a) the pre-exponential factor and activation energy for the hydrolysis of t-butyl chloride are 2.1xE16 sec-1 and 102 kJ mol-1, respectively. calculate the values of delta S and delta H at 286 K for the reaction.
b)the pre-exponential factor and activation energy for the gas-phase cycloaddition of maleic anyhyrdride and cycopentadiene are 5.9xE7 M-1S-1 and 51 kJ mol-1 respectively. calculate values of delta S and delta H at 293 K for the reaction.
Answers
Answer:
Delta S = 356.64J/mol and Delta H = 99620J/mol
Explanation:
The detailed steps and appropriate use of the arrehenius equation and necessary susbtitution were made as shown in the attached file.
Related Questions
Which specific gravity is generally used for calculation of the volume occupied by the aggregate in Portland cement concrete, and why?
Answers
Answer: BULK RELATIVE DENSITY.
WHY?
BULK RELATIVE DENSITY GIVES A BETTER UNDERSTANDING OF THE QUALITY OF THE MATERIAL.
Explanation:Bulk relative density is a type of Specific gravity which is often used in determining the volume occupied by the aggregates in various mixtures containing aggregate including Portland cement concrete, bituminous concrete etc this mixtures are proportioned based on an absolute volume basis. Bulk relative density is considered because of its ability to give a better understanding of the materials which makes up the mixture.
An air-standard Otto cycle has a compression ratio of 6 and the temperature and pressure at the beginning of the compression process are 520 deg R and 14.2 lbf/in^2, respectively. The heat addition per unit mass of air is 600 Btu/lb. Determine (a) the maximum temperature, in deg R.
Answers
Answer:
The maximum temperature of the cycle is 1065⁰R
Explanation:
The maximum temperature in degree Rankin can be obtained using the formula below;
[tex]\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}[/tex]
Where;
T₂ is the maximum temperature of the cycle
T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R
V₁/V₂ is the compression ratio = 6
[tex]T_2 = T_1(\frac{V_1}{V_2})^{0.4}[/tex]
[tex]T_2=T_1(6)^{0.4}[/tex]
[tex]T_2=520^0R(6)^{0.4}[/tex]
T₂ = 1064.96 ⁰R
Therefore, the maximum temperature of the cycle is 1065⁰R
Fictional Corp is looking at solutions for their new CRM system for the sales department. The IT staff already has a fairly heavy workload, but they do not want to hire any additional IT staff. In order to reduce the maintenance burden of the new system, which of the following types of CRM should they choose to meet these needs?
a. IaaS
b. PaaS
c. SaaS
d. DBaaS
Answers
Answer:
SaaS
Explanation:
Software as a service (SaaS) is also called software on demand, it involves a third party that centrally hosts the software and provides it to the end user.
All aspects of hosting is handled by the third party: application, data, runtime, middleware, operating system, server, virtualization, storage and networking are all handled by the provider.
This is an ideal software service for Fictional corp, as there will be no need to hire additional IT staff to maintain the new CRM software.
Do you think the mining process is faster when you know in advance that the land must be restored? Explain.
Answers
Answer: No, the mining process isn't faster when you know in advance that the land must be restored.
Explanation:
The mining process would be slower when mining companies know in advance that the land must be restored because they have to be more careful about their impact on the environment & avoid taking unnecessary damage to ruin the land which they know they must restore.
So, it's evident how this would slow down the mining process. If there were no obligations to restore land, the mining companies would just come at the mining space and run their mining process fast without big concerns for the consequences.
But the caution definitely slows them down.
People often have different points of view. If I think mining process is faster when you know in advance that the land must be restored, My answer is No.
This is because even without miners been aware that the land had to be restored before mining, they still carelessly remove minerals without taking into cognizant the damage it will do to the land.
There are different processes of mining. They include;
UndergroundSurfacePlacer In-situThe type of mining method used is usually based on the kind of resource that is needed, the deposit's location etc.
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A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of the curve is 35 degrees and the PI is at station 482 + 72. What is the station of the PTand how many feet have to be cleared from the lane's shoulder edge to provide adequate stopping sight distance?
Answers
Answer:
The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.
Explanation:
From table 3.5 of Traffic Engineering by Mannering
R_v=835
R=835+(10ft/2)= 840 ft.
Now T is given as
T=R tan(Δ/2)
Here Δ is the central angle of curve given as 35°
So
T=R tan(Δ/2)
T=840 x tan(35/2)
T=840 x tan(17.5)
T=264.85
Now
STA PC=482+72-(2+64.85)=480+07.15
Also L is given as
L=(π/180)RΔ
Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.
L=(π/180)RΔ
L=(π/180)x840 x35
L=512.87 ft
STA PT=480+07.15+5+12.87=485+20.02
Now Ms is the minimum distance which is given as
[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\[/tex]
Here R_v is given as 835
SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering
So Ms is
[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft[/tex]
Now for the clearance from the inside lane
Ms=Ms-lane length
Ms=26.92-5= 21.92 ft.
So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.
All brake lights are dimmer than normal. Technician A says that bad bulbs could be the cause. Technician B says that high resistance in the brake switch could be the cause. Which technician is correct?a. Technician A onlyb. Technician B onlyc. Both Technician A and Bd. Neither Technician A nor B
Answers
Answer:
All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.
Explanation:
According to Technician A
When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .
According to Technician B
When a high resistance inserted in series circuit the voltage across each resistance is reduced and this cause the light glow dimly.
Formula of resistance in series circuit
Rt=r1+r2+r3......
P1.30 shows a gas contained in a vertical piston– cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of thepiston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2
Answers
In the process of analyzing a thermodynamic system it is important to identify what system is being worked on and the processes and properties if the system
The magnitude of the force acting on the shaft, is approximately 1,336.5 N
The reason the value for the force magnitude acting on the shaft is correct is as follows:
The known parameters are:
The cross-sectional area of the shaft, Aₐ = 0.8 cm²
The required gas pressure in the cylinder, P = 3 bar
The mass of the piston, m₁ = 24.5 kg
The mass of the shaft, m₂ = 0.5 kg
The diameter of the piston, D = 10 cm
The atmospheric pressure, Pₐ = 1 bar
Required:
The magnitude of the force F acting on the shaft
Solution:
The force due to the gas in the cylinder, [tex]\mathbf{F_{gas}}[/tex], is given as follows;
[tex]F_{gas}[/tex] = 3 bar × π × (10 cm)²/4 = 2,359.19449 N
The force due to the atmosphere, [tex]\mathbf{F_{atm}}[/tex], is given as follows;
[tex]F_{atm}[/tex] = 1 bar × ((π × (10 cm)²/4) - 0.8 cm²) ≈ 777.4 N
The force due to the piston and shaft, [tex]\mathbf{F_{ps}}[/tex], is given as follows;
[tex]F_{ps}[/tex] = (24.5 kg + 0.5 kg) × 9.81 m/s² = 245.25 N
The magnitude of the force acting on the shaft, F = [tex]F_{gas}[/tex] - ([tex]\mathbf{F_{atm}}[/tex] + [tex]\mathbf{F_{ps}}[/tex])
∴ F = 2,359.19449 N - (777.4 N + 245.25 N) ≈ 1,336.5449 N
The magnitude of the force acting on the shaft, F ≈ 1,336.5 N
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Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan starts in front of toll booth A, goes through toll booth B, and ends after passing toll booth C. Let dAB and dBC be the distance between A-B, and B-C.
a. Suppose dAB = dBc = 10 km. What is the end-to-end delay if the caravan travels together (i.e., the first car must wait for the last car after passing each toll booth)?
b. Repeat a), but assume the cars travel separately (i.e., not waiting for each other).
c. Repeat a) and b), but suppose dAB = dBC =100 km
d. Still suppose dAB = dBC = 100 km. Suppose toll booth B takes 10 minute to pass each car (A and C still takes 1 minute per car). Where is the first car when the second car passes B?
e. Under the assumption of d), what is the maximum value of dBC such that the first car has passed C when the second car passes B?
Answers
Answer:
A. 36 minutes
B. 120 minutes
C.
i. 144 minutes
ii. 984 minutes
D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.
E. 98.33km
Explanation
A.
Given
dAb = 10km
dBc = 10km
Propagation Speed = 100km/hr
Delay time = 1 minute
Numbers of cars = 8
Number of tolls = 3
Total End to End delay = Propagation delay + Transition delay
Calculating Propagation Delay
Propagation delay = Total Distance/Propagation speed
Total distance = 10km + 10km = 20km
So, Propagation delay = 20km/100km/hr
Propagation delay = 0.2 hour
Translation delay = delay time* numbers of tolls * numbers of cars
Transitional delay = 1 * 3 * 8
Transitional delay = 24 minutes
Total End delay = 24 minutes + 0.2 hours
= 24 minutes + 0.2 * 60 minutes
= 24 minutes + 12 minutes
= 36 minutes
B.
Total End to End delay = Propagation delay + Transition delay
Calculating Propagation Delay
Propagation delay = Total Distance/Propagation speed
Total distance = 10km + 10km = 20km
So, Propagation delay = 20km/100km/hr
Propagation delay = 0.2 hour
Translation delay = delay time* numbers of tolls ------ Cars traveling separately
Transitional delay = 1 * 3
Transitional delay = 3 minutes
Total End delay for one car = 3 minutes + 0.2 hours
= 3 minutes + 0.2 * 60 minutes
= 3 minutes + 12 minutes
= 15 minutes
Total End delay for 8 cars = 8 * 15 = 120 minutes
C.
Given
dAb = 100km
dBc = 100km
Propagation Speed = 100km/hr
Delay time = 1 minute
Numbers of cars = 8
Number of tolls = 3
i. Cars travelling together
Total End to End delay = Propagation delay + Transition delay
Calculating Propagation Delay
Propagation delay = Total Distance/Propagation speed
Total distance = 100km + 100km = 200km
So, Propagation delay = 200km/100km/hr
Propagation delay = 2 hours
Translation delay = delay time* numbers of tolls * numbers of cars
Transitional delay = 1 * 3 * 8
Transitional delay = 24 minutes
Total End delay = 24 minutes + 2 hours
= 24 minutes + 2 * 60 minutes
= 24 minutes + 120 minutes
= 144 minutes
ii. Cars travelling separately
Total End to End delay = Propagation delay + Transition delay
Calculating Propagation Delay
Propagation delay = Total Distance/Propagation speed
Total distance = 100km + 100km = 200km
So, Propagation delay = 200km/100km/hr
Propagation delay = 2 hours
Translation delay = delay time* numbers of tolls ------ Cars traveling separately
Transitional delay = 1 * 3
Transitional delay = 3 minutes
Total End delay for one car = 3 minutes + 2 hours
= 3 minutes + 2 * 60 minutes
= 3 minutes + 120 minutes
= 123 minutes
Total End delay for 8 cars = 8 * 123 = 984 minutes
D.
Distance = 100km
Time = 1 min/car
Car 1 is 1 minute ahead of car 2 --- at toll A and B
If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes
Time delay = 11 - 10 = 1 minute
Distance = time * speed
= 1 minute * 100km/hr
= 1 hr/60 * 100 km/hr
= 100/60
= 1.67km
E.
Given
Distance = 100km
Distance behind = 1.67
Maximum value of dBc = 100km - 1.67km = 98.33km
The maximum distance that can be reached is 98.33km
What properties should the head of a carpenter’s hammer possess? How would you manufacture a hammer head?
Answers
Properties of Carpenter's hammer possess
Explanation:
1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.
2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.
3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush the rock and shape the metal.It is not suitable for heavy work.
How hammer head is manufactured :
1.Hammer head is produced by metal forging process.
2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.
3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.
4.The third process is repeated for several times.
A flow is described by velocity field V=ai+bxj where a = 2 (m/s) and b = 1 (1/s) , coordinates are measured in meters. a) Obtain the equation for the streamline passing through the point (2,5). b) At t=2s , what are the coordinates of the particle that passed through point (0,4) at t=0. c) At t=3s , what are the coordinates of the particle that passed through point (1,4.25) 2 seconds earlier.
Answers
Answer:
Explanation:
The concept of differential in rate of change is applied to solve the problem.
First is to compare the equation given V=ai+bxj , where u = a and v = bx.
the detailed steps and appropriate integration and substitution is as shown in the attached file.
The streamline equation and the coordinates of the point are
A) [tex]x^2-4y+16[/tex]
B) (4, 12)
C) (3, 11.25)
Given that, [tex]V=ai+bxj[/tex], [tex]a=\frac{2m}{s}[/tex] and [tex]b=1s^{-1}[/tex]
Putting the value, we get
[tex]V=2i+1.xj[/tex]
From this equation, we get
[tex]u=2\frac{m}{s}[/tex] and [tex]v=x\frac{m}{s}[/tex]
Obtain the equation for the streamline passing through the point (2,5), streamline equation
[tex]\frac{dx}{u}=\frac{dy}{v}[/tex]
Putting the value, we get
[tex]\frac{dx}{2}=\frac{dy}{x}[/tex]
rearrange it and do integration:
[tex]\int xdx=\int 2dy[/tex]
[tex]\frac{x^2}{2}=2y+c -------(i)[/tex]
C is the integration constant.
[tex]\frac{2^2}{2}=(2\times5)+C[/tex]
[tex]C=-8[/tex]
Put the value of C in equation (i), we get
[tex]\frac{x^2}{2}=2y-8[/tex]
[tex]y=\frac{x^2}{4}+4[/tex]
[tex]x^2-4y+16=0[/tex]
b) [tex]u=2\frac{m}{s}[/tex]
[tex]\frac{dx}{dt}=2[/tex]
Rearrange and integrate
[tex]\int dx=\int 2dt[/tex]
[tex]x=2t+C_1[/tex]
Put x=0 and t=0
[tex]0=0+C_1[/tex]
[tex]C_1=0[/tex]
Put the value of constant we get
[tex]x=2t[/tex]
at [tex]t=2s[/tex]
[tex]x=2\times2= 4 --------(2)[/tex]
Now, [tex]v=x \frac{m}{s}[/tex]
[tex]\frac{dy}{dt}=x[/tex]
Rearrange and integrate:
[tex]\int dy=\int xdt[/tex]
[tex]y=xt+C_2[/tex]
Putting the value of x, y and t we get the value of constant.
[tex]4(0\times0)+C_2[/tex]
[tex]C_2=4[/tex]
Putting the value of constant we get
[tex]y=xt+4[/tex]
Put the value of [tex]t=2[/tex] and [tex]x=4[/tex] we get
[tex]y=(4\times2)+4[/tex]
[tex]= 8+4[/tex]
[tex]y= 12 -------(3)[/tex]
From (2) and (3)
we get the coordinates of the particle that passed through point [tex](0,4)[/tex] at [tex]t=0[/tex].
After [tex]t=2s[/tex] as [tex](4, 12)[/tex]
c) first we have to find the integration constant through initial condition then put the value of time to get the coordinates.
From above we have:
(1) [tex]x=2t+C_1[/tex] and
(2) [tex]y=xt+C_2[/tex]
Given [tex]t=2s[/tex] and [tex]P(1, 4.25)[/tex]
Putting the value in first equation, we get
[tex]1=(2\times2)+C_1[/tex]
[tex]C_1 =-3[/tex]
Put the value of time [tex]t=3s[/tex] we get
[tex]x=(2\times3)-3=3 ---------(a)[/tex]
Now, putting the value in second equation we get
[tex]4.25=(1\times2)+C_2[/tex]
[tex]C_2 =2.25[/tex]
Then [tex]y=xt+2.25[/tex]
Put the value of time [tex]t=3s[/tex] we get
[tex]y=(3\times3)+2.25=11.25 -------(b)[/tex]
From (a) and (b) we get the coordinate of the point after [tex]t=3s[/tex] is [tex](3, 11.25)[/tex].
Therefore, streamline equation and the coordinates of the point are
A) [tex]x^2-4y+16[/tex]
B) (4, 12)
C) (3, 11.25)
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Define the following terms in your own words: (a) elastic strain, (b) plastic strain, (c) creep strain, (d) tensile viscosity, (e) recovery, and (f) relaxation.
Answers
Answer:
Detailed Answer is given below,
Explanation:
(a) Elastic strain
Elastic deformation is defined as the limit for the deformation values up to which the object will bounce and return to the original shape when the load is removed.
When the object is subjected to an external load, it deforms. If the load exceeds the load corresponding to the elastic limit of the object, then, after removing the load, the object cannot return to its original geometric specification.
(b) Plastic strain
Plastic Strain is a deformation that cannot be recovered after eliminating the deforming force. In other words, if the applied tension is greater than the elastic limit of the material, there will be a permanent deformation.
(c) Creep strain
The deformation of the material at a higher temperature is much more than at the normal temperature known as creep.
Subsequently, the deformation produced in heated material is called creep deformation. Creep deformation is a function of temperature.
(d) Tensile viscosity
Viscosity is a main parameter when measuring flux fluids, such as liquids, semi-solids, gases and even solids. Brookfield deals with liquids and semi-solids. Viscosity measurements are made together with the quality and efficiency of the product. Any person involved in the characterization of the flow, in research or development, quality control or transfer of fluids, at one time or another is involved with some kind of viscosity measurement.
(e) Recovery
Recovery is a process whereby deformed grains can reduce their stored energy by eliminating or reorganizing defects in their crystalline structure. These defects, mainly dislocations, are introduced by plastic deformation of the material and act to increase the elastic limit of a material.
(f) Relaxation
Stress relaxation occurs in polymers when they remain tense for long periods of time.
These alloys have very good resistance to stress relaxation and, therefore, are used as spring materials.
Stress relaxation is a gradual reduction of stress over time in constant tension.
A constant torque of 5 Nm is applied to an unloaded motor at rest at time t ¼ 0. The motor reaches a speed of 1800 rpm in 3 s. Assuming the damping to be negligible, calculate the motor inertia.
Answers
Answer:
The motor inertia is 7.958 X 10⁻² kg.m²
Explanation:
To determine the motor inertia, the following formula applies.
Neglecting the damping effect,
[tex]T = \frac{J}{9.55}.\frac{\delta n}{\delta T}[/tex]
Where;
T is the constant torque applied to the motor = 5Nm
J is the motor inertia = ?
δn is the change in angular speed of the motor = 1800 r/min
δT is change in time of the unloaded motor from rest = 3 sec
[tex]J = \frac{9.55* \delta T* T}{\delta n}[/tex]
[tex]J = \frac{9.55* 3* 5}{1800}[/tex] = 0.07958 kg.m² = 7.958 X 10⁻² kg.m²
Therefore, the motor inertia is 7.958 X 10⁻² kg.m²
A firm has 62 employees. During the year, there are seven first-aid cases, three medicaltreatment injuries, an accident in which an injured employee was required to work 1 week in restricted work activity, a work-related illness in which the employee lost 1 week of work, a work-related illness in which the employee lost 6 weeks of work, and a fatality resulting from an electrocution. Calculate the total incidence rate, the number-of-lostworkdays rate, and the LWDI.
Answers
Answer:
The answers to the question are
11.2967.746.45Explanation:
The number of employees in the firm = 62
Number of first-aid cases = 7
Number of medical treatment injuries = 3
Injury resulting restricted work activity = 1
Illness resulting in one week loss work day = 1
Illness resulting in loss of 6 weeks of work =1
Incident resulting in fatality = 1
Total incidence rateTotal hours worked = 40×62×50 = 124000 Hrs
Where 200,000 is the number of hours worked by Full time employees numbering 100 that work for 40 Hours a week for 50 weeks in a year
Number of recordable incident = Those incident that results in lost work days, death, restricted ability to work, or transfer to another task or more severe injury treatment beyond first aid
Therefore number recordabe incident = 7
Recordable Incident Rate is calculated by
IR = (Number of recordable incident Cases X 200,000)÷(Number of Employee labor hours worked)
= (7×200000)/124000= 11.29
The number-of-lostworkdays rateThe Lost Workday Rate is given by
Lost Workday Rate = (Total number of days lost to injury or illness)÷( Cumulative number of hours employees worked) × 200000
Total number of days lost to injury or illness = 42 days
Lost Workday Rate = (42/124000) × 200000 = 67.74
LWDI. Lost Work Day InjuryLWDI = (Number of incident resulting in lost workdays and restricted activity) ×200000 /(Total number of hours worked by all employees in one year)
LWDI = LWD×200000/ EH = 4×200000/124000 = 6.45
LWDI = 6.45
A motor keep a Ferris wheel (with moment of inertia 6.8 × 107 kg · m 2 ) rotating at 12 rev/hr. When the motor is turned off, the wheel slows down (because of friction) to 9.6 rev/hr in 17 s. What was the power of the motor that kept the wheel rotating at 12 rev/hr despite friction? Answer in units of W.
Answers
Answer:
Power of the motor that kept the wheel rotating at 12 rev/hr despite friction is 342.79W.
Explanation:
Pls refer to the attached file. The explanation is long to pen down here.
An op-amp is connected in an inverting configuration with R1 = 1kW and R2 = 10kW, and a load resistor connected at the output, RL = 1kW.
a. Find the values of v1, i1, i2, vO, iL, and iO.
b. Determine the voltage gain (vO/vI), current gain (iL/iI), and power gain (PO/PI).
Answers
Answer:
View Image
Explanation:
You didn't provide me a picture of the opamp.
I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...
no current will go in the inverting(-) and noninverting(+) side of the opampV₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting sideSince no current is going into the + and - side of the opamp, then
i₁ = i₂
Since V₊ is connected to ground (0V) then V₋ must also be 0V.
V₊ = V₋ = 0
Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.
You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at its rated conditions. Calculate the per-phase current that has to be carried by the cables to the power electronics converter and the step-up transformer located at the base of the tower.
Answers
To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are
[tex]V = 690V[/tex]
[tex]P_{real} = 2.3MW[/tex]
Real power in 3 phase
[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]
Now the Phase Voltage is,
[tex]V_{ph} = \frac{V}{\sqrt{3}}[/tex]
[tex]V_{ph} = \frac{690}{\sqrt{3}}[/tex]
[tex]V_{ph} = 398.37V[/tex]
The current phase would be,
[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]
Rearranging,
[tex]I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}[/tex]
Replacing,
[tex]I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}[/tex]
[tex]I_{ph}= 2.26kA/phase[/tex]
Therefore the current per phase is 2.26kA
Consider the following incomplete code segment, which is intended to print the sum of the digits in num. For example, when num is 12345, the code segment should print 15, which represents the sum 1 + 2 + 3 + 4 + 5.
int num = 12345;
int sum = 0;
/* missing loop header */
{
sum += num % 10;
num /= 10;
}
System.out.println(sum);
Which of the following should replace /* missing loop header */ so that the code segment will work as intended?
while (num > 0)
A
while (num >= 0)
B
while (num > 1)
C
while (num > 2)
D
while (num > sum)
E
Answers
Answer:
A) while (num >= 0)
Explanation:
To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.
A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.
B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0
C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result
D) The same problem than C
E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result
Loops are program statements that are used to carry out repetitive and iterative operations
The missing loop header is (a) while (num > 0)
To calculate the sum of the digits of variable num, the following must be set to be true
The loop header must be set to keep repeating the loop operations as long as the value of variable num is more than 0
To achieve this, we make use of while loop,
And the loop condition (as described above) would be num > 0
Hence, the true option is (a) while (num > 0)
Read more about loops at:
https://brainly.com/question/19344465
What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?
Answers
Answer:
a.The current is cut in half
b.the current stays the same
c.the current doubles
d.the current becomes zero
Explanation:
The current doubles
Determine the following for a south facing surface at 30� slope in
Gainesville, FL (Latitude = 29.68�N, Longitude = 82.27�W) on September 21 at
noon solar time (Assuming a ground reflectivity of 0.2):
A. Zenith Angle
B. Angle of Incidence
C. Beam Radiation
D. Diffuse Radiation
E. Reflected Radiation
F. Total Radiation
G. Local Time (note Gainesville has daylight savings on Sep 21)
Answers
Answer:
z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m², Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM
Explanation:
A. Zenith Angle:
As we know that,
Zenith angle=z=90⁰-α=L(latitude)=29.68⁰
Another way to do it is to find α first,
At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as
sinα=cosL
α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°
B. Angle of incidence:
angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°
i=cos⁻¹(sin92.32°)=0.32°
C. Beam Radiation:
First we need to calculate extra terrestrial radiations
Iext.=1353[1+0.034cos(360n/365)]
where n=264
=1345 W/m²
Now,
Beam Radiation=CIext⁻ⁿ
where n=0.1/sin60.32°
Beam Radiation = 1097.2 W/m²
D. Diffude Radiation:
difuse radiation = Id = 0.0921ₙcos²(β/2)
where β=30°
Id = 94.2 W/m²
E. Reflected Radiations:
Ir=pIn(sinα+0.092)sin²(β/2)
= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)
= 14.1W/m²
F. Total Radiation:
total radiation = beam radiation + diffuse radiation + reflected raddiation
= 1205.4 W/m²
G. Local Time:
LST= ST-ET-(lₓ-l(local))4min/₀
= 12:00-7.9min-(75°-82.27°)4min/₀
=12:21PM
Local time
LDT=LST+=12:21+1:00=1:21PM
The two pond system is fed by a stream with flow rate 1.0 MGD (million gallons per day) and BOD (nonconservative pollutant) concentration of 20 mg/L. The rate of decay of BOD is 0.3/day. The volume of the first pond is 5 million gallons and the second is 3 million. Assuming complete mixing within the pond, find the BOD concentration leaving each pond.
Answers
Answer: First pond= 8.0 mg/L
second pond = 4.2 mg/L
Explanation:
Cn/Co = [ 1/ [ 1 + (k*t/n)]]
for the first pond
where Co into the first pond is 20mg/L,
Cn = 20*[ 1/ [ 1+ ((1 x 5)/0.3)]]
Cn = 8 mg/L
into the second pond we calculate the BOD leaving the pond of volume 3million liters
we have Cn = 4.2 mg/L
a. For a 200g load acting vertically downwards at point B’, determine the axial load in members A’B’, B’C’, B’D’, C’D’ and C’E’.
b. Repeat this for the load hung at C’, D’ and F’
Answers
Answer:
attached below
Explanation:
An 80-percent-efficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a page 101rate of 1.5 ft3/s through a constant-diameter pipe. The free surface of the pool is 80 ft above that of the lake. Determine the mechanical power used to overcome frictional effects in piping. Answer: 2.37 hp
Answers
To calculate the mechanical power used to overcome friction in piping for the described pump operation, calculate the hydraulic power needed, then subtract the pump's output power (based on its efficiency) from its total input.
Explanation:The question involves determining the mechanical power used to overcome frictional effects in piping for an 80-percent-efficient pump with a power input of 20 hp, which is pumping water at a rate of 1.5 ft3/s from a lake to a pool 80 ft above. Firstly, calculate the total hydraulic power required to move the water to the desired height, then calculate the power output of the pump based on its efficiency. Finally, subtract this value from the total input power to find the power used to overcome friction.
Steps to Solve:Calculate the hydraulic power needed using the formula P = ρghQ, where ρ is the density of water, g is the acceleration due to gravity, h is the height difference, and Q is the flow rate.Calculate the efficient power output of the pump (Actual Power Output) using its efficiency and the input power.Subtract the pump's output power from its input to find the power lost to friction.Through such calculations, one can find that the mechanical power used to overcome frictional effects in piping is 2.37 hp.
explain the four functional blocks on an oscilloscope and describe the major controls within each block
Answers
Answer:
The cathode ray oscilloscope (CRO) consists of a set of blocks. Those are vertical amplifier, delay line, trip circuit, time base generator, horizontal amplifier, cathode ray tube (CRT) and power supply. The CRO block diagram is shown in attached figure.
The function of each CRO block is mentioned below,
Vertical amplifier amplifies the input signal, which will be displayed on the CRT screen.
Delay line provides a certain amount of delay to the signal, which is obtained at the output of the vertical amplifier. This delayed signal is then applied to the CRT vertical deflection plates.
Trigger circuit produces a trigger signal to synchronize the horizontal and vertical deviations of the electron beam.
Time base generator produces a sawtooth signal, which is useful for horizontal deviation of the electron beam.
Horizontal amplifier amplifies the sawtooth signal and then connects it to the CRT horizontal deflection plates.
Power supply produces high and low voltages. The high negative voltage and the low positive voltage apply to CRT and other circuits respectively.
Cathode ray tube (CRT)
it is the main important block of CRO and consists mainly of four parts. Those are electronic guns, vertical deflection plates, horizontal deflection plates and fluorescent display.
The electron beam, which is produced by an electron gun, is deflected both vertically and horizontally by a pair of vertical deflection plates and a pair of horizontal deflection plates, respectively. Finally, the deflected beam will appear as a point on the fluorescent screen.
In this way, CRO will display the input signal applied on the CRT screen. So, we can analyze the signals in the time domain using CRO.
Explanation:
The oscilloscopes which is widely used for analysis purpose of circuits is divided into four main groups: the horizontal and vertical controls, the input controls and the activation controls.
Found in the front panel section marked Horizontal, the oscilloscope's horizontal controls allow users to adjust the horizontal scale of the screen. This section includes the control of the horizontal delay (displacement), as well as the control that indicates the time per division on the x-axis. The first control allows users to scan through a time range, while the latter allows users to approach a particular time range by decreasing the time per division.
Meanwhile, the oscilloscope's vertical controls are usually found in a section specifically marked as Vertical. The controls found in this section allow users to adjust the vertical appearance of the screen and include the control that indicates the number of volts per division on the axis and the grid of the screen. Also in this section is the control of the vertical displacement of the waveform, which translates the waveform up or down on the screen.
Signal activation helps provide a usable and stable display and allows users to synchronize the oscilloscope acquisition in the waveform of interest. The oscilloscope trigger controls allow users to choose the vertical trigger level, as well as the desired trigger capability. Common types of activation include fault activation, edge activation and pulse width activation.
Useful for identifying random errors or failures, the activation of faults allows users to fire at a pulse or event whose width is less than or greater than a specific period of time. This activation mode allows users to capture errors or technical problems that do not occur very frequently, which makes them very difficult to see.
The most famous trigger mode, edge tripping occurs when the voltage exceeds a set threshold value. This mode allows users to choose between shooting on a falling or rising edge.
Although pulse width activation is comparable to fault activation when users search for pulse width, it is, however, more general since it allows users to fire pulses of specified width. Users can also select the polarity of the pulses to be activated and set the horizontal position of the trigger. This allows users to see what happened during pre-shot or post-shot.
The input panels of an oscilloscope usually include two or four analog channels. They are usually numbered and have a button associated with each channel that allows users to activate and deactivate them. This section may also include a selection that allows users to specify the DC or AC coupling. Selecting the DC coupling implies that the entire signal will be input. The AC pairing, on the other hand, blocks the DC component and focuses the waveform around zero volts. Operators can also identify the probe impedance of the channels through a selection button. In adding, the input panels permit users to select the type of sampling to be used.
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 power factor leading, are supplied by a balanced, three-phase, 480-volt source. (a) Draw the power triangle for each load and for the combined load. (b) Determine the power factor of the combined load and state whether lagging or leading. (c) Determine the magnitude of the line current from the source. (d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity? Give your answer in Ω. (e) Compute the magnitude of the current in each capacitor and the line current from the source.
Answers
Answer:
(a) attached below
(b) [tex]pf_{C}=0.85[/tex] [tex]lagging[/tex]
(c) [tex]I_{C} =32.37 A[/tex]
(d) [tex]X_{C} =49.37[/tex] Ω
(e) [tex]I_{cap} =9.72 A[/tex] and [tex]I_{line} =27.66 A[/tex]
Explanation:
Given data:
[tex]P_{1}=15 kW[/tex]
[tex]S_{2} =10 kVA[/tex]
[tex]pf_{1} =0.6[/tex] [tex]lagging[/tex]
[tex]pf_{2}=0.8[/tex] [tex]leading[/tex]
[tex]V=480[/tex] [tex]Volts[/tex]
(a) Draw the power triangle for each load and for the combined load.
[tex]\alpha_{1}=cos^{-1} (0.6)=53.13[/tex]°
[tex]\alpha_{2}=cos^{-1} (0.8)=36.86[/tex]°
[tex]S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA[/tex]
[tex]Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99[/tex] ≅ [tex]20kVAR[/tex]
[tex]P_{2} =S_{2}*pf_{2} =10*0.8=8 kW[/tex]
[tex]Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99[/tex] ≅ [tex]-6 kVAR[/tex]
The negative sign means that the load 2 is providing reactive power rather than consuming
Then the combined load will be
[tex]P_{c} =P_{1} +P_{2} =15+8=23 kW[/tex]
[tex]Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR[/tex]
(b) Determine the power factor of the combined load and state whether lagging or leading.
[tex]S_{c} =P_{c} +jQ_{c} =23+14j[/tex]
or in the polar form
[tex]S_{c} =26.92<31.32[/tex]°
[tex]pf_{C}=cos(31.32) =0.85[/tex] [tex]lagging[/tex]
The relationship between Apparent power S and Current I is
[tex]S=VI^{*}[/tex]
Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.
(c) Determine the magnitude of the line current from the source.
Current of the combined load can be found by
[tex]I_{C} =S_{C}/\sqrt{3}*V[/tex]
[tex]I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A[/tex]
(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω
[tex]Q_{C} =3*V^2/X_{C}[/tex]
[tex]X_{C} =3*V^2/Q_{C}[/tex]
[tex]X_{C} =3*(480)^2/14*10^3[/tex] Ω
(e) Compute the magnitude of the current in each capacitor and the line current from the source.
Current flowing in the capacitor is
[tex]I_{cap} =V/X_{C} =480/49.37=9.72 A[/tex]
Line current flowing from the source is
[tex]I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A[/tex]
The electric potential difference between the ground and a cloud in a particular thunderstorm is 3.0 ✕ 109 V. What is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?
Answers
Answer:
The question is incomplete, below is the complete question "The electric potential difference between the ground and a cloud in a particular thunderstorm is [tex]3.0*10^{9}V[/tex]. What is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?"
Answer:
[tex]U=3.0*10^{9}eV[/tex]
Explanation:
data given,
Potential difference,V=3.0*10^9V
charge on an electron, q=1e.
Recall that the relationship between potential difference (v), charge(Q) and the potential energy(U) is expressed as
[tex]U=qV[/tex]
from the question, we asked to determine potential energy given the charge and the potential difference.
Hence if we substitute values into the equation, we arrive at
[tex]U=qV\\U=3.0*10^{9}*1e\\U=3.0*10^{9}eV[/tex]
Hence the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud is [tex]3.0*10^{9}eV[/tex]
The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way concrete joist framing system. d) an unreinforced floor and roof framing system
Answers
Answer:
a) the two-way concrete joist framing system
Explanation:
A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.
You are designing a three-story office building (Occupancy B) with 20,000 square feet per floor. What types of construction will you be permitted to use under the IBC if you do not install sprinklers?
Answers
Answer:
not provide the sprinklers, then the type of construction will be Type II B under IBC
Explanation:
given data
3 story office building = 20,000 square feet per floor
solution
we know when sprinkler is provide in high rise building to resist fire
and it provide in building as floor area exceed allowable permissible area of building as IBC
so IBC for Type II B allowable area = 19000 square feet per floor
and type III B allowable area = 23000 square feet per floor
so when we design the building by type III B construction, the sprinklers require to provide
but not provide the sprinklers, then the type of construction will be Type II B under IBC
The annual average insolation (energy of sunlight per unit area) striking a fixed solar panel in Buffalo, New York, is 200 W*m^−2, while in Phoenix, Arizona, it is 270 W*m^−2. In each location, the solar panel converts 15% of the incident energy into electricity. Average annual electricity use in Buffalo is 6000 kW*h at an average cost of $0.15 kW*h, while in Phoenix it is 11,000 kW*h at a cost of $0.09 kW*h.
1. In each city, what area of solar panel is needed to meet the average electrical needs of a residence?
Answers
Answer:
The area of solar panel needed in Buffalo is 22.83 m²
The area of solar panel needed in Phoenix is 31 m²
Explanation:
FOR BUFFALO:
First we, calculate the the annual power requirement by dividing the energy by the time of 1 year.
Therefore,
Power Requirement = P = 6000 KWhr/(1 yr)(365 days/yr)(24 hr/day)
P = 684.93 W
Now, to calculate the area we use the formula:
Area = A = P/(Annual Insolation)(Conversion Factor)
A = 684.93 W/(200 W/m²)(0.15)
A = 22.83 m²
FOR PHOENIX:
First we, calculate the the annual power requirement by dividing the energy by the time of 1 year.
Therefore,
Power Requirement = P = 11000 KWhr/(1 yr)(365 days/yr)(24 hr/day)
P = 1255.7 W
Now, to calculate the area we use the formula:
Area = A = P/(Annual Insolation)(Conversion Factor)
A = 1255.7 W/(270 W/m²)(0.15)
A = 31 m²
A composite plane wall consists of a Li = 125 mm thick layer of insulation (ki = 0.05 W/m.K) and a Ls = 25 mm thick layer of siding (ks = 0.10 W/m.K). The inner temperature of the insulation is 20 C. The outer temperature of the siding is -10 C. Steady state conditions apply.
Determine the temperature at the interface between the two layers, in C, and the rate of heat transfer through the wall, in W per m2 of surface area.
Answers
Answer:
Interface temperature = -7.27C
Rate of heat transfer = 10.91W/m2
Explanation:
The steps and the application of fourier's law of heat conduction is as shown in the attached file.
For the given composite wall, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².
Explanation:To determine the temperature at the interface between the two layers, we can use the formula for 1-D steady-state heat conduction:
q = (T2 - T1) / ((L1 / k1) + (L2 / k2))
where q is the heat transfer rate per unit area, T1 and T2 are the temperatures of the inner and outer surfaces, L1 and L2 are the thicknesses of the insulation and siding layers, and k1 and k2 are the thermal conductivities of the insulation and siding layers, respectively.
Plugging in the given values:
q = (-10 - 20) / ((125 / 0.05) + (25 / 0.10)) = -30 / (2500 + 250) = -30 / 2750 = -0.0109 W/m².
The negative sign indicates heat transfer from the siding to the insulation.
To find the temperature at the interface, we can use the formula:
T_interface = T1 + (q * (L1 / k1))
Plugging in the values:
T_interface = 20 + (-0.0109 * (125 / 0.05)) = 20 - 2.725 = 17.275 °C.
Therefore, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².
When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Complete the sentences to best explain the difference.
Answers
Answer:
Due to differences in the nature of the adhesive force and cohesive forces in the molecules of the individual substances and the glass tube.
Explanation:
To understand why this is so, a deep understanding of adhesive and cohesive force is required.
First, what is adhesive force?
Note: Adhesive and cohesive forces are best discussed under macroscopic level.
Adhesive force is the Intermolecular forces that exist between atoms of different molecules e g the forces explain when unlike charges stick together.
Cohesive force on the other hand is the Intermolecular force that exist between atoms of the same molecules e.g the force between hydrogen bonding.
Hence to explain the case scenario above, the adhesive force between the water molecules and the glass molecules is higher than the cohesive force between the water molecules. Hence the high rise.
For the case of the Hexane, the cohesive forces between the molecules hexane is far greater then the adhesive force between the glass molecules and the hexane molecules.