Answer:
Printing press , lenses etc,
Explanation:
Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.
Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.
where can I find solved problems of advanced soil structure interaction?
The seismic response of nuclear power plant structures is often calculated using lumped parameter methods. A finite element model of the structure is coupled to the soil with a spring-dashpot system used to represent the interaction process. The parameters of the interaction model are based on analytic solutions to simple problems which are idealizations of the actual problems of interest. The objective of the work reported in this paper is to compare predicted responses using the standard lumped parameter models with experimental data. These comparisons are shown to be good for a fairly uniform soil system and for loadings that do not result in nonlinear interaction effects such as liftoff. 7 references, 7 figures.
A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a true average compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. State the relevant hypotheses, and describe the type I and type II errors in the context of this problem.
Answer:
Null Hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.
Type I error (false positive) - this occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.
Type II error (false-negative)- his occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2 in actual scenario but the measurements show an incorrect reading of True average compressive strength less than 1300 KN/m2.
Explanation:
The hypothesis for this piece of information is as follows -
Null Hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.
Alternate hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is less than or equal to 1300 KN/m2
Type I error (false positive) - this occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.
Type II error (false-negative)- his occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2 in actual scenario but the measurements show an incorrect reading of True average compressive strength less than 1300 KN/m2.
identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
Answer:
L = Henry
C = Farad
Explanation:
The electrical parameter represented as L is the inductance whose unit is Henry(H).
The electrical parameter represented as C is the inductance whose unit is Farad
Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :
To obtain :
At resonance, Inductive reactance = capacitive reactance
Equate the inductive and capacitive reactance
Inductive reactance(Xl) = 2πFL
Capacitive Reactance(Xc) = 1/2πFC
Inductive reactance(Xl) = Capacitive Reactance(Xc)
2πFL = 1/2πFC
Multiplying both sides by F
F * 2πFL = F * 1/2πFC
2πF²L = 1/2πC
Isolating F²
F² = 1/2πC2πL
F² = 1/4π²LC
Take the square root of both sides to make F the subject
F = √1 / √4π²LC
F = 1 /2π√LC
Hence, the proof.
How do guest room hotel smoke alarms work and differ then regular home versions?
Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room
Steam enters a turbine with a pressure of 30 bar, a temperature of 400 oC, and a velocity of 160 m/s. Saturated vapor at 100 oC exits with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 350 K. Determine the rate at which entropy is produced within the turbine per kg of steam flo
Answer:
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
Explanation:
By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:
Principle of Mass Conservation
[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)
First Law of Thermodynamics
[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)
Second Law of Thermodynamics
[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)
By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:
[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)
[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)
Where:
[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.
[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.
[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.
[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.
[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.
[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.
[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.
[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.
[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.
By property charts for steam, we get the following information:
Inlet
[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
Outlet
[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]
If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:
[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]
[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]
[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]
[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]
[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]
The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.
A resistor, inductor, and capacitor are in parallel in a circuit where the frequency of operation can vary. The R, L, and C values are such that at the frequency omega subscript 0, the magnitude of all the impedances are equal to each other. If the frequency of operation approaches zero, which element will dominate in determining the equivalent impedance of this parallel combination?
a. The inductor.
b. The capacitor.
c. The resistor.
d. Insufficient information provided.
Answer:
Option A is correct
Explanation:
As we know
Inductive Susceptance = ½(pi)*f*L
Or Inductive Susceptance is inversely proportional to the frequency
Likewise conductive Susceptance = 2 (pi)*f*C
Conductive Susceptance is directly proportional to the frequency
When the frequency will reach the value zero, then the Inductive Susceptance will become infinite
Hence, inductor will dominate in determining the equivalent impedance of this parallel combination
Option A