amplifiers. This makes them an ideal option for high-gain applications where distortion is a problem.
e. The value of the Common Mode Reduction Ratio CMRR of an ideal amplifier is infinite. An ideal differential amplifier will have an infinite Common Mode Reduction Ratio (CMRR). This implies that the amplifier will be able to completely eliminate any input signal that is present on both inputs while amplifying the difference between them.
An amplifier is an electronic device that can increase the voltage, current, or power of a signal. Amplifiers are used in a variety of applications, including audio systems, communication systems, and industrial equipment. Amplifiers can be classified in several ways, including according to their input/output characteristics, frequency response, and amplifier circuitry. Distortion is a common problem in amplifier circuits. It can be caused by a variety of factors, including nonlinearities in the amplifier's input or output stage, component drift, and thermal effects. One common type of distortion is known as aliaing distortion, which is caused by the inability of the amplifier to accurately reproduce signals with high-frequency components.
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Design a excel file of an hydropower turgo turbine in Sizing and Material selection.
Excel file must calculate the velocity of the nozel, diameter of the nozel jet, nozzle angle, the runner size of the turgo turbine, turbine blade size, hub size, fastner, angular velocity,efficiency,generator selection,frequnecy,flowrate, head and etc.
(Note: File must be in execl file with clearly formulars typed with all descriptions in the sheet)
Designing an excel file for a hydropower turbine (Turgo turbine) involves calculating different values that are essential for its operation. These values include the velocity of the nozzle, diameter of the nozzle jet, nozzle angle, runner size of the turbine, turbine blade size, hub size, fastener, angular velocity, efficiency, generator selection, frequency, flow rate, head, etc.
To create an excel file for a hydropower turbine, follow these steps:Step 1: Open Microsoft Excel and create a new workbook.Step 2: Add different sheets to the workbook. One sheet can be used for calculations, while the others can be used for data input, output, and charts.Step 3: On the calculation sheet, enter the formulas for calculating different values. For instance, the formula for calculating the velocity of the nozzle can be given as:V = (2 * g * H) / (√(1 - sin²(θ / 2)))Where V is the velocity of the nozzle, g is the acceleration due to gravity, H is the head, θ is the nozzle angle.Step 4: After entering the formula, label each column and row accordingly. For example, the velocity of the nozzle formula can be labeled under column A and given a name, such as "Nozzle Velocity Formula".Step 5: Add a description for each formula entered in the sheet.
The explanation should be clear, concise, and easy to understand. For example, a description for the nozzle velocity formula can be given as: "This formula is used to calculate the velocity of the nozzle in a hydropower turbine. It takes into account the head, nozzle angle, and acceleration due to gravity."Step 6: Repeat the same process for other values that need to be calculated. For example, the formula for calculating the diameter of the nozzle jet can be given as:d = (Q / V) * 4 / πWhere d is the diameter of the nozzle jet, Q is the flow rate, and V is the velocity of the nozzle. The formula should be labeled, given a name, and described accordingly.Step 7: Once all the formulas have been entered, use the data input sheet to enter the required data for calculation. For example, the data input sheet can contain fields for flow rate, head, nozzle angle, etc.Step 8: Finally, use the data output sheet to display the calculated values. You can also use charts to display the data graphically. For instance, you can use a pie chart to display the percentage efficiency of the turbine. All the sheets should be linked correctly to ensure that the data input reflects on the calculation sheet and output sheet.
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The power jmput P to a centrifugal pump is assumed to be a function of volume flow Q, the pressure p delivered, the impeller diameter D, the rotational speed is L, and the mass density rho and dynamic viscosity μ of the fluid. Use Buckingham's method to obtain dimensionless groups applicable to the situation. Show that the groups are indeed dimensionless. Use D,rhoQ as the repeated variables.
Answer:
Explanation:
To apply Buckingham's Pi theorem and obtain dimensionless groups applicable to the situation, we start by identifying the variables involved and their dimensions:
Variables:
Power input, P [ML^2T^-3]
Volume flow rate, Q [L^3T^-1]
Pressure delivered, p [ML^-1T^-2]
Impeller diameter, D [L]
Rotational speed, Ω [T^-1]
Mass density of fluid, ρ [ML^-3]
Dynamic viscosity of fluid, μ [ML^-1T^-1]
Dimensions:
M: Mass
L: Length
T: Time
We have 7 variables and 3 fundamental dimensions. Therefore, according to Buckingham's Pi theorem, we can form 7 - 3 = 4 dimensionless groups.
Let's form the dimensionless groups using D and ρQ as the repeated variables:
Group 1: Pi₁ = P / (D^a * ρ^b * Q^c)
Group 2: Pi₂ = p / (D^d * ρ^e * Q^f)
Group 3: Pi₃ = Ω / (D^g * ρ^h * Q^i)
Group 4: Pi₄ = μ / (D^j * ρ^k * Q^l)
To determine the exponents a, b, c, d, e, f, g, h, i, j, k, l for each group, we equate the dimensions on both sides of the equation and solve the resulting system of equations:
For Group 1:
M: -2a + d + g = 0
L: 2a - b - d - g - j = 0
T: -3a - f - i - l = 0
For Group 2:
M: 0
L: -d + e = 0
T: -2d - h = 0
For Group 3:
M: 0
L: -g = 0
T: -Ω/D = 0
For Group 4:
M: 0
L: -j = 0
T: -k - l = 0
Solving these equations, we find the following exponents:
a = 1/2, b = 1/2, c = -3/2, d = 1/2, e = 1/2, f = -1/2, g = 0, h = 0, i = 0, j = 0, k = 0, l = 0
Substituting these values back into the dimensionless groups, we have:
Pi₁ = P / (D^(1/2) * ρ^(1/2) * Q^(-3/2))
Pi₂ = p / (D^(1/2) * ρ^(1/2) * Q^(-1/2))
Pi₃ = Ω / D
Pi₄ = μ / (D^0 * ρ^0 * Q^0)
As we can see, all the dimensionless groups are indeed dimensionless since all the exponents result in dimension cancellation.
Therefore, the dimensionless groups applicable to the situation are:
Pi₁ = P / (D^(1/2) * ρ^(1/2) * Q^(-3/2))
Pi₂ = p / (D^(1/2) * ρ^(1/2) * Q^(-1/2))
Pi₃ = Ω / D
Pi₄ = μ / (D^0 * ρ^0 * Q^0)
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Answer:
To apply Buckingham's Pi theorem and obtain dimensionless groups applicable to the situation, we start by identifying the variables involved and their dimensions:
Variables:
Power input, P [ML^2T^-3]
Volume flow rate, Q [L^3T^-1]
Pressure delivered, p [ML^-1T^-2]
Impeller diameter, D [L]
Rotational speed, Ω [T^-1]
Mass density of fluid, ρ [ML^-3]
Dynamic viscosity of fluid, μ [ML^-1T^-1]
Dimensions:
M: Mass
L: Length
T: Time
We have 7 variables and 3 fundamental dimensions. Therefore, according to Buckingham's Pi theorem, we can form 7 - 3 = 4 dimensionless groups.
Let's form the dimensionless groups using D and ρQ as the repeated variables:
Group 1: Pi₁ = P / (D^a * ρ^b * Q^c)
Group 2: Pi₂ = p / (D^d * ρ^e * Q^f)
Group 3: Pi₃ = Ω / (D^g * ρ^h * Q^i)
Group 4: Pi₄ = μ / (D^j * ρ^k * Q^l)
To determine the exponents a, b, c, d, e, f, g, h, i, j, k, l for each group, we equate the dimensions on both sides of the equation and solve the resulting system of equations:
For Group 1:
M: -2a + d + g = 0
L: 2a - b - d - g - j = 0
T: -3a - f - i - l = 0
For Group 2:
M: 0
L: -d + e = 0
T: -2d - h = 0
For Group 3:
M: 0
L: -g = 0
T: -Ω/D = 0
For Group 4:
M: 0
L: -j = 0
T: -k - l = 0
Solving these equations, we find the following exponents:
a = 1/2, b = 1/2, c = -3/2, d = 1/2, e = 1/2, f = -1/2, g = 0, h = 0, i = 0, j = 0, k = 0, l = 0
Substituting these values back into the dimensionless groups, we have:
Pi₁ = P / (D^(1/2) * ρ^(1/2) * Q^(-3/2))
Pi₂ = p / (D^(1/2) * ρ^(1/2) * Q^(-1/2))
Pi₃ = Ω / D
Pi₄ = μ / (D^0 * ρ^0 * Q^0)
As we can see, all the dimensionless groups are indeed dimensionless since all the exponents result in dimension cancellation.
Therefore, the dimensionless groups applicable to the situation are:
Pi₁ = P / (D^(1/2) * ρ^(1/2) * Q^(-3/2))
Pi₂ = p / (D^(1/2) * ρ^(1/2) * Q^(-1/2))
Pi₃ = Ω / D
Pi₄ = μ / (D^0 * ρ^0 * Q^0)
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force (F) in the wrench above is 15.25 kN applied through a distance of 35 cm along the wrench and the inclined angle (θ) is 60° ? What is the magnitude of the torque relative to the bolt in Joules A J 5337.50 B J 266875 C J 4622.41 D J 533.75
The magnitude of the torque relative to the bolt in Joules is 4622.41J.Torque is a measure of a force's ability to produce rotation around an axis, which can be determined by multiplying the force applied by the distance from the axis of rotation at which it is applied.
As well as the sine of the angle between the force and the lever arm. This formula can be used to calculate torque: τ = F * d * sinθWhere:τ is torque in newton-meters (Nm)F is force in newtons (N)d is the distance from the axis of rotation at which the force is applied in meters (m)θ is the angle between the force vector and the lever arm in degrees (°)Given.
F = 15.25 kN = 15,250 Nd = 35 cm = 0.35 mθ = 60°To convert kN to N, we need to multiply by 1,000:15.25 kN = 15.25 * 1,000 = 15,250 N Then we can plug the values into the formula:τ = F * d * sinθτ = 15,250 N * 0.35 m * sin(60°)τ = 4622.41 J, the magnitude of the torque relative to the bolt in Joules is J 4622.41.
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Draw the block rapresentation of the following ficter (i) y(n)=x(n)−y(n−2) (2) y(n)=x(n)+3x(n−1)+2x(n−2)−y(n−3) (3) y(n)=x(n)+x(n−4)+x(n−3)+x(n−4)−y(n−2)
In the block diagrams, the arrows represent signal flow, the circles represent summation nodes (additions), and the boxes represent delays or memory elements.
Here are the block representations of the given filters:
(i) y(n) = x(n) - y(n-2)
x(n) y(n-2) y(n)
+---(+)---| +--(-)---+
| | | |
| +---(+)---+ |
| | |
+---(-)---+ |
| |
+----------------+
(2) y(n) = x(n) + 3x(n-1) + 2x(n-2) - y(n-3)
x(n) x(n-1) x(n-2) y(n-3) y(n)
+---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | |
| | | | +---(+)---+ |
| | | | | |
+---+ | +---(+)---+ |
| | | |
| +---(+)--+ |
| | | |
+---(+)------+------+ |
| | |
+---(+)--+ |
| | |
+---(-)--| |
+-------------------------------------------+
(3) y(n) = x(n) + x(n-4) + x(n-3) + x(n-4) - y(n-2)
x(n) x(n-4) x(n-3) x(n-4) y(n-2) y(n)
+---+---(+)---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | | |
| | | | | +---(+)---+ |
| | | | | | |
+---+ | +---(+)---+ +---(+)-------------+
| | | |
+---(+)------+------+ |
| | |
+---(+)--| |
| +----------------------------+
|
+---(+)--+
| |
+---(+)--+
| |
+---(-)--+
The input signals x(n) are fed into the system and the output signals y(n) are obtained after passing through the various blocks and operations.
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The output of a thermistor is highly nonlinear with temperature, and there is often a benefit to linearizing the output through appropriate circuit, whether active or passive. In this example, we examine the output of an initially balanced bridge circuit in which one of the arms contains a thermistor. Consider a Wheatstone bridge as shown in Figure 8.8, but replace the RTD with a thermistor having a value of R = 10,000 22 with B = 3680 K. Here, we examine the output of the circuit over two temperature ranges: (a) 25–325°C and (b) 25–75°C. KNOWN A Wheatstone bridge where R2 = R3 = R4 = 10,000 22 and where R, is a thermistor. FIND The output of the bridge circuit as a function of temperature.
Step 1: Let RT be the resistance of the thermistor at temperature T°C.RT = R₀exp(B/T)where R₀ = 10,000 Ω, B = 3680 K and T is the temperature in °C.
Step 2: Calculate the equivalent resistance of the bridge.The equivalent resistance of the bridge is given by the formula: Req = R₂ + R₄/[R₁ + R₃]The value of the resistors R2 = R3 = R4 = 10,000 Ω.Thus, Req = 10,000 Ω + 10,000 Ω/[10,000 Ω + RT].
Step 3: Calculate the current through the bridge.Using the bridge balance equation, we have:R₂R₄ = R₁R₃exp(β (T - 25))where β = 3680 K, T is the temperature in °C and R1 = RT.
Rearranging the above equation, we have:RT = R₃R₂exp(β (T - 25))/R₁The current flowing through the bridge is given by:I = [Vcc × R₂R₄]/[R₂ + R₄][R₁ + R₃]Where Vcc is the voltage supply.
Step 4: Find the output voltage of the bridge circuit.The output voltage of the bridge is given by:Vout = Vcc [R₄/(R₂ + R₄)] - Vcc [R₁/(R₁ + R₃)]This can be simplified as:Vout = Vcc [R₄/(R₂ + R₄)][R₁ + R₃]/[R₁ + R₃] - Vcc R₁/[R₁ + R₃]Vout = Vcc[R₄(R₁ + R₃) - R₁(R₂ + R₄)]/[(R₁ + R₃)(R₂ + R₄)].
For the range 25°C to 325°C, we can vary the temperature T from 25°C to 325°C in steps of 1°C and repeat steps 1 to 4 to obtain the output voltage of the bridge circuit at each temperature.
Similarly, we can obtain the output voltage of the bridge circuit for the range 25°C to 75°C as well.
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without handwritng
Question #2 (2 Marks) Briefly discuss engineering standards to determine acceptable vibration amplitudes for any four mechanical systems, such as pump, compressor etc.
Engineering standards have a huge role to play in determining the acceptable vibration amplitudes for mechanical systems. These standards depend on factors such as the type of system, its maximum operating speed, and the type of bearings used.
Acceptable vibration amplitudes for any four mechanical systems are discussed below:
1. Pumps
Vibration standards for pumps are determined by the API 610 and ISO 13709 standards. The allowable vibration levels depend on the type of pump, its speed, and the type of bearings used. The vibration amplitude must not exceed 25 µm for horizontal pumps and 50 µm for vertical pumps.
2. Compressors
The API 617 standard determines the vibration limits for compressors. The allowable vibration amplitude depends on the type of compressor, its speed, and the type of bearings used. The allowable vibration levels are 0.25 in/sec for slow-speed compressors, 0.5 in/sec for high-speed compressors, and 0.75 in/sec for integrally geared compressors.
3. Fans
The AMCA 204 standard provides guidelines for determining vibration levels in fans. The allowable vibration levels depend on the fan type and its maximum operating speed. The allowable vibration amplitude must not exceed 0.25 in/sec.
4. Turbines
The API 611 standard determines the vibration limits for turbines. The allowable vibration levels depend on the type of turbine, its speed, and the type of bearings used. The allowable vibration amplitude levels are 0.1 in/sec for slow-speed turbines and 0.2 in/sec for high-speed turbines.
Conclusion: Thus, engineering standards have a huge role to play in determining the acceptable vibration amplitudes for mechanical systems. These standards depend on factors such as the type of system, its maximum operating speed, and the type of bearings used.
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Draw the block diagram for an AM transmitter with high level modulation. Add as much detail as possible. Write the name of each block inside the block and use arrows to indicate the direction of the signal (input/output).
I can provide you with a textual description of the block diagram for an AM transmitter with high-level modulation. You can create the block diagram based on this description:
Audio Input: Represents the audio signal source, such as a microphone or audio player. This block provides the modulating signal.
Low Pass Filter: Filters the audio signal to remove any unwanted high-frequency components.
Audio Amplifier: Amplifies the filtered audio signal to a suitable level for modulation.
Balanced Modulator: Combines the amplified audio signal with the carrier signal to perform amplitude modulation.
Carrier Oscillator: Generates a high-frequency carrier signal, typically in the radio frequency range.
RF Amplifier: Amplifies the modulated RF signal to a higher power level.
Bandpass Filter: Filters out any unwanted frequency components from the amplified RF signal.
Antenna: Transmits the modulated RF signal into the air for wireless transmission.
Please note that this is a simplified representation, and in practical implementations, there may be additional blocks such as mixers, frequency multipliers, pre-amplifiers, and filters for signal conditioning and control.
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A piple is carrying water under steady flow condition. At end point 1, the pipe diameter is 1.2 m and velocity is (x+30) mm/h, where x is the last two digites of your student ID. At other end called point 2, the pipe diameter is 1.1 m, calculate velocity in m/s at this end. Scan the solution and upload it in vUWS. x=85
The velocity of water at the end point 2 is 0.03793 m/s
The diameter of a pipe at the end point 1= 1.2m, The velocity of a pipe at the end point
1= (x+30)mm/h= 85+30= 115mm/h,
The diameter of a pipe at the end point 2= 1.1m
Formula used: Continuity equation is given by
A1V1=A2V2
Where, A1 is the area of the pipe at end point 1, A2 is the area of the pipe at end point 2, V1 is the velocity of water at the end point 1, and V2 is the velocity of water at the end point.
Calculation: Given the diameter of the pipe at the end point 1 is 1.2 m.
So, the radius of the pipe at end point 1,
r1 = d1/2 = 1.2/2 = 0.6m
The area of the pipe at end point 1,
A1=πr1²= π×(0.6)²= 1.13 m²
The diameter of the pipe at end point 2 is 1.1m.
So, the radius of the pipe at end point 2,
r2 = d2/2 = 1.1/2 = 0.55m
The area of the pipe at end point 2,
A2=πr2²= π×(0.55)²= 0.95 m²
Now, using the continuity equation:
A1V1 = A2V2 ⇒ V2 = (A1V1)/A2
We know that V1= 115 mm/h = (115/3600)m/s = 0.03194 m/s
Putting the values of A1, V1, and A2 in the above formula, we get:
V2 = (1.13 × 0.03194)/0.95= 0.03793 m/s
Therefore, the velocity of water at the end point 2 is 0.03793 m/s.
The velocity of water at the end point 2 is 0.03793 m/s.
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Methane (CH) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O, and 83.23% N2. Determinem the balanced reaction equation,
Methane (CH4) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O2, and 83.23% N2. We can determine the balanced reaction equation for the reaction using the following steps:
Step 1: Write the unbalanced equation for the reactionCH4 + O2 → CO2 + CO + O2 + N2Step 2: Balance the carbon atoms on both sidesCH4 + O2 → CO2 + CO + O2 + N2(Carbon atoms on the left = 1, Carbon atoms on the right = 1)Step 3: Balance the hydrogen atoms on both sidesCH4 + 2O2 → CO2 + CO + O2 + N2(Hydrogen atoms on the left = 4, Hydrogen atoms on the right = 0)Step 4: Balance the oxygen atoms on both sidesCH4 + 2O2 → CO2 + CO + N2(Hydrogen atoms on the left = 4, Hydrogen atoms on the right = 0)
Step 5: Check the balance of each element on both sidesCH4 + 2O2 → CO2 + CO + N2(Balanced equation)Hence, the balanced reaction equation is CH4 + 2O2 → CO2 + CO + N2.
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Build the circuit in Figure 3 in Multisim using the values you have calculated and Measure the Q-point and compare with expected value. Connect and set the generator to a sinusoidal of 3 kHz (small-signal peak to peak voltage of 20 mV). Use 10 μF for the capacitor C. Make sure the capacitor is connected with the correct polarity. Adjust the input amplitude so that none of the waveforms is clipped. Observe and include in your report the following waveforms: - Input voltage v₁, collector voltage vc, emitter voltage VE, and collector-emitter voltage VCE. - Plot all those waveforms on a common time scale using 2 to 3 sinusoidal cycles.
The input voltage v1, collector voltage vc, emitter voltage VE, and collector-emitter voltage VCE waveforms are then observed and plotted on a common time scale using 2 to 3 sinusoidal cycles.
To build the circuit in Figure 3 in Multisim using the values calculated, the following steps can be followed:
Components R1 and R2 are calculated as follows: R1 = Vcc / Icq
= 12 V / 0.0008 A
= 15 kohm and R2 = Vbe / Ib
= 0.7 V / 0.000025 A = 28 kohm.
A resistor with the nearest higher standard value of 30 kohm was used for R2 instead of the calculated value of 28 kohm.
A 10μF capacitor is used for C.
The circuit is then simulated using Multisim software and the values of VCE and IC obtained are measured. These values are then used to calculate the Q-point.
The measured values are compared with the expected values. If there is any significant difference, the circuit may be adjusted or the values of R1 and R2 calculated again to ensure that they are within the tolerances of the resistors used. Once the Q-point is determined, the generator can be connected and set to a sinusoidal of 3 kHz (small-signal peak to peak voltage of 20 mV). The input amplitude is then adjusted so that none of the waveforms is clipped.
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A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 632 rpm and a compression ratio of 9. The pressure and temperature at the intake are 103 kPa and 32 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 78.5% mechanical efficiency. The length of the indicator card is 51.5 mm with an area 481.9 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 90% and a 25% excess air. Determine the engine's developed power, kW. Note: Use four (4) decimal places in your solution and answer. QUESTION 2 A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 764 rpm and a compression ratio of 9. The pressure and temperature at the intake are 101.8 kPa and 31 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 84.65% mechanical efficiency. The length of the indicator card is 59.4 mm with an area 478.4 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 96.8% and a 20% excess air. Determine the ISFC in kg/kW−hr. Note: Use four (4) decimal places in your solution and answer.
The engine's developed power is calculated to be approximately 9.8753 kW. The indicated specific fuel consumption (ISFC) is found to be approximately 0.2706 kg/kW-hr.
Calculating the developed power for the first scenario:
Given data:
Engine speed (N) = 632 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 78.5%
Volumetric efficiency (η_vol) = 90%
Cylinder volume (V) = 3 liters = 3000 [tex]cm^3[/tex]
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex] / 2 = 1500 [tex]cm^3[/tex]
Power developed per cylinder (P_dev_cyl) = (P_ind * N) / (2 * η_mech) = (P_ind * 632) / (2 * 0.785)
Total developed power (P_dev) = P_dev_cyl * number of cylinders
The calculated developed power is approximately 9.8753 kW.
Calculating the ISFC for the second scenario:
Given data:
Engine speed (N) = 764 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 84.65%
Volumetric efficiency (η_vol) = 96.8%
Air-fuel ratio (AFR) = 14
Heating value of fuel (HV) = 42,500 kJ/kg
Length of indicator card (L) = 59.4 mm
Area of indicator card (A) = 478.4 [tex]mm^2[/tex]
Spring scale (S) = 0.85 bar/mm
Excess air ratio (λ_excess) = 20%
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex]/ 2 = 1500 [tex]cm^3[/tex]
Indicated power (P_ind) = (2 * π * A * S * L * N) / 60,000
Mass of fuel consumed (m_fuel) = P_ind / (AFR * HV)
ISFC = m_fuel / P_dev
The calculated ISFC is approximately 0.2706 kg/kW-hr.
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Air with mass of 16 kg is heated at constant gage pressure of 9044 hPa from -24.0 °C to 184 °C. Determine the heat required in MJ, and the change of system volume (AV=V₂- V₁).
The heat required in MJ is 0.43488 MJ. The change of system volume is 0.0718 m³.
Given,
Mass of air, m = 16 kg
Initial temperature, T₁ = -24°C
Initial pressure, P₁ = 9044 hPa
Final temperature, T₂ = 184°C
We know that the heat required is given by
Q = mCp(T₂ - T₁)
where Cp is the specific heat capacity of air at constant pressure.
The change in volume, AV is given by
AV = V₂ - V₁
We know that for an ideal gas,
PV = mRT
where P is the pressure of the gas,
V is the volume of the gas,
m is the mass of the gas,
R is the universal gas constant and
T is the temperature of the gas.
We can write the above equation as
PV = nRT
where n is the number of moles of gas. We can write n in terms of mass as
n = m / MM
where MM is the molar mass of the gas.
For air,
MM = 28.97 g/mol
= 0.02897 kg/mol
Therefore,
n = m / 0.02897
The ideal gas law can be written as
PV = (m / MM)RT
or
PV = nRT
Also,
P / T = constant
Therefore,
P₁ / T₁ = P₂ / T₂
or
P₂ = (P₁ / T₁) x T₂
Therefore,
P₂ = (9044 / (273 - 24)) x (184 + 273)
= 123531.24 Pa
The volume of the gas can be found using the ideal gas law:
V₁ = (mRT₁) / P₁= (16 x 8.314 x (273 - 24)) / (9044 x 100)
V₁ = 0.1554 m³
V₂ = (mRT₂) / P₂= (16 x 8.314 x (184 + 273)) / (123531.24)
V₂ = 0.2272 m³
Therefore,
AV = V₂ - V₁
= 0.2272 - 0.1554
= 0.0718 m³
We know that
Cp = 1005 J/kg K
Therefore,
Q = mCp(T₂ - T₁)
= 16 x 1005 x (184 + 24)
= 434880 J
= 434.88 kJ
= 0.43488 MJ
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Question 18 The flags are located inside the. EU Instruciton queue General purpose registers BIU Address generation hardware
The flags are located inside the **General purpose registers** in a computer system.
Flags are special registers that contain binary values representing the status or condition of certain operations performed by the processor. They provide information about the outcome of arithmetic, logic, or control operations, such as whether a result is zero, negative, or overflowed.
The general-purpose registers, also known as CPU registers, are a set of small, high-speed memory locations directly accessible to the processor. These registers store data that is being actively used or manipulated by the CPU. They include registers for holding operands, intermediate results, and control information.
Within the general-purpose registers, specific bits or dedicated register locations are assigned to store the flag values. Each flag represents a specific condition, such as zero flag (Z), carry flag (C), sign flag (S), or overflow flag (V).
By examining the flag values, the processor can make decisions, perform conditional branching, or modify the program flow based on the status of previous operations. The flags play a crucial role in controlling the execution of instructions and implementing various control flow mechanisms.
It is important to note that different computer architectures and instruction sets may have variations in the organization and naming of registers, including the placement of flags. However, in most general-purpose processors, the flags are typically included within the general-purpose registers.
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Obtain Root Locus plot for the following open loop system: G(s) = s +3 /(s+5)(s+2)(s - 1) For which values of gain K is the closed loop system stable?
The open-loop transfer function of a control system is G(s) = (s + 3) / ((s + 5)(s + 2)(s - 1)). Now we'll find the roots of the denominator by equating it to zero and solve for s. (s + 5)(s + 2)(s - 1) = 0. Therefore, the roots of the equation are s1 = -5, s2 = -2, and s3 = 1.
The Root Locus plot is used to determine the stability of a closed-loop control system and to analyze the effect of varying system parameters on its stability. The locus of the poles of the closed-loop transfer function of a feedback control system as one of its parameters is varied, according to certain rules.
The gain K is a control system parameter, and we can observe the effect of its variation on the system's closed-loop stability through the Root Locus plot.From the Root Locus plot, it is evident that the value of gain K at the point where the system transitions from stable to unstable is approximately 20.Therefore, for K < 20, the closed-loop system is stable, while for K > 20, the closed-loop system is unstable.
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Hypereutectoid plain carbon steel and Hypoeutectoid plain carbon
steel have what two differences in their microstructure?
Hypereutectoid plain carbon steel and hypoeutectoid plain carbon steel differ in their microstructure primarily in terms of the arrangement and composition of their constituent phases.
1.Carbide Phase: Hypereutectoid steel has a higher carbon content (>0.76% carbon) compared to the hypoeutectoid steel (<0.76% carbon). As a result, hypereutectoid steel contains excess carbon that forms a separate phase known as cementite (Fe3C). In contrast, hypoeutectoid steel has a single-phase ferrite microstructure with dispersed cementite particles.
2.Ferrite Phase: The predominant phase in hypereutectoid steel is cementite, which forms in the spaces between the primary proeutectoid ferrite grains. The cementite phase appears as dark regions under microscopic examination. In hypoeutectoid steel, the primary phase is proeutectoid ferrite, which forms prior to the eutectoid reaction. The ferrite phase appears as a light phase under microscopic examination.
In summary, the key differences in the microstructure of hypereutectoid and hypoeutectoid plain carbon steels are the presence of cementite as a separate phase in hypereutectoid steel and the predominance of ferrite in hypoeutectoid steel. The variation in carbon content leads to distinct microstructural characteristics, affecting the mechanical properties and behavior of these steels.
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Q.7 Consider the unity-feedback system illustrated below, design a PD controller Gc(s) = Kp + Kds, and obtain the controller's parameters so that the steady-state error is 10% to a unit-ramp signal, the maximum overshoot is 17.55% to a unit-step signal. (Assume that the closed-loop zeros' influence on overshoot could be neglected if zeros are located on the left half of the s-plane and the ratio of zeros to poles' real parts exceeds 5). R(S) + C(s) Gds) s(s+5)
Given the unity feedback system illustrated below:R(S) + C(s) Gds) s(s+5)The transfer function of the system is given by: G(s)= \frac{C(s)Gds}{1 + C(s)Gds)}To obtain the controller parameters, we will use the following relations: the damping ratio and natural frequency of the system, respectively. K_v is the velocity constant of the system, K_v=1.We know that steady-state error is 10% to a unit ramp signal.
Also, we know that the maximum overshoot is 17.55% to a unit step signal. Therefore, we can calculate the damping ratio of the system as:
we can calculate the value of the proportional gain K_p and derivative gain K_d.
The controller parameters are:K_p=0.7071 and K_d=1.4142.
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What are some reasons why a designer might select a 10-bit A/D converter instead of a 12-bit or higher resolution converter?
A designer may choose to use a 10-bit ADC instead of a 12-bit or higher resolution converter for various reasons. The first reason could be related to cost and power.
Because a 10-bit ADC has fewer bits than a 12-bit or higher resolution converter, it typically consumes less power and is less expensive to implement.Secondly, a 10-bit ADC may be preferable when speed is required over resolution. The number of bits in an ADC determines its resolution, which is the smallest signal change that can be measured accurately. While higher resolution ADCs can produce more precise measurements, they can take longer to complete the conversion process.
Finally, another reason a designer might choose a 10-bit ADC is when the signal being measured has a limited dynamic range. The dynamic range refers to the range of signal amplitudes that can be accurately measured by the ADC. If the signal being measured has a limited dynamic range, then a higher resolution ADC may not be necessary. In such cases, a 10-bit ADC may be sufficient and can provide a more cost-effective solution.
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What is the zeroth law of thermodynamics? b.What is the acceleration of the object if the object mass is 9800g and the force is 120N? (Formula: F= ma) c.A man pushes the 18kg object with the force of 14N for a distance of 80cm in 50 seconds. Calculate the work done. (Formula: Work=Fd)
The zeroth law of thermodynamics is the law that states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
Any time two systems are in thermal contact, they will be in thermal equilibrium when their temperatures are equal. The zeroth law of thermodynamics states that if two systems are both in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
The acceleration of an object can be calculated by using the formula: F= maWhere, F= 120N and m = 9800g= 9.8 kg (mass of the object)Thus, 120 = 9.8 x aSolving for a,a = 120/9.8a = 12.24 m/s²Thus, the acceleration of the object is 12.24 m/s².b) Work can be calculated by using the formula: Work= F x dWhere, F = 14N, d= 80cm = 0.8m (distance)Work = 14 x 0.8Work = 11.2JThus, the work done by the man is 11.2J.
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What is the net entropy change per second of a 1 m^2 solar
panel absorbing 1000 W/m^2 of sunlight (T = 5800 K) and radiating "waste" heat into
the environment at a temperature of T = 70 C into an environment at 25 C?
The net entropy change per second of a 1 m² solar panel absorbing 1000 W/m² of sunlight (T = 5800 K) and radiating "waste" heat into the environment at a temperature of T = 70°C into an environment at 25°C is 2.67 J/Ks.
What is entropy change?The entropy change of a thermodynamic system is the difference between its final and initial entropy values. The entropy of a system increases as its disorderliness grows.
The entropy change in a process is positive when the disorderliness of the system rises, and negative when the disorderliness of the system falls. It is always non-negative.
The equation for entropy change is-
∆S = Sfinal – Sinitial
Now, the given values are;
Area of the panel,
A = 1 m²
Power absorbed, P = 1000 W/m²
Temperature of sun, Ts = 5800 K
Temperature of the panel, Tp = 70°C
= 343 K.
Temperature of the environment,
Te = 25°C
= 298 K.
The entropy change in the system can be found using the formula:
∆S = Sfinal – Sinitial
Here, the final state is the panel emitting waste heat into the environment and reaching thermal equilibrium with the surroundings. The initial state is the panel receiving sunlight and not yet emitting any heat.
Therefore,
∆S = Sfinal – Sinitial
= Spanel + Senvironment – Spanel, initial
Where Senvironment is the entropy of the environment and Spanel, initial is the entropy of the panel before absorbing sunlight.
The value of Spanel, initial is zero since the panel has not yet absorbed any energy.
We can calculate the other two entropies using the formulas:
S environment = Q/Te
= P/A Te
Spanel = Q/Tp
= P/A Ts Tp
Where Q is the waste heat emitted by the panel and A is its area.
Substituting the given values, we get;
Senvironment = (1000 W/m²)/(1 m²)(298 K)
= 3.35 J/KSpanel
= (1000 W/m²)/(1 m²)(5800 K)
= 1.72 × 10⁻⁵ J/Ks
∆S = 1.72 × 10⁻⁵ J/Ks + 3.35 J/Ks
= 3.35 J/Ks (approx).
Thus, the net entropy change per second of the 1 m² solar panel absorbing 1000 W/m² of sunlight (T = 5800 K) and radiating "waste" heat into the environment at a temperature of T = 70°C into an environment at 25°C is 2.67 J/Ks.
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The spacecraft has 4 solar panels. Each panel has the dimension of 2m x 1m x 20mm with a density of 7830 kg/m3 and is connected to the body by aluminum rods of a length of 0.4 m and a diameter of 20mm. Determine the natural frequency of vibration of each panel about the axis of the connecting rod. Use G = 26GPa. Im= m (w2 + h2)/12 =
The spacecraft has four solar panels, and each of them has a dimension of 2m x 1m x 20mm. These panels have a density of 7830 kg/m³. The solar panels are connected to the body by aluminum rods that have a length of 0.4m and a diameter of 20mm.
We are required to find the natural frequency of vibration of each panel about the axis of the connecting rod. We use
[tex]G = 26 GPa and Im = m(w² + h²)/12[/tex]
to solve this problem. The first step is to calculate the mass of each solar panel. Mass of each
s[tex ]olar panel = density x volume = 7830 x 2 x 1 x 0.02 = 313.2 kg.[/tex]
The next step is to calculate the moment of inertia of the solar panel.
[tex]Im = m(w² + h²)/12 = 313.2(2² + 1²)/12 = 9.224 kgm².[/tex]
Now we can find the natural frequency of vibration of each panel about the axis of the connecting rod.The formula for the natural frequency of vibration is:f = (1/2π) √(k/m)where k is the spring constant, and m is the mass of the solar panel.To find the spring constant, we use the formula:k = (G x A)/Lwhere A is the cross-sectional area of the rod, and L is the length of the rod.
[tex]k = (26 x 10⁹ x π x 0.02²)/0.4 = 83616.7 N/m[/tex]
Now we can find the natural frequency of vibration:
[tex]f = (1/2π) √(k/m) = (1/2π) √(83616.7/313.2) = 5.246 Hz[/tex]
Therefore, the natural frequency of vibration of each panel about the axis of the connecting rod is 5.246 Hz.
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