Answer:
Information technology is important in our lives because it helps to deal with every day's dynamic things. Technology offers various tools to boost development and to exchange information. Both these things are the objective of IT to make tasks easier and to solve many problems.
what are the main subsystem of GSM
network
Explanation:
Network and Switching Subsystem (NSS)
Base-Station Subsystem (BSS)
Mobile station (MS)
Operation and Support Subsystem (OSS)
Trình bày sự khác nhau của Dây chuyền đẳng nhịp đồng nhất, dây chuyền đẳng nhịp không đồng nhất, cho ví dụ minh họa
Explain the 11 sections that a typical bill of quantity is divided into
Answer:
The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.
The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of hydrogen that pass per hour (in kg/h) through a 4.0-mm thick sheet of palladium having an area of 0.25 m^2 at 500°C. Assume a diffusion coefficient of 6.0 x 10^-8 m^2/s, that the concentrations at the high- and low-pressure sides of the plate are 3.5 and 0.25 kg/m^3 (kilogram of hydrogen per cubic meter of palladium), and that steady-state conditions have been attained. (clearly show your solution step by step, pay attention to units otherwise you will lose points!)
Side milling cutter is an example of ______ milling cutter.
Answer:
special type
Explanation:
As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.
anxiety: a. is never normal. b. is common of many psychological disorders c. is identical to fear d. is a modern development, unlikely to have roots in human history
Answer:
B
Explanation:
Anxiety is very common especially nowadays but it's especially common in psychological disorders
beacuse thye want them to hav egoood and thye wn thme tto
Answer:
I don't understand the question
Problem 89:A given load is driven by a 480 V six-pole 150 hp three-phase synchronous motor with the following load and motor data. Determine the voltage E necessary for this operating condition. Note: assume that the rotational loss torque is negligible. Load: Tൌ0.05∗????ଶ????mMotor: Eൌ400 V; Xௗൌ1ΩAnswer: Eതൌ400∠-17.36° V
Answer:
[tex]E_f=400<-17.4volts[/tex]
Explanation:
From the question we are told that:
Load [tex]V=480[/tex]
Poles [tex]p=6[/tex]
Power [tex]P=150hp[/tex]
3-Phase
Load:
[tex]Tl=0.05*\omega_s^2Nm[/tex]
Motor:
[tex]Ef=400V\\\\X_d=1ohm[/tex]
Generally the equation for Synchronous speed is mathematically given by
[tex]N_s=\frac{120F}{p}=\frac{120*60}{6}[/tex]
[tex]N_s=1200rpm\\\\N_s=125.66 rads/sec[/tex]
Therefore
[tex]Tl=0.05*\omega_s2Nm[/tex]
With
[tex]\omega=N_s[/tex]
We have
[tex]Tl=0.05*(125.66)^2Nm[/tex]
[tex]T_l=789.52 Nm[/tex]
Therefore
Load Power
[tex]P_l=T_l*\omega_s\\\\P_l=789.52*125.66[/tex]
[tex]P_l=9922watts[/tex]
Generally the equation for Load Power is mathematically given by
[tex]P_l=\frac{\sqrt{3}*E_f.V_t}{x_d}*sin\theta\\\\9922=\frac{\sqrt{3}*480*400}{1}*sin\theta[/tex]
[tex]\theta=17.4 \textdegre3[/tex]
Therefore
Voltage
[tex]E_f=400<-17.4volts[/tex]
Functional and nonfunctional requirements documents are used to _____.
define the financial budget of a system
define the purpose of a system
facilitate communication between the users and a system
help exercise control over the inner workings of the firm.
Answer:
Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance
Explanation:
Tech A says that the voltage regulator controls the strength of the rotor s magnetic field. Tech B says that the voltage regulator is installed between the output terminal of the alternator and the positive terminal of the battery. Who is correct?
Answer:
Voltage Regulator
Technician A is correct.
Explanation:
Technician B is not correct. The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B. Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct. The computer can also be used to control the output of the alternator by controlling the field current.
An incompressible viscous fluid flows through a pipe with a flow rate of 1 mL/s. The pipe has a uniform diameter D0 and a length L0. A pressure difference of P0 between the ends of the pipe is required to maintain the flow rate. What would be the flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0
Answer:
[tex]Q_2 = 32[/tex] mL/s
Explanation:
Given :
The flow is incompressible viscous flow.
The initial flow rate, [tex]Q_1[/tex] = 1 mL/s
Initial diameter, [tex]D_1= D_0[/tex]
Initial length, [tex]L_1=L_0[/tex]
The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]
We know for a viscous flow,
[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]
[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]
[tex]$Q \propto \Delta P \times D^4$[/tex]
[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]
[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]
∴ [tex]Q_2 = 32[/tex] mL/s
The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s
We are given;
Initial flow rate; Q1 = 1 mL/s
Initial uniform diameter; D0
Initial Length; L0
Initial Pressure difference; P0
Relationship between pressure, flow rate and diameter for vicious flow is given by;
Q1/Q2 = (P1/P2) × (D1/D2)⁴
Where;
Q1 is initial flow rate
Q2 is final flow rate
P1 is initial pressure difference
P2 is final pressure difference
D1 is initial diameter
D2 is final diameter
We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;
P2 = 2P0
D2 = 2D0
Thus;
1/Q2 = (P0/2P0) × (D0/2D0)⁴
>> 1/Q2 = ½ × (½)⁴
1/Q2 = 1/32
Q2 = 32 mL/s
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A one electron species, Xm, where m is the charge of the one electron species and X is the element symbol, loses its one electron from its ground state when it absorbs 7.84×10−17 J of energy. Using the prior information, the charge of the one electron species is?
Answer:
c +5
Explanation:
we have difference in energy =
2.18x10⁻¹⁸ x z² / n²
now n = 1
amount of energy absorbed Δdelta = 7.84×10−17 J
7.84×10⁻¹⁷ = 2.18x10⁻¹⁸ x z²
we divide through by 2.18x10⁻¹⁸
z² = 7.84×10⁻¹⁷ / 2.18x10⁻¹⁸
z² = 35.9633
z = √35.9633
z = 5.9969
≈ 6
charge = atomic number 6 - number of electrons available in the element 1
= 6-1 = 5
from the calculations above, the charge of the one electron specie would be c +5
The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 41:1. The primary coil is plugged into a standard 120-V outlet. The current in the secondary coil is 1.2 x 10-3 A. Find the power consumed by the air filter.
Answer:
5.9 watts
Explanation:
The secondary voltage is the primary voltage multiplied by the turns ratio:
(120 V)(41) = 4920 V
The power is the product of voltage and current:
(4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W
The power consumed is about 5.9 watts.
định khoản nghiệp vụ sau : tạm ứng cho nhân viên A đi công tác bằng tiền mặt 50.000
Answer:
report on a fight you have witnessed
The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Sieve analysis Sieve Size No. 4 (4.75 mm) No. 10 (2.00 mm) No. 40 (0.425 mm) No. 200 (0.075 mm) Percent passing by weight 80 60 30 10 Atterberg limits Liquid limit (LL) Plastic limit (PL 31 25
(a) Classify this soil according to USCS system, providing the group symbol for it. Show how you arrive at the final classification.
(b) According to USCS system, what is a group name for this soil?
(c) Is this a clean sand? If not, explain why.
Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
A) Classifying the soil according to USCS system
( using 2nd image attached below )
description of sand :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Use a truth table to verify the first De Morgan law ¬(p ∧ q) ≡ ¬p ∨ ¬q.
Answer:
p q output ¬(p ∧ q)
0 0 1
0 1 1
1 0 1
0 0 0
p q output ¬p ∨ ¬q
0 0 1
0 1 1
1 0 1
0 0 0
Explanation:
We'll create two separate truth tables for both sides of the equation, and see if they match.
The expressions in the question use AND, OR and NOT operators.
The AND operation needs both inputs to be 1 to return a 1.The OR operation needs at least 1 of the inputs to be 1 to return a 1. The NOT operation takes a 1 and turns it into a 0, or takes a 0 and turns it into a 1.Let's start with ¬(p ∧ q)
NOT (0 AND 0) = NOT (0) = 1NOT (0 AND 1) = NOT (0) = 1NOT (1 AND 0) = NOT (0) = 1NOT (1 AND 1) = NOT (1) = 0Now let's move on to the second expression ¬p ∨ ¬q
NOT(0) OR NOT(0) = 1 OR 1 = 1NOT(0) OR NOT(1) = 1 OR 0 = 1NOT(1) OR NOT(0) = 0 OR 1 = 1NOT(0) OR NOT(0) = 0 OR 0 = 0Therefore we can say the two expressions are equivalent.
Attached the truth table to verify the first De Morgan's law ¬(p ∧ q) ≡ ¬p ∨ ¬q:
What is the explanation of the truth table?As you can see from the attached truth table, the truth values for ¬(p ∧ q) and ¬p ∨ ¬q are the same for all combinations of p and q, confirming the validity of the first De Morgan's law.
De Morgan's law is a fundamental principle in propositional logic.
It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the individual propositions.
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Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
how he would plan and manage projects in the future?
Select an answer:
project highlights
major changes and risks
summary of schedule and cost performance
summary of project management effectiveness
Answer:
Dalton
The part of the close-out report that would describe how he would plan and manage projects in the future is:
summary of project management effectiveness
Explanation:
The Project Close-out Report is a project management document, which identifies the variances from the baseline plans. These variances are specified in terms of project performance, project cost, and schedule. The project close-out report records the completion of the project and the subsequent handover of project deliverables to others. The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
1. Is the team developing confidence that the solution space has been fully explored?
2. Are there alternative diagrams and alternative ways to decompose the problem?
3. Have external sources been thoroughly pursued, and everyone’s ideas been accepted and integrated in the process?
4. All of the above
Answer:
While reflecting on the solutions and the process of concept generation, the development team takes a look at some critical questions such as:________.
4. All of the above
Explanation:
The team must explore its solution space, including some external sources. Then, it must integrate its findings with the ideas of team members, ensuring the consideration of all possible ways to decompose the problem. This is because employing a structured process to concept generation enables the team to come up with creative solutions to design concepts.
A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the water velocity is 3 m/s, how high will the water rise in the vertical leg relative to the level of the water surface of the stream?
Answer:
[tex]h=0.46m[/tex]
Explanation:
From the question we are told that:
Velocity of water [tex]V=3m/s[/tex]
Height=?
Generally, the equation for Water Velocity is mathematically given by
[tex]V=\sqrt{2gh}[/tex]
Therefore Height h is given as
[tex]h=\frac{v}{2g}[/tex]
[tex]h=\frac{3^2}{2*9.81}[/tex]
[tex]h=0.46m[/tex]
Flat plate collector can provide temperature upto_____
Answer:
80⁰C
Explanation:
80°C.
Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities
Several applications are listed below. Determine the relative importance of the resilience and toughness of the materials chosen for each application. Sort the items based on whether resilience is most important, toughness is most important, or both are equally important.
a. a non-critical spring that is used repeatedly
b. a high tension music wire
c. a one use safety device that absorbs impact energy
d. a burst panel designed to rupture at
e. a certain pressure
d. a structural member in a building
Answer:
they are important together, but if you want to use just one future you must think about which one is first needed. and then try to learn for economical so don't use more money
1025 steel wire is stretched with a stress of 70 MPa at room temperature 20 C. If th length is held constant, to what temperature in 'C and 'F must the wire be heated to reduce the stres to 17 MPa?
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A Rankine power generation cycle operates with steam as the working fluid. The turbine produces 100 MW of power using 89 kg/s of steam entering at 700C and 5MPa. The steam exits the turbine at 0.10135 MPa. Saturated liquid water exits the condenser and is pumped back to 5Mpa before it is fed to an isobaric boiler. a. Draw a schematic of the cycle. Number the streams and label any constraints across the units. b. The turbine operates adiabatically but not reversibly. What is the temperature of the steam exiting the turbine
To avoid a possible collision with a manned airplane, you climb your unmanned aircraft to yield the right of way. In doing so, your unmanned aircraft reached an altitude greater than 600 feet AGL. To whom must you report the deviation?
Answer:
you have to report to the federal aviation administration upon request
Explanation:
An unmanned aircraft can also be a drone it is an aircraft that does not have passengers or a human pilot.
Such an aircraft could be fully autonomous but most times they have a human pilot who controls it remotely.
Now you have to report this deviation to the federal aviation administration. This body is responsible for the enforcement of regulations that have to do with the manufacture, operation and maintenance of aircrafts. The body makes sure there is an efficient and safe system for a navigation and they also control air traffic.
Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address
lessons learned on a project?
Select an answer:
Add lessons learned as a topic in status meetings
Review past lessons learned so a new one does not have to be created,
Create lessons learned at the end of the project.
Brainstorm lessons learned at the beginning of a project
Answer: Create lessons learned at the end of the project.
Explanation:
Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.
The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.
. Bơm kiểu piston tác dụng đơn có áp suất p=0,64 Mpa và lưu lượng Q=3,5 l/s. Xác định tốc độ quay của trục bơm và công suất của bơm nếu biết đường kính piston D=150 mm; bán kính tay quay R=60 mm; hiệu suất thể tích của bơm là 0=0,94; hiệu suất chung của bơm b=0,80.
Answer:
not understand language
1) (30 pts ) Oxygen (O2) flows through a pipe, entering at at 4 m/sec at 10000 kPa, 227oC. For a pipe inside diameter of 3.0 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas
Complete Question
Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted
Answer:
[tex]H=9.91kJ/sec[/tex]
Explanation:
From the question we are told that:
Velocity [tex]v=4 m/sec[/tex]
Pressure [tex]P=1000kPa[/tex]
Temperature [tex]T=227 \textdegree C[/tex]
Diameter [tex]d=3cm=>0.03m[/tex]
Generally the equation for volumetric Flow Rate is mathematically given by
[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]
[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]
[tex]V_r=0.002827m^3/s[/tex]
Generally the equation for mass Flow Rate is mathematically given by
[tex]m_r=\frac{PV_r}{RT}[/tex]
[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]
[tex]m_r=0.019kg/sec[/tex]
Generally the equation for mass Flow Rate is mathematically given by
Using gas Table for enthalpy Value
[tex]T=500K=>h=520.75kg[/tex]
Therefore
[tex]H=mh[/tex]
[tex]H=0.019*520.75[/tex]
[tex]H=9.91kJ/sec[/tex]
Theo Anh / Chị, để đáp ứng yêu cầu phát triển nền kinh tế thị trường định hướng Xã hội Chủ nghĩa ở Việt Nam trong bối cảnh thời đại hiện nay, cần chú trọng giải quyết những vấn đề gì ?
A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance it slides in 10seconds if the kinetic coefficient is 0.4.
Answer:
21 m
Explanation:
Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.
Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N
The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N
The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N
If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.
Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.
For the block to be in contact with the surface, the vertical forces on the block must balance.
Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,
N = mg + F₃" (since both the weight and the resultant vertical force act downwards)
N = mg + F₃"
Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N
So,
N = mg + F₃"
N = 35 kg × 9.8 m/s² + 8.82 N
N = 343 N + 8.82 N
N = 351.82 N
So, the net horizontal force F = F₃' - f.
F = 155.41 N - 0.4 × 351.82 N
F = 155.41 N - 140.728 N
F = 14.682 N
Since F = ma, where a = acceleration of block,
a = F/m = 14.682 N/35 kg = 0.42 m/s²
To find the distance the block moved, x we use the equation
x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²
Substituting the values of the variables into the equation, we have
x = ut + 1/2at²
x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²
x = 0 m + 1/2 × 0.42 m/s² × 100 s²
x = 0.21 m/s² × 100 s²
x = 21 m
So, the distance moved by the block is 21 m.
Ô tô có khối lượng m (kg) đặt tại trung tâm h . Khoảng cách từ h tới 2 bánh xe hai bên của a (m) và b (m) , khoảng cách vết bánh xe AB = L ( m) . Ô tô không bị trượt ngang và đang quay vòng trên đoạn đường có góc nghiêng aphal , bán kính quay vòng r ( m ), vận tốc xe v ( m/s ). Tính chiều cao trọng tâm lớn nhất để xe không bị lật ngang .
Answer:
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Explanation:
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