Explanation:
Network and Switching Subsystem (NSS)
Base-Station Subsystem (BSS)
Mobile station (MS)
Operation and Support Subsystem (OSS)
A structure is designed using 4 circular columns. Due to a quirky design, the four columns will all carry different loads of 1800 N, 2100 N, 2275 N, and 2200 N. A factor of safety of 5 is used to design the columns. The diameter of each of the columns is supposed to be 50 cm, at most. Determine the maximum height of the structure (i.e. the column height) so that the structure will not fail. Assume that all columns may be modeled as Euler columns for your analysis. Assume a pinned-pinned boundary condition for your analysis, and assume the elastic modulus of the column material is 10 MPa.
Answer:
5.16 M
Explanation:
Loads ; 1800N, 2100N, 2275N, 2200N
safety factor = 5
diameter of each column = 50 cm = 0.5 m
Elastic modulus = 10 MPa
Calculate the max height of structure
moment of inertia for a circular section ( I ) = πd^4 / 64
lets represent the required maximum height of the column as L
Applying Euler column theory
The bucker load of the column = ( attached below )
attached below is the remaining solution
Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Report the final temperature in units of K and using three significant digits.
Answer:
[tex]T_2=315.69k[/tex]
Explanation:
Initial Temperature [tex]T_1=500K[/tex]
Initial Pressure [tex]P_1=1000kPa[/tex]
Final Pressure [tex]P_2=200kPa[/tex]
Generally the gas equation is mathematically given by
[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]
Where
n for [tex]CO=1.4[/tex]
Therefore
[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]
[tex]T_2=315.69k[/tex]
True or false: You can create a network with two computers.
Answer
True
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]
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
An apple, potato, and onion all taste the same if you eat them with your nose plugged
Answer:
I didn't understand your question or is it a fun fact
Trình bày sự khác nhau của Dây chuyền đẳng nhịp đồng nhất, dây chuyền đẳng nhịp không đồng nhất, cho ví dụ minh họa
P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diameter is 3 cm. Estimate (a) the mass flow rate through the nozzle and (b) the Mach number at the throat.
Answer:
a) [tex]m=0.17kg/s[/tex]
b) [tex]Ma=0.89[/tex]
Explanation:
From the question we are told that:
Pressure [tex]P=60kPa[/tex]
Diameter [tex]d=3cm[/tex]
Generally at sea level
[tex]T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4[/tex]
Generally the Power series equation for Mach number is mathematically given by
[tex]\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}[/tex]
[tex]\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]Ma=0.89[/tex]
Therefore
Mass flow rate
[tex]\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}[/tex]
[tex]\rho=0.848kg/m^3[/tex]
Generally the equation for Velocity at throat is mathematically given by
[tex]V=Ma(r*T_0\sqrt{T_e}[/tex])
Where
[tex]T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}[/tex]
[tex]T_e=248[/tex]
Therefore
[tex]V=0.89(1.4*288\sqrt{248})\\\\V=284[/tex]
Generally the equation for Mass flow rate is mathematically given by
[tex]m=\rho*A*V[/tex]
[tex]m=0.84*\frac{\pi}{4}*3*10^{-2}*284[/tex]
[tex]m=0.17kg/s[/tex]
‘Politics and planning are increasingly gaining prominence in contemporary urban and regional planning debates’. Using relevant examples, discuss this assertion reflecting on the critical success factors for the successful implementation of the land reform program in South Africa.
Answer:
The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output
I hope it is helpful
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:
wiwhwnwhwwbbwbwiwuwhwhehehewhehehheheheehehehehhehehwh
Explanation:
jwhwhwhwhwhwwhhahwhahahwh
what is the term RF exiciter?
If a corporation is socially responsible, it will develop and implement a sustainability plan and communicate it to stakeholders.
True
False
Answer:
True
Explanation:
All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.
A steam turbine receives steam at 1.5MPa and 220oC, and exhausts at 50kPa, 0.75 dry. Neglecting heat losses and changes in kinetic and potential energy, estimate the work output per kg steam.
If when allowance is made for friction, radiation and leakages losses, the actual work of that estimated in (a), calculate the power output of the turbine when consuming 600kg of steam per minute.
Answer:
Can you make friend with me ?
determine if the fluid is satisfied
what is geo technical
The end of the industrial robotic arm extends along the path (r = 2 + 2 cos (5t)) m. At the instant ( 0 = 0.8t ) radians. When the arm is located at (t = 0.85) second Determine the velocity and acceleration of the object A at this instant.
Answer:
v = 8.95 rad / s, a = 22.3 rad / s²
Explanation:
This is an exercise in kinematics, where we must use the definitions of velocity and acceleration
v = dr / dt
we perform the derivative
v = 0+ 2 (-sin 5t) 5
v = -10 sin 5t
we calculate for t = 0.85 t,
remember angles are in radians
v = -10 sin (5 0.85)
v = 8.95 rad / s
acceleration is defined by
a = dv / dt
we perform the derivatives
a = -10 (cos 5t) 5
a = - 50 cos 5t
we calculate for t = 0.85 s
a = -50 are (5 0.85)
a = 22.3 rad / s²
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.
Learn more about truth table at:
https://brainly.com/question/28605215
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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
Read more about vicious flow at; https://brainly.com/question/2684299
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.
State three types of maintenance.
Answer:
Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:
Explanation:
Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.
Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.
Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.
Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.
Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).
Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domestic wastewater from an influent BOD concentration of 250 mg/L to an effluent concentration of 10 mg/L. X (MLVSS) = 3000 mg/L, and the kinetics are first order and not variable order. The first order equation you must use to calculate the specific substrate utilization rate is q = K S where S is the effluent BOD concentration and K is the first order BOD degradation rate constant. The value of K is 0.04 L/(day*mg). What is the required reactor volume in MG (millions of gallons)? All the choices below are in units of MG.
0.4
1.0
0.2
4.8
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
which type of clectrical circuit is represented by this diagram?
Answer:
parallel
Explanation:
All components in this circuit are tied in parallel. Each component experiences the same voltage from one terminal to the other. It is a parallel circuit.
Microsoft Project là phần mềm có sẵn trong bộ Office 365, đúng (True) hay sai (False)?
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
Question 1. If a fiber weight 3.0 g and composite specimen weighing 4.g. The composite specimen weighs 2.0 g in water. If the specific gravity of the fiber and matrix is 2.4 and 1.3, respectively, find the 1. Theoretical density of composite 2. Experimental density 3. Void fraction
Answer:
Explanation:
From the given information:
weight of fiber [tex]w_f[/tex] = 3.0 g
weight of composite specimen [tex]w_c[/tex] = 4.0 g
specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g
specific gravity of fiber [tex]S_f[/tex] = 2.4
specific gravity of matrix [tex]S_m[/tex] = 1.3
The weight of the matrix = weight of the composite - the weight of fiber
⇒ (4.0 - 3.0) g
= 1.0 g
The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]
[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]
[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]
[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]
The experimental density [tex]\rho _{ce}[/tex] is determined by using the equation:
[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]
[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]
The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]
[tex]= \dfrac{1.980-1.620}{1.980}[/tex]
= 0.1818
8- Concentration polarization occurs on the surface of the.......
a- cathode.
b- anode.
C- both
d-ption 4
Explanation:
Concentration overpotential, ηc,
I hope it helps you
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.
A 2-stage dcv that has an internal pilot does not work well (if at all) on
Answer:
i really font onow why tbh eot you
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 input sin(20) is sampled at 20 ms intervals by using impulse train sampling: i. Construct the input and sampled signal spectra.
Solution :
Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]
[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]
[tex]$=\frac{1}{20}$[/tex]
= 0.05 kHz
[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex] rad/s
We know that,
FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]
The sampled signal is :
[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]
So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
[tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
