Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of 60 ft/s. The ratio of the exit area to the inlet area is 8. Assume that The air is the ideal gas model for the air and ignoring heat transfer, determine
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
first step : calculate the value of h2 using the equation below
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
If an elevator repairer observes that cables begin to fray after 15 years, what process might he or she use to create a maintenance schedule for their replacement? fallacious reasoning reductive reasoning inductive reasoning deductive reasoning
Answer:
inductive reasoning
Explanation:
Inductive reasoning is one of the type of reasoning method in which generalized consequences are derived from limited observations. By observing few data, general conclusions are drawn. The conclusions drawn are false in inductive reasoning. In the given situation, the conclusion drawn by the elevator repairer has been drawn by inductive reasoning. His observation of some cables led him to draw the conclusion about all the cables. The result of the reasoning is false.
Which of the following would make a column more likely to buckle?
Increase its ultimate compressive strength
Increase its length
Increase its Young’s modulus
Increase its cross-sectional area
Increase its area moment of inertia
True or false: Increasing a material’s ultimate compressive strength makes it less likely to crush under its own weight (assuming all other things equal).
True
False
True or false: The area moment of inertia of an object only depends on its shape, not its size.
True
False
Answer:
1) B: Increase its length
2) True
3) True
Explanation:
1) Columns are compressive members and are subjected to primarily compressive stresses.
Now, there is what we call slender Ness ratio in columns which is basically used to check the ability of a column to resist buckling.
The formula is;
Slenderness ratio = Effective length of column/radius of gyration
Thus, the longer the column the more the Slenderness ratio and the more likely it is to buckle.
Thus, increasing the length is what makes columns likely buckle.
2) Compressive strength is the ability of a material to withstand loads that may reduce size or make the material crush under load.
Now, increasing the compressive strength simply means more ability to withstand loads that may lead to crushing under load.
Ultimate compressive strength is the maximum amount of compressive stress that a material can take before it crushes under load. Thus, increasing the ultimate compressive strength means it is less likely to crush under its own weight.
3) There are different factors that affect moment of inertia and they are;
- the mass of the body
- axis of rotation of the body
- shape and size of the body.
However, for area moment of inertia, what is most relevant to us is the shape of the body in question since we are dealing with area and not how big it is.
Elliptic curve cryptography is considered as the latest and probably the one with a future. Having seen RSA in earlier modules, in which ways do YOU think elliptic cryptography is more advanced than RSA. You may read other material or get this information from the internet to answer this question. But make sure to provide necessary references when you do cite others.
Answer:
The answer is below.
Explanation:
Some of the ways, how I think elliptic cryptography is more advanced than RSA are the following:
1. ECC - Elliptic Curve Cryptography uses smaller keys for the same level of security, particularly at greater levels of security.
2. ECC can work well and at a faster rate on a small-capacity device compared to RSA
3. It uses offer speedier SSL handshakes that enhance security
4. It offers fast signatures
5. It allows signatures to be computed in two stages, which enables lower latency than inverse throughput.
6. Relatively quick encryption and decryption
On the inner surface of the cylinder, the third principal stress is not zero. The third principal stress acts in the radial direction. Determine the third principal stress. Include the proper sign for your value.
Answer:
σ1, σ2 and σ3.
Explanation:
The three principal stresses are labelled or written as σ1, σ2 and σ3. The σ1 is the maximum principal stress or most tensile stress, σ2 is the intermediate principal stress and σ3 is the minimum or most compressive principal stress. Principal stresses are the maximum and minimum extensional stresses present in an object. The principal directions have no shear stresses connected with them.
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
122 cm
182 cm
82 cm
22 cm
108.2 cm
Answer:
The new length of the rod is 182 cm.
Explanation:
Given that a rod that was originally 100-cm-long experiences a strain of 82%, to determine what is the new length of the rod, the following calculation must be performed:
100 x 1.82 = X
182 = X
Therefore, the new length of the rod is 182 cm.
A piece of coaxial cable has 75ohms Characteristics impedance and a nominal capacitance of 69pF/m. what is its inductance per meter ? If the diameter of the inner conductor is 0.584mm, and the dielectric constant of the insulation is 2.23, what is the outer diameter of the conductor ?
Answer:
A) L = 0.388 μm
B) D = 3.78 mm
Explanation:
We are given;
Characteristics impedance; Z_o = 75 ohms
Nominal capacitance; C = 69pF/m = 69 × 10^(-12) F/m
dielectric constant; k = 2.23
Inner diameter of conductor; d = 0.584 mm
A) Now formula for Characteristics impedance is given as;
Z_o = √(L/C)
Where L is the inductance per metre.
Making L the subject, we have;
L = (Z_o)²C
L = 75² × 69 × 10^(-12)
L = 0.388 × 10^(-6) m
L = 0.388 μm
B) To get the outer diameter, we will use the formula;
Z_o = (138/√k) log(D/d)
Where;
D is outer diameter.
Thus, Plugging in the relevant values;
75 = (138/√2.23) × log (D/0.584)
log (D/0.584) = 0.81158611538
(D/0.584) = 10^(0.81158611538)
D/0.584 = 6.48016576435
D = 6.48016576435 × 0.584
D = 3.78 mm
An ideal gas within a piston-cylinder assembly undergoes a Carnot refrigeration cycle. The isothermal compression occurs at 325 K from 2 bar to 4 bar. The isothermal expansion occurs at 250 K. Determine:
a. the coefficient of performance
b. the heat transfer to the gas during the isothermal expansion, in kj per kmol of gas
c. the magintude of the net work input, in kj per kmol of gas.
Answer:
a) [tex]\mu=3.3[/tex]
b) [tex]Q=1440.7KJ/Kmol[/tex]
c) [tex]W=1872.9KJ/Kmol[/tex]
Explanation:
From the question we are told that:
Initial Temperature [tex]T_1=325k[/tex]
initial Pressure [tex]P_1=2 bar[/tex]
Final Pressure [tex]P_2=4 bar[/tex]
iso-thermal expansion [tex]T_2=250k[/tex]
a)
Generally the equation for Coefficient of performance is mathematically given by
[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]
[tex]\mu=\frac{250}{325-250}[/tex]
[tex]\mu=3.3[/tex]
b)
Generally the equation for Heat Expansion is mathematically given by
[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]
Where
R=Gas constant
[tex]R=8.314462618[/tex]
Therefore
[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]Q=1440.7KJ/Kmol[/tex]
c)
Generally the equation for work input is mathematically given by
[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]
[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]W=1872.9KJ/Kmol[/tex]
The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol
Given Data:
T1 = 325KP1 = 2 barP2 = 4 barT2 = 250KIsothermal expansion occurs at 250K.
a) The coefficient of performanceThis is calculated as
COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]
b) Heat Transfer in isothermal expansion[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]
Therefore; In isothermal process du = 0
R = 8.314 AkJ/K.mol
Q = 8.314 * 250 In(4/2)
Q = 1440.71kJ/K.mol
c) Work InputW[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]
The work input is 1872.92kJ/K.mol
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A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion
and gear are 20 and 50. The module is 5 mm, the pressure angle is 20o
and the face width is 45 mm. The
rotational speed of pinion is 300 rev/min. Both the pinion and the gear material are Nitralloy 135 Grade2 with a hardness of 277 Brinell. The quality standard number Qv is 5 and installation is open gearing
quality. Find the AGMA bending and contact stresses and the corresponding factors of safety for a
pinion life of 109
cycles and a reliability of 0.98
Answer:
mark me as a brainleast
Explanation:
209781
Steam enters a nozzle operating at steady state at 20 bar, 2808C, with a velocity of 80 m/s. The exit pressure and temperature are 7 bar and 1808C, respectively. The mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energy, determine (a) the exit velocity, in m/s. (b) the inlet and exit flow areas, in cm2
Answer:
A) 600.8 m/s
B) (i) 22.5 cm^2 (ii) 7.11 cm^2
Explanation:
Given data :
P1 = 20 bar , T1 = 2808°C
P2 = 7 bar , T2 = 1808°C
mass flow rate = 1.5 kg/s
Using the superheated vapor region in Table A-4
h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg
h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg
A) calculate exit velocity ( m/s )
given that we are to neglect heat transfer and potential energy
V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )
= [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )
= 600.8 m/s
B) calculate the inlet and exit flow areas ( cm^2 )
i) Inlet flow area
A1 = ( m * v1 ) / V1
= ( 1.5 * 0.1200 )/ 80 = 22.5 cm^2
ii) exit flow area
A2 = ( m * v2 ) / V2
= ( 1.5 * 0.2847 ) / 600.8 = 7.11 cm^2
Given : x² + 200x = 166400 The current park is a square, and the addition will increase the width by 200 meters to give the expanded park a total area of 166,400 square meters To Find : the side length of the current square park. Solution: x² + 200x = 166400 => x(x + 200) = 166400 166400 = 320 * 520 => (320)(320 + 200) = 166400 => x = 320 side length of the current square park. = 320 m Learn More: Which expression is a possible leading term for the polynomial ... brainly.In/question/13233517
Answer:
320 m
Explanation:
To find the side length of the current park, x, we solve the quadratic equation for the area of the park
x² + 200x = 166400
x² + 200x - 166400 = 0
We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x
So, we re-write the expression as
x² + 200x - 166400 = 0
x² + 520x - 320x - 166400 = 0
We write out the factors of the expression,
x² + 520x - 320x - 320 × 520 = 0
Factorizing the expression, we have
x(x + 520) - 320(x + 520) = 0
(x + 520)(x - 320) = 0
x + 520 = 0 or x - 320 = 0
x = -520 or x = 320
Since x is not negative, we take the positive answer.
So, x = 320 m
zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available supply voltage of 15 V can varyas much as ±10%. For this diode, what is the value of VZ0?For a nominal load resistance RL of 1 k and a nominal zenercurrent of 10 mA,what current must flow in the supply resistorR? For the nominal value of supply voltage, select a valuefor resistor R, specified to one significant digit, to provideat least that current. What nominal output voltage results?For a ±10% change in the supply voltage, what variationin output voltage results? If the load current is reduced by50%, what increase in VO results? What is the smallest valueof load resistance that can be tolerated while maintainingregulation when the supply voltage is low? What is the lowestpossible output voltage that results? Calculate values for theline regulation and for the load regulation for this circuit usingthe numerical results obtained in this problem.
Answer:
[tex]V_z=9.1v[/tex]
[tex]V_{zo}=8.74V[/tex]
[tex]I=10mA[/tex]
[tex]R=589 ohms[/tex]
Explanation:
From the question we are told that:
Zener diode Voltage [tex]V_z=9.1-V[/tex]
Zener diode Current [tex]I_z=9 .A[/tex]
Note
[tex]rz = 40\\\\IZK= 0.5 mA[/tex]
Supply Voltage [tex]V_s=15[/tex]
Reduction Percentage [tex]P_r= 50 \%[/tex]
Generally the equation for Kirchhoff's Voltage Law is mathematically given by
[tex]V_z=V_{zo}+I_zr_z[/tex]
[tex]9.1=V_{z0}+9*10^{-3}(40)[/tex]
[tex]V_{zo}=8.74V[/tex]
Therefore
[tex]At I_z-10mA[/tex]
[tex]V_z=V_{z0}+I_zr_z[/tex]
[tex]V_z=8.74+(10*10^{-3}) (40)[/tex]
[tex]V_z=9.1v[/tex]
Generally the equation for Kirchhoff's Current Law is mathematically given by
[tex]-I+I_z+I_l=0[/tex]
[tex]I=10mA+\frac{V_z}{R_l}[/tex]
[tex]I=10mA+\frac{9.1}{0}[/tex]
[tex]I=10mA[/tex]
Therefore
[tex]R=\frac{15V-V_z}{I}[/tex]
[tex]R=\frac{15-9.1}{10*10^{-3}}[/tex]
[tex]R=589 ohms[/tex]
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more ________________ than bypass and absorb the energy of the transient.
Answer:
I think ( MOV Metal oxide varistors )
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more MOV Metal oxide varistors than bypass and absorb the energy of the transient.
In heavy traffic areas you should wave pedestrians across the street if there is no crosswalk
In heavy traffic areas, you should wave pedestrians across the street if there is no crosswalk: False.
What is a crosswalk?A crosswalk can be defined as the marked or specially paved part of a road that is characterized by heavy traffic, so as to enable pedestrians have right of way to cross the street because drivers are required by traffic law to stop for them.
However, a driver or other road users in heavy traffic areas shouldn't wave pedestrians across the street if there is no crosswalk
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Please help me:
Use the Node analysis to find the power of all resistors
The main water line into a tall building has a pressure of600kPaat 5mbelow ground level. Apump brings the pressure up so the water can be delivered at 200kPaat the top floor 150maboveground level. The volumetric flow rate is 0.01m3/s. Consider the liquid water as an incompressiblesubstance. Assume steady state operation, negligible changes in kinetic energy, and the water temperature remains constant. Determine the required pumpingpower [kW, hp].
Answer:
Required pump power = - 11.205 kW
Explanation:
Pi = 600 kPa
Zi ( initial height ) = - 5m
P2 = 200 kPa
Ze ( top floor above ground level ) = 150m
Flowrate = 0.01 m^3/s = 10 kg/s
Temperature = 10°C ( assumed value ) remains constant
calculate the value of pumping power required ( i.e. work done )
Applying the energy equation
Hi + 1/2(vi)^2 + gZi = He + 1/2(Ve)^2 + gZe + W
given negligible internal/kinetic energy difference in "h's = the pv terms
W = ( Pi Vi - Pe V2 ) + g( Zi - Z2 )
= ( (600 * 0.001) - (200 * 0.001) ) + (9.81/1000) ( - 5 - 150 )
= - 1.1205 Kg/kJ
required pump power = flowrate * -1.1205
= 10 * ( - 1.1205 ) = - 11.205 kW
a video inspection snake is use
Answer:
very good thx
Explanation:
Apart from the type of emergency, what factors affect the decision on weather to evacuate or shelter in place?
Answer:
. 1. Type of building
2. Location of emergency
3. Extent of emergency
Explanation:
1. The kind of building in which people find themselves could be a factor that can be used to make this decision. a lot of buildings can be easily affected by disasters such as explosions or tornadoes the extent of the effect is dependent on how the building is constructed. in some situations it is better to shelter in, while in others it is best to evacuate.
2. Another factor to be considered is the location or area where this is happening it is good to consider this so that people can be safely moved given that help can be easily accessed or if best to stay in.
3. The last is the extent of what is happening. The risk involved is one way of making the choice to evacuate or to stay.
The factors that affect the decision on whether to evacuate or shelter in place include:
Type of building.Location of emergency.Extent of emergencyIt should be noted that the building where an individual lives play a vital role during emergencies. Buildings that have poor foundations can easily be affected during emergencies.
Another factor that should be considered is the location where the emergency is taking place. Lastly, the extent of the emergency can determine if the person should stay or not.
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Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
True
False
Answer:
True ❤️
-Solid by solid can make Cylindrical wire doubles Strengths in tension
For an applied transformer with a primary winding of 220, a secondary winding of 113 turns, a core of 45 cm2 cross-section and a rated supply voltage of 220 V, frequency f = 50 Hz, number system Core style K = 1,2.
one. Calculated from the sensor in the core transformer.
NS. Calculate the rated capacity of the transformer application.
C. Calculating no-load voltage and current levels.
NS. If the power point is reduced by 7% from the voltage level; voltage scortation feature.
e. If use the core machine variable on random form; powerpress same as the host variable on; compare the capacity of the random transformer with the capacity of the transformer application.
Explanation:
https://www.chegg.com/homework-help/questions-and-answers/iron-core-wish-design-transformer-part-dc-power-supply-moderately-high-current-rating-volt-q53186459?trackid=83dbc34cec2e&strackid=3efc9f324415
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?
Answer:
[tex]V_{pp}=2V[/tex]
Explanation:
Source Voltage [tex]V= 120V[/tex]
Frequency [tex]f=60Hz[/tex]
Peak output voltage [tex]Vp=15V[/tex]
Peak Output Voltage with filter [tex]V_p'=14V[/tex]
Generally the equation for Peak to peak voltage is mathematically given by
[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]
Therefore
[tex]V_{pp}=2(V_p-v_p')[/tex]
[tex]V_{pp}=2(15-14)[/tex]
[tex]V_{pp}=2V[/tex]
Determine the resistance values for a voltage divider that must meet the following specifications:_______.
a. The current drawn from the source under unloaded condition is not to exceed 5 mA.
b. The sourcevoltage is to be 10 V, and the required outputs are to be 5 V and 2.5 V.
c. Sketch the circuit.
d. Deter-mine the effect on the output voltages if a load is connected to each tap one at a time.
Answer:
i) when circuit is unloaded : R1 + R2 = 2kΩ.
ii) when 5V output voltage is applied : R1 = 1 kΩ , R2 = 1 kΩ
iii) when 2.5 v output voltage is applied : R1 = 1500 Ω, R2 = 500 Ω
iv) when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
Explanation:
Current drawn from source under loaded condition ≤ 5 mA
source voltage = 10 v , required output = 5 v , 2.5 v
attached below is the sketch of the circuit
Resistance values
i) when the circuit is unloaded
Req = R1 + R2 = 2 kΩ ( Req = Vs / I = 10 / 5*10^-3 = 2 kΩ )
ii) when output voltage = 5 v
we will apply voltage divider rule
R1 = 1 kΩ ,
R2 = 1 kΩ
iii) When the output voltage = 2.5 v
applying voltage divider rule
R1 = 1500 Ω
R2 = 500 Ω
iv) when the load is connected to each tap one at a time
i.e. when the resistance are in parallel
when: R1 = 1 kΩ , R2 = 1 kΩ is connected in parallel output voltage < 5 V
When : R1 = 1500 Ω, R2 = 500 Ω is connected in parallel output voltage > 2.5V
attached below is the detailed solution to the given problem
A steam turbine takes in steam at a temperature of 400 Celsius and releases steam to the condenser at a temperature of 120 Celsius. If the turbine takes 500 kJ of heat in each cycle, what is the maximum amount of work that could be generated by the turbine in each cycle
Answer:
[tex]W= 208 KJ[/tex]
Explanation:
From the question we are told that:
Take in Temperature [tex]T_1=400C=>673K[/tex]
Take Out Temperature [tex]T_2=120+>393K[/tex]
Heat [tex]Q=500kJ[/tex]
Generally the equation for Carnot Engine Efficiency is mathematically given by
[tex]n_c=\frac{T_1-T_2}{T_1}[/tex]
[tex]n_c=\frac{673-393}{673}[/tex]
[tex]n_c=0.4[/tex]
Where
[tex]n_c=\frac{W}{Q}[/tex]
[tex]W = Q*n_c[/tex]
[tex]W= 500 * 0.4[/tex]
[tex]W= 208 KJ[/tex]
A force measuring instrument comes with a certificate of calibration that identifies two instrument errors and assigns each an uncertainty at 95% confidence over its range. Provide an estimate of the instrument design-stage uncertainty.
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Answer:
[tex]U=\pm 0.382N[/tex]
Explanation:
From the question we are told that:
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Generally the equation for Stage Uncertainty is mathematically given by
[tex]U=\sqrt{u_0^2+u_T^2}[/tex]
Where
[tex]u_0=Zero\ order\ uncertainty[/tex]
[tex]u_0=\pm 0.5*0.25[/tex]
[tex]u_0=\pm=0.125[/tex]
And
u_T=Total instrumental Uncertainty
[tex]u_T=\sqrt{l_e^2+h_e^2}[/tex]
Where
l_e=Error of linearity
h_e=Error due to hysteresis
Hence
[tex]u_T=\sqrt{0.20^2+0.30^2}[/tex]
[tex]u_T=\pm 0.36[/tex]
Therefore
[tex]U=\sqrt{(0.125)^2+0.36^2}[/tex]
[tex]U=\pm 0.382N[/tex]
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.20 kg; the mass of the liquid is 1.85 kg. Balance D reads 3.10 kg and balance E reads 7.50 kg. The volume of block A is 4.15 × 10−3 m3.
a) What is the density of the liquid?
b) What will the balance D read if block A is pulled up out of the liquid?
c) What will the balance E read if block A is pulled up out of the liquid?
Answer:
a) [tex]m_e= 3.05 Kg[/tex]
b) [tex]\rho=1072.3kg/m^3[/tex]
c) [tex]m_e= 3.05 Kg[/tex]
Explanation:
From the question we are told that:
Beaker Mass [tex]m_b=1.20[/tex]
Liquid Mass [tex]m_l=1.85[/tex]
Balance D:
Mass [tex]m_d=3.10[/tex]
Balance E:
Mass [tex]m_e=7.50[/tex]
Volume [tex]v=4.15*10^{-3}m^3[/tex]
a)
Generally the equation for Liquid's density is mathematically given by
[tex]m_e=m_b+m_l+(\rho*v)[/tex]
[tex]\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}[/tex]
[tex]\rho=1072.3kg/m^3[/tex]
b)
Generally the equation for D's Reading at A pulled is mathematically given by
m_d = mass of block - mass of liquid displaced
[tex]m_d=m- (\rho *v )[/tex]
[tex]m=3.10+ (1072.30 *4.15*10^{-3}m^3 )[/tex]
[tex]m=18.10kg[/tex]
c)
Generally the equation for E's Reading at A pulled is mathematically given by
[tex]m_e=m_b+m_l[/tex]
[tex]m_e = 1.20 + 1.85[/tex]
[tex]m_e= 3.05 Kg[/tex]
Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance heater with 84.5% efficiency supplies heat to the pot. How many minutes will it take to boil 6.03 kg of water?
Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)
[tex]t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}[/tex]
t = 6179.1 s = 102.9 min = 1.7 h
A moving-coil instrument, which gives full-scale deflection with 0.015 A has a copper coil having resistance of 1.5 Ohm at 15°C and a temperature coefficient of 1/234.5 at 0 degree C in series with a swamp resistance 3.5 Ohm having a negligible temperature coefficient. Determine the resistance of shunt required for a full-scale deflection of 20 A and the resistant required for a full-scale deflection of 250 V. If the instrument reads correctly at 15°C, determine the percentage error in each case when the temperature is 25°C.
Answer: check answers in pictures (2 p)
The resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.
What is resistance?Resistance is a measure of an electrical circuit's resistance to current flow. Resistance is measured in ohms, which is represented by the Greek letter omega.
Full-scale deflection current = 0.015 A
Full-scale deflection current for 20 A = 20/0.015 = 1333.33 times full-scale deflection current
The current through the meter coil is given by:
Ic = Im * (Rm / (Rm + Rs))
At full scale, Ic = 1333.33 * 0.015 A = 20 A, Rm = 1.5 Ohm, and Rs = 3.5 Ohm.
Therefore:
20 A = Im * (1.5 Ohm / (1.5 Ohm + 3.5 Ohm))
Im = 20 A * (3.5 Ohm / 5 Ohm)
Im = 14 A
The current through the shunt resistor is therefore:
Ish = Im - Ic
Ish = 14 A - 20 A
Ish = -6 A
Since the shunt resistor is in parallel with the meter, its resistance can be calculated using the following formula:
Rs = Rm * (Im / Ish - 1)
Substituting the values, we get:
3.5 Ohm = 1.5 Ohm * (14 A / (-6 A) - 1)
3.5 Ohm = 1.5 Ohm * (-1.333 - 1)
3.5 Ohm = 1.5 Ohm * (-2.333)
Rs = 1.5 Ohm * (14 A / (-6 A) - 1) / (-2.333)
Rs = 2.528 Ohm
Therefore, the resistance of the shunt required for a full-scale deflection of 20 A is 2.528 Ohm.
Now, let's find the resistance required for a full-scale deflection of 250 V:
Full-scale deflection current = 0.015 A
Resistance required for full-scale deflection of 250 V = 250 V / 0.015 A
Resistance required for full-scale deflection of 250 V = 16666.67 Ohm
This resistance is in parallel with the meter coil, so the total resistance in the circuit will be:
Rtotal = Rm || Rext
Rtotal = (Rm * Rext) / (Rm + Rext)
Substituting the values, we get:
Rtotal = (1.5 Ohm * 16666.67 Ohm) / (1.5 Ohm + 16666.67 Ohm)
Rtotal = 1.43 Ohm
This resistance is in series with the meter coil, so the total resistance in the circuit will be:
Rtotal = Rm + Rext + Rs
Substituting the values we get:
1.43 Ohm = 1.5 Ohm + 16666.67 Ohm + Rs
Rs = 1.43 Ohm - 1.5 Ohm - 16666.67 Ohm
Rs = -16666.74 Ohm
Thus, the calculated value of Rs is negative, it means that the shunt resistor must be used instead of a series resistor to achieve a full-scale deflection of 250 V.
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A investor will invest in mutual fundwith a probability of 0.6, will invest in government fundwith a probability of 0.3, and will invest in both fundswith a probability of 0.15. Find the probability that the investor will invest in either mutual fundor government fund.
Answer:
0.75
Explanation:
From this question above we have the following information
A = probability of investment in mutual fund= 0.6
B = probability of investment in government fund = 0.3
C = probability of investing in both the mutual fund and the government= 0.15
Where to find the probability of this investor investing in either of these two
= Prob(a) + prob(b) - prob(c)
= 0.6 + 0.3 - 0.15
= 0.9 - 0.15
= 0.75
Explain ROLAP, and list the reasons you would recommend its use in the relational database environment.
Answer:
ROLAP is a branch of OLAP that is used to contain Relational database ( RDB ). which is a very fast database ( quick process of queries )
Very fast to access and also fast in processing queries provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesExplanation:
ROLAP ( Relational On-line Analytical processing ) is a branch of OLAP that is used to contain Relational database ( RDB ).
Advantages of ROLAP ( reasons for the use of ROLAP )
Very fast to access ( fast in processing queries )provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesthe integral of In3x²/×⁵
If the integral is
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx[/tex]
substitute u = ln(3x ²) and du = 6x/(3x ²) dx = 2/x dx.
Then x ² = exp(u)/3 and x ⁴ = exp(2u)/9.
The integral is transformed to
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = \int \frac{\ln(3x^2)}{2x^4} \times \dfrac2x \,\mathrm dx \\\\ = \int \frac{u}{2\times\dfrac{e^{2u}}9}\,\mathrm du \\\\ = \frac92 \int ue^{-2u}\,\mathrm du[/tex]
Integrate by parts:
[tex]f = u \implies \mathrm df = \mathrm du \\\\ \mathrm dg = e^{-2u}\,\mathrm du \implies g = -\dfrac12 e^{-2u}[/tex]
[tex]\displaystyle \int ue^{-2u}\,\mathrm du = fg - \int g\,\mathrm df \\\\ = -\dfrac12 ue^{-2u} + \displaystyle \frac12 \int e^{-2u}\,\mathrm du \\\\ = -\frac12 ue^{-2u} - \frac14 e^{-2u} + C[/tex]
Then
[tex]\displaystyle \frac92 \int ue^{-2u}\,\mathrm du = -\frac94 ue^{-2u} - \frac98 e^{-2u} + C[/tex]
which in terms of x would be
[tex]\displaystyle \int \frac{\ln(3x^2)}{x^5}\,\mathrm dx = -\frac94\times\frac{\ln(3x^2)}{9x^4} - \frac98 \times \frac1{9x^4} + C \\\\ = \boxed{-\frac{\ln(3x^2)}{4x^4}-\frac1{8x^4}+C}[/tex]
