Answer:
The value of [tex]\mathbf{BOD_5 \ at \ 15^0 \ C \ is \simeq 130.14 \ mg/L}[/tex]
Explanation:
From the given information:
To find the BOD5 (in mg/L) at 15° C; we need to know the ultimate carbonaceous BOD and the reaction rate coefficient at 15° C.
So, to start with the calculation of the ultimate carbonaceous BOD by using the formula:
[tex]BOD_t = L_o (1-e^{-k*t})[/tex]
where;
[tex]BOD _t[/tex] = Biochemical oxygen demand at t days
[tex]L_o[/tex] = the ultimate carbonaceous
∴
[tex]150= L_o (1-e^{-0.20*5})[/tex]
[tex]L_o =\dfrac{150}{ (1-e^{-0.20*5})}[/tex]
[tex]L_o =\dfrac{150}{ (1-0.3679)}[/tex]
[tex]L_o =237.30 \ mg/L[/tex]
Thus, the ultimate carbonaceous BOD = 237.30 mg/L
For the reaction rate coefficient; we use the formula:
[tex]k_{\tau} = k_{20} \times \theta^{\tau - 20}[/tex]
where;
[tex]k_{\tau}[/tex] = rate of the reaction constant at various temperature (T) = 15
[tex]k_{20}[/tex] = rate of the reaction constant at standard laboratory = 0.20
[tex]\theta[/tex] = constant = 1.047
∴
[tex]k_{15} =0.20 \times \theta^{15 - 20}[/tex]
[tex]k_{15} =0.20 \times 1.047^{-5}[/tex]
[tex]k_{15} =0.20 \times0.7948[/tex]
[tex]k_{15} = 0.1590 / day[/tex]
Thus, at 15° C, the reaction constant (k) = 0.1590 / day
Finally, the BOD5 (in mg/L) at 15° C can be calculated by using the formula:
[tex]BOD_t = L_o (1 - e^{-k*t})[/tex]
[tex]BOD_t = 237.30 \times (1 - e^{-0.1590*5})[/tex]
[tex]BOD_t = 237.30 \times (1 -0.45158)[/tex]
[tex]BOD_t = 237.30 \times (0.54842)[/tex]
[tex]\mathbf{BOD_5 \simeq 130.14 \ mg/L}[/tex]
The amount of FORCE that moves electrons is measured in
A commercial grade cubical freezer, 3 m on a side, has a composite wall consisting of an exterior sheet of 6.35-mm-thick plain carbon steel, an intermediate layer of 100-mm-thick cork insulation, and an inner sheet of 6.35-mm-thick aluminum alloy (2024). Adhesive interfaces between the insulation and the metallic strips are each characterized by a thermal contact resistance of = 2.5 x 10–4 m3 · K / W. What is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are 22°C and −6°C, respectively?
Answer:
589.5 W
Explanation:
Given that
the thickness of the carbon steel and insulation is L(a) = L(c) = 0.00635 m
Width of the freezer, W = 3m
Thermal contact resistance, R(t.c) = 0.00025 m²
T(s.i) = -6°C = 267 K
T(s.o) = 22°C = 295 K
See attachment for calculation
2. A freeway is being designed for a location in rolling terrain. The expected free-flow speed is 55 mph. During the peak hour, it is expected there will be a directional peak-hour volume of 2700 vehicles and 18% heavy vehicles. The PHF is expected to be 0.88. If a LOS no worse than D is desired, determine the necessary number of lanes.
Answer:
Number of necessary lanes = 3
Explanation:
Expected free flow speed = 55 mph
Volume of vehicles expected during peak-hour = 2700
PHF = 0.88
18% = 0.18 for heavy vehicles ( Pt )
Et ( for a rolling terrain ) = 3
Detailed solution attached below
First we calculate the Heavy-vehicle adjustment factor before determining the number of lanes
Contains tires, wheels, engine, transmission, and drive axle assembly.
CRM software programs enable sales people to track of their customer contact information, needs, sales status, and other useful information. True False
Answer:
The correct answer is True
Explanation:
Most business organizations especially those that are into sales, call centers, even individuals that are into marketing use CRM (Customer Relationship Management) software to drive their business forward by enabling them to interact more with customers and strive to provide better customer service. This is achievable because The software (CRM) which is a business tool stores customers' information ranging from names, contact info, sales, and other important information based on the need of the salespeople (organization) to track, meet and respond to customers needs. CRM benefits range from getting to know more about prospective or existing customers to building a better relationship with customers so as to retain them and allow better communication between customers in a bid to meet their demands.
What is 22 centimeters in inches
Answer:
8.66142
Explanation:
Answer:
8.66
Explanation:
hope it helps
In a non air-entrained concrete mixture you should assume this % of air by volume for the purposes of concrete mixture proportioning: _______ %
Answer:200%
Explanation:
so it can melt to use it later?
I drive safe
Determine a safe speed to travel base on
Write a Verilog module that describes 8-bit right shifter register with control signals A(arithmetic), L(logical), CLR(clear) and LD(load). Input signals are CLK, LD and 8-bit input D. Output signal is 8-bit output Q. CLR signal will clear Q. LD signal will load input D to Q. A signal will do arithmetic right shift of Q, while L signal will do logical right shift of Q.
What do you know about love?
Some measurements on a blood sample at 37 ˚C (98.6 ˚F) indicate a shearing stress of 0.540 N/m2 for a corresponding rate of shearing strain of 198 s-1. Determine (a) the apparent viscosity of the blood and (b) its ratio to the viscosity of water at the same temperature.
Answer:
[tex]0.00273\ \text{Pa s}[/tex]
3.944
Explanation:
[tex]\tau[/tex] = Shearing stress = [tex]0.54\ \text{N/m}^2[/tex]
[tex]\dot{\gamma}[/tex] = Rate of shearing strain = [tex]198\ \text{s}^{-1}[/tex]
(a) Viscosity is given by
[tex]\mu_b=\dfrac{\tau}{\dot{\gamma}}\\\Rightarrow \mu_b=\dfrac{0.54}{198}\\\Rightarrow \mu_b=0.00273\ \text{Pa s}[/tex]
The apparent viscosity of the blood is [tex]0.00273\ \text{Pa s}[/tex]
(b) Viscosity of water at 37 ˚C = 0.0006922 Pa s = [tex]\mu_w[/tex]
The ratio is
[tex]\dfrac{\mu_b}{\mu_w}=\dfrac{0.00273}{0.0006922}\\\Rightarrow \dfrac{\mu_b}{\mu_w}=3.944[/tex]
The required ratio is 3.944
A specimen of some metal having a rectangular cross section 9.18 mm x 13.2 mm is pulled in tension with a force of 3310 N, which produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.
Answer: 0.0000084
Explanation:
Given data:
Cross sectional area = 9.18mm*13.22mm
Force N = 3310
Elastic modulus = 79 GPa
Solution:
Area = 9.18mm * 13.22mm
= 121.36mm^2
= π(d)^2
= π(0.121)^2
= 0.05m^2
Stress = F/A
= 3310N/0.05m^2
= 66200Nm^2
Strain = Stress/ modulus of elastic
= 66200/7.9*10^10
= 0.0000084
6-137 Consider a Carnot heat-engine cycle executed in a steady-flow system using steam as the working fluid. The cycle has a thermal efficiency of 30%, and steam changes from saturated liquid to saturated vapor at 275oC during the heat addition process. If the mass flow rate of the steam is 3 kg/s, determine the net power output of this engine, in kW.
Answer:
1417.05 kw
Explanation:
From the question,
The Mass flow rate = 3kg
The Saturated temperature = 275⁰c
The Thermal efficiency = 30%
The enthalpy of vaporization is 1574.5kj/kg
Rate of heat transfer = 3x1574.5 = 4723.5 kW
Thermal efficiency of engine =
0.30 x 4723.5
= 1417.05 kW
This answer is the net power output of this engine, in kW.
Thank you!
A signal processor has a reliability of 0.90. Because of the low reliability a redundant signal processor is to be added. However, a signal splitter must be added before the signal processors, and a comparator must be added after the signal processors. Each of these new components has a reliability of 0.95. Does adding a redundant signal processor increase the system reliability
Answer:
0.9732
Explanation:
A signal splitter must be placed before a processor and a comparator be placed after a processor.
This contains a series combination of a splitter, processor and comparator. Let R1 represent the reliability of the first subsystem and R2 be the reliability of the added redundant system.
Given: R(processor) = 0.9, R(new components) = 0.95 (i.e. splitter, new processor, comparator)
R1 = series combination of splitter, processor and comparator
R1 = R(splitter) × R(processor) × R(comparator) = 0.95 × 0.9 × 0.95 = 0.81225
R2 = series combination of splitter, processor and comparator
R2 = R(splitter) × R(new processor) × R(comparator) = 0.95 × 0.95 × 0.95 = 0.8574
R2 is the redundant system which is connected in parallel with R1, hence:
R(system) = 1 - (1 - R1)(1 - R2) = 1 - (1 - 0.81225)(1 - 0.8574)
R(system) = 0.9732
Which type of engineer finds his or her skills needed in the electronics, architectural, medical, and aerospace industries?
nuclear engineer
chemical engineer
biomedical engineer
mechanical engineer
Answer:
mechanical engineer
Explanation:
What type of engineer construct the infrastructure necessary for roads air fields and water ports a construction engineer be environmental engineer see structural engineer D transportation engineer
Answer:
Construction engineers often focus on a specific type of construction project. Some of these specialties are: Building - commercial housing or business building. Electrical - electrical systems. Mechanical - plumbing, heating or air conditioning. Highway or Heavy - bridges, airports, highways or water waste.
Explanation:
Assume that the density of fresh water is 998 kg/m3 , and the density of salt water is 1025 kg/m3 . The gate weighs 0.448 kN/m (into the page). The gate is 5 m wide, into the page. Determine the height of the fresh water at which the gate opens. (15 points)
Answer:
hello your question lacks the required diagram attached below is the diagram showing the resultant forces at the gate
answer: 1.975 m
Explanation:
Given data :
density of fresh water = 998 kg/m^3
density of salt water = 1025 kg/m^3
gate weight = 0.448 kN/m ≈ 100 Ib/m
attached below is the detailed solution required
g What is the difference between surface erosion and bulk degradation, and for what application is each best suited
Explanation:
The main difference between surface erosion and bulk degradation is that in surface erosion as the name implies that the material or polymer begins to erode or degrade only from the external surface until it reaches the interior of the material. However, in bulk degradation, the material undergoes complete degradation from within and outside the material.
Surface erosion is best applied in pharmaceuticals in terms of drug delivery into the body, while bulk degradation is best applied in plastics which may degrade in any form.
6-44Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.
Answer:
The correct answer will be "8640 kWh".
Explanation:
The given values are:
Rate (Ql)
= 5040 kJ/h
As we know,
⇒ [tex]COP = \frac{Cooling}{heat \ required}[/tex]
On putting the values, we get
⇒ [tex]=\frac{5040 \ kJ/h}{1 \ kW}[/tex]
⇒ [tex]=1.4[/tex]
Now,
⇒ [tex]Q = Ql + W[/tex]
[tex]= 1 + 1.4[/tex]
[tex]=2.4 \ kW[/tex]
[tex]=2.4\times 3600[/tex]
[tex]=8640 \ kWh[/tex]
Para un intercambiador de calor encargado de precalentar pulpa de fruta, se utiliza agua caliente que entra a 180°C y sale a 78°C, mientras que la pulpa de fruta entra a 3°C y sube su temperatura hasta 55°C. Realizar los esquemas de perfil de temperaturas para un intercambiador de calor que funcione en paralelo y en contracorriente. Además, calcular LMTD.
Answer:
La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.
La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.
Explanation:
De la teoría de Transferencia de Calor tenemos que un intercambiador de calor en paralelo presenta las siguientes dos características:
1) Tanto el fluido caliente como el fluido frío entran por el mismo lado.
2) Tanto el fluido caliente como el fluido frío salen por el mismo lado.
Mientras que el intercambiador de calor en contracorriente tiene que:
1) El fluido caliente y el fluido frío entran por lados opuestos.
2) El fluido caliente y el fluido frío salen por lados opuestos.
A continuación, anexamos los esquemas de perfil de cada intercambiador.
Ahora, la Diferencia Media Logarítimica de Temperatura ([tex]\Delta T_{lm}[/tex]), medida en grados Celsius, queda definida como sigue:
[tex]\Delta T_{lm} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}} }[/tex] (Eq. 1)
Donde [tex]\Delta T_{1}[/tex] y [tex]\Delta T_{2}[/tex] son las diferencias de temperatura de los fluidos en cada extremo del intercambiador, medido en grados Celsius.
Procedemos a determinar esas diferencias y la Diferencia Media Logarítimica de Temperatura para cada configuración:
Intercambiador en paralelo
[tex]\Delta T_{1} = 180\,^{\circ}C-3\,^{\circ}C[/tex]
[tex]\Delta T_{1} = 177\,^{\circ}C[/tex]
[tex]\Delta T_{2} = 78\,^{\circ}C - 55\,^{\circ}C[/tex]
[tex]\Delta T_{2} = 23\,^{\circ}C[/tex]
[tex]\Delta T_{lm} = \frac{177\,^{\circ}C-23\,^{\circ}C}{\ln \frac{177\,^{\circ}C}{23\,^{\circ}C} }[/tex]
[tex]\Delta T_{lm} \approx 75.466\,^{\circ}C[/tex]
La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.
Intercambiador en contracorriente
[tex]\Delta T_{1} = 180\,^{\circ}C-55\,^{\circ}C[/tex]
[tex]\Delta T_{1} = 125\,^{\circ}C[/tex]
[tex]\Delta T_{2} = 78\,^{\circ}C-3\,^{\circ}C[/tex]
[tex]\Delta T_{2} = 75\,^{\circ}C[/tex]
[tex]\Delta T_{lm} = \frac{125\,^{\circ}C-75\,^{\circ}C}{\ln \frac{125\,^{\circ}C}{75\,^{\circ}C} }[/tex]
[tex]\Delta T_{lm} \approx 97.881\,^{\circ}C[/tex]
La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.
5. Saturated liquid R-134a enters an expansion valve at 40 C. The refrigerant leaves the expansion valve at 6 C. What is the specific volume at the exit
Answer:
The specific volume of R-134a at the exit of the expansion valve is 0.014655 cubic meters per kilogram.
Explanation:
From Thermodynamics, we know that expansion valves are devices that work at steady state, in which heat and work interactions are usually non-existent and changes in gravitational potential and translational kinetic energy. From First Law of Thermodynamics and the Principle of Mass Conservation we get that expansion valve is represented by the following model:
[tex]\dot H_{in}-\dot H_{out} = 0[/tex] (Eq. 1)
[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (Eq. 2)
Where:
[tex]\dot H_{in}[/tex], [tex]\dot H_{out}[/tex] - Rates of change of inlet and output enthalpies, measured in kilowatts.
[tex]\dot m_{in}[/tex], [tex]\dot m_{out}[/tex] - Mass flow rates at inlet and outlet, measured in kilograms per second.
From definition of enthalpy and (Eq. 2), we simplify (Eq. 1) as follows:
[tex]u_{in} + P_{in}\cdot \nu_{in} = u_{out} + P_{out}\cdot \nu_{out}[/tex] (Eq. 3)
Where:
[tex]u_{1}[/tex], [tex]u_{2}[/tex] - Specific internal energies at inlet and outlet, measured in kilojoules per kilogram.
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Inlet and outlet pressures, measured in kilopascals.
[tex]\nu_{1}[/tex], [tex]\nu_{2}[/tex] - Inlet and outlet specific volumes, measured in cubic meters per kilogram.
Now we get the following information from property charts:
Inlet (Saturated Liquid)
[tex]T = 40\,^{\circ}C[/tex]
[tex]P = 1017.1\,kPa[/tex]
[tex]\nu = 0.0008720\,\frac{m^{3}}{kg}[/tex]
[tex]u = 107.39\,\frac{kJ}{kg}[/tex]
[tex]h = 108.28\,\frac{kJ}{kg}[/tex]
[tex]x = 0[/tex]
Outlet (Liquid-Vapor Mix)
[tex]T = 6\,^{\circ}C[/tex]
[tex]P = 362.23\,kPa[/tex]
[tex]\nu = 0.014655\,\frac{m^{3}}{kg}[/tex]
[tex]u = 102.968\,\frac{kJ}{kg}[/tex]
[tex]h = 108.28\,\frac{kJ}{kg}[/tex]
The specific volume of R-134a at the exit of the expansion valve is 0.014655 cubic meters per kilogram.
Calculate the drain current in an NMOS transistor if Kn = 400 μA/V2, VT N = 0.5 V, λ = 0.02 V−1, VGS = 5 V, and VDS = 6 V. (Round off the answer to two decimal places.)
Answer:
Drain current = 4050 μA
Explanation:
Determine the drain current in an NMOS transistor
Given data :
Kn = 400 μA/V^2
Vt = 0.5 v
λ = 0.02 V^−1
Vgs = 5 V
Vds = 6 v
Id = ?
The drain current can be calculated using the formula below
[tex]I_{D} = \frac{1}{2} U_{n} C_{OX} \frac{W}{L} ( V_{GS} - V_{t} ) ^2[/tex]
= [tex]\frac{1}{2} Kn ( Vgs - Vt)^2[/tex]
= 1/2 ( 400 μA/V^2) ( 5 - 0.5 )^2
= (200 μA/V^2) * (20.25 v^2)
= 4050 μA
(a)A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300MPa (43,500psi). If the specimen radius is 5.0mm (0.20in.) and the support point separation distance is 15.0mm (0.61in.), would you expect the specimen to fracture when a load of 7500N (1690lbf) is applied? Justify your answer.
Answer:
The specimen would not fracture having a flexural strength ( 300 MPa ) > ( 286.5 MPa )
Explanation:
The expression to represent the stress on a cylindrical specimen by a three-point transverse bending test
б = [tex]\frac{F_{f}L }{\pi R^3}[/tex] ---------------- ( 1 )
where : R = radius, Ff = flexural load,
L = distance between loads ( two loads )
Given data:
7500 N = flexural load
(15 * 10^-3) m = distance between loads
(5 * 10^-3) m = radius
input the given values into equation 1
б = 286.5 * 10^6 N/m^2 = 286.5 MPa
Note : the flexural strength of the cylindrical specimen is 300MPa
Therefore the specimen would not fracture having a flexural strength ( 300 MPa ) > ( 286.5 MPa )
If you needed to calculate the mass flow rate rather than volumetric flow rate through the Venturi meter how would you go about it
Answer:
Follows are the solution to this question:
Explanation:
Its difference in power supply will be smaller whenever the pressure transducer is being performed. Its transducer sensitivity(s) is often used to calculate its number of changes occurring at voltage output of ([tex]v_o[/tex]) when the supply tension ([tex]v_s[/tex]) changes with the measured pressure ([tex]p_m[/tex]) and the rated tension of transducers ([tex]p_s[/tex]).
[tex]v_o= (s) V_s (\frac{P_m}{P_s})[/tex]
In a hydroelectric power plant, water enters the turbine nozzles at 880 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity to which water can be accelerated by the nozzles before striking the turbine blades. The maximum velocity is m/s.
Answer:
39.5 m/s
Explanation:
To solve this question, we would use Bernoulli's equation
P₁/ρ + V₁²/2 + gz₁ = P₂/ρ + V₂²/2 + gz₂,
where V₁ = 0 and z₁ = z₂, so that we have
P₁/ρ = P₂/ρ + V₂²/2
making V₂² subject of formula, we have that
V₂² = 2[P₁/ρ - P₂/ρ]
Then we finally go ahead and substitute the values such that
V₂² = 2 [(880000/1000) - (100000/1000)]
V₂² = 2 (880 - 100)
V₂² = 2(780)
V₂² = 1560
V₂ = √1560
V₂ = 39.5 m/s
Determine the minimum radius of a horizontal curve required for a highway if the design speed is 70 mph and the superelevation rate is 0.08.
Answer:
1819.4 ft
Explanation:
Let us assume the coefficient of side friction is 0.10
The radius of curve is given by the formula:
[tex]R=\frac{V^2}{g(f_s+\frac{e}{100} )}[/tex]
Where R is the minimum radius of a horizontal curve, V is the speed, f is coefficient of side friction and e/100 = super elevation rate, g =acceleration due to gravity
Given that:
V = 70 mph = (70 * 1.467) ft/s = 102.69 ft/s, g =32.2 ft/s², e/100 = 0.08, f= 0.1
[tex]R=\frac{V^2}{g(f_s+\frac{e}{100} )}=\frac{102.69^2}{32.2(0.1+0.08)}=1819.4\ ft[/tex]
the minimum radius of a horizontal curve is 1819.4 ft
A high strength waste having an ultimate CBOD of 1,000 mg/L is discharged to a river at a rate of 2 m3 /s. The river has an ultimate CBOD of 10 mg/L and is flowing at a rate of 8 m3 /s. Assuming a reaction rate coefficient of 0.1/day, calculate the ultimate and 5-day CBOD of the waste at the point of discharge (0 km) and 20 km downstream. The river is flowing at a velocity of 10 km/day.
Answer:
Attached below is the detailed solution
answer:
The Ultimate CBOD of waste water at (0) km and 20 Km
= 81.84 mg/l , 67 mg/l
Explanation:
Given Data :
UBOD of waste stream = 1000 mg/l
Volumetric flow rate of waste water = 2 m^3 / s
UBOD of River = 10 mg/l
Volumetric flow rate of river = 8 m^3 / s
reaction rate coefficient = 0.1/day
Velocity of river = 10km/day
The Ultimate CBOD of waste water at (0) km and 20 Km
= 81.84 mg/l , 67 mg/l
It is known that the kinetics of recrystallization for some alloy obey the Avrami equation, and that the value of n in the exponential is 2.4. If, at some temperature, the fraction recrystallized is 0.30 after 100 min, determine the rate of recrystallization at this temperature.
Answer:
rate = 7.580 × [tex]10^{-3}[/tex] /min
Explanation:
given data
value of n = 2.4
fraction recrystallized = 0.30
time = 100 min
solution
we will get here rate of recrystallization at this temperaturewe use here avrami equation that is
y = 1 - exp (-k [tex]t^n[/tex] ) ................1
we get hete first k
k = [tex]-\frac{ln (1-y)}{t^n}[/tex]
put heer value n is 2.4 and y = 0.30 and t is 100
k = [tex]-\frac{ln (1-0.30)}{100^{2.4}}[/tex]
k = 5.65 × [tex]10^{-6}[/tex]
now we get here [tex]t^{0.5}[/tex]
value of t at y 0.5
[tex]t^{0.5}[/tex] = [tex][\frac{-ln(1-y)}{k}]^{1/n}[/tex]
[tex]t^{0.5}[/tex] = [tex][\frac{-ln(1-0.5)}{5.65\times 106{-6}}]^{1/2.4}[/tex]
[tex]t^{0.5}[/tex] = 131.923 min
and
rate = 1 ÷ [tex]t^{0.5}[/tex]
rate = 1 ÷ 131.923
rate = 7.580 × [tex]10^{-3}[/tex] /min
what ia the value of the wod HEX begining at address 414 if the machine is Little Endian? Big endian
Answer:
hello your question is incomplete attached below is the complete question
Given you have a . 24 bit CPU The contents of memory are Address 412 413 Value0x120x34 Ox56 0x78 0x9A OxBC OxDEOxF1 0x23 414 419 418 420 415 416417 What is the value of the word in HEX beginning at address 414 if the machine is Little Endian? If it is Big Endian? Write your answer in HEX . Hex digits that are alphabetical should be written in uppercase For example A and not a Make sure that Ox appears at the beginning of your answer Do not put whitespace in your answer
Answer: Little Endian: 0x9A7856.
Big Endian: 0x56789A.
Explanation:
Given data :
24 bit CPU ( size of each word = 24 bit )
next we have to determine how many bytes are contained in each 24 bit word
= 24 / 8 = 3 bytes
with this information a word in Hex beginning at 414 will end at 416
when we use the Little Endian representation the last byte of the binary representation which is 416 will be stored first followed by 415 and then 414
hence the word beginning at 414 using little Endian = 0x9A7856.
For Big Endian the process is vice versa
Hence the word beginning at 414 using Big Endian = 0x56789A.
19. Which symbol in the figure is used to represent a carbon block lightning arrestor?
A. D
ОВ. А
C. B
O D.C
Answer:
C
Explanation:
