Answer:
radiative heat loss substantially increases as the wall temperature declines
Explanation:
The body's heat loss due to convection is ...
(2 W/m^2·K)((32 -20)K) = 24 W/m^2
__
The body's heat loss due to radiation in the summer is ...
[tex]\epsilon\sigma(T_b^4-T_w^4)\quad\text{where $T_b$ and $T_w$ are body and wall temperatures ($^\circ$K)}\\\\0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-300.15^4)\,\text{W/m$^2$}\\\\\approx 28.3\,\text{W/m$^2$}[/tex]
The corresponding heat loss in the winter is ...
[tex]0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-287.15^4)\,\text{W/m$^2$}\\\\\approx 95.5\,\text{W/m$^2$}[/tex]
Then the total of body heat losses to surroundings from convection and radiation are ...
summer: 24 +28.3 = 52.3 . . . W/m^2
winter: 24 +95.5 = 119.5 . . . W/m^2
__
It is reasonable that a person would feel chilled in the winter due to the additional radiative loss to the walls in the winter time. Total heat loss is more than doubled as the wall temperature declines.
Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp
Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.
Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²
Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec
= 0.2000856 m³ /sec
Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K
Gas constant R=287 J/kg K
Then;
1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
When checking the resistance of a dual voltage wye motor, there should be ____ resistance readings. 1) twelve 2) six 3) three
Answer:
1) twelve
Explanation:
The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.
how does a TV'S screen work
Answer:
A TVS screen works when the pixels are switched on electronically using liquid crystals to rotate polarized light.
Explanation:
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
Answer:
No, the velocity profile does not change in the flow direction.
Explanation:
In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in.2 and a temperature of 60F at the beginning of compression. The maximum temperature in the cycle is 5200R. Based on this model, calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]
[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]
[tex]V_1-0.16V_1= 36.55714291[/tex]
[tex]0.84 V_1 =36.55714291[/tex]
[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]
[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]
[tex]V_1 = 0.02518 \ ft^3[/tex]
the mass in air ( lb) can be determined by using the formula:
[tex]m = \dfrac{P_1V_1}{RT}[/tex]
where;
R = 53.3533 ft.lbf/lb.R°
[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
[tex]v_{r1} =158.58[/tex]
[tex]u_1 = 88.62 Btu/lb[/tex]
At state of volume 2; the relative volume can be determined as:
[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]
[tex]v_{r2} = 158.58 \times 0.16[/tex]
[tex]v_{r2} = 25.3728[/tex]
The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
[tex]v_{r3} = 0.1828[/tex]
[tex]u_3 = 1098 \ Btu/lb[/tex]
To determine the relative volume at state 4; we have:
[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]
[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]
[tex]v_{r4} =1.1425[/tex]
The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]
[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]
[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]
[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (410.08)[/tex]
[tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
[tex]W = 4 \times N' \times W_{net[/tex]
where ;
[tex]N' = \dfrac{2400}{2}[/tex]
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.
Given the following data:
Diameter of bore = 3.7 inStroke length = 3.4 inClearance volume = 16% = 0.16Speed of 2400 RPM.Initial temperature = 60 F to R = 519.67 R. Initial pressure = 14.5 [tex]lbf/in^2[/tex] to [tex]lbf/ft^2[/tex] = 2088 [tex]lbf/ft^2[/tex] Maximum temperature = 5200 R.Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]
How to calculate the net work per cycle.First of all, we would determine the volume, mass and specific energy as follows:
[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]
For the mass:
[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]
M = 0.0019 lb.
At a temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]
For the relative volume at the second state, we have:
[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]
Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.
At a maximum temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]
For the relative volume at state 4, we have:
[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]
Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.
Now, we can calculate the net work per cycle by using this following formula:
[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]
W = 0.7792 Btu.
How to calculate the power developed.In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:
[tex]W=4N'W_{net}[/tex]
But;
[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]
Substituting the given parameters into the formula, we have;
[tex]W=4 \times 20 \times 0.7792[/tex]
W = 62.336 Btu/sec.
In horsepower:
W = 88.20 hp.
Read more on net work here: https://brainly.com/question/10119215
At steady state, a refrigerator whose coefficient of performance is 3 removes energy by heat transfer from a freezer compartment at 0 degrees C at the rate of 6000 kJ/hr and discharges energy by heat transfer to the surroundings, which are at 20 degrees C. a) Determine the power input to the refrigerator and compare with the power input required by a reversible refrigeration cycle operating between reservoirs at these two temperatures. b) If electricity costs 8 cents per kW-hr, determine the actual and minimum theoretical operating costs, each in $/day
Answer:
(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost = 0.234 $/day
Explanation:
Solution
Given that:
The coefficient of performance is =3
Heat transfer = 6000kJ/hr
Temperature = 20°C
Cost of electricity = 8 cents per kW-hr
Now
The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.
Thus
(A)The coefficient of performance is given below:
COP = Heat transfer from freezer/Power input
3 =6000/P
P =6000/3
P= 2000
P = 2000 kJ/hr = 2000/(60*60) kW
= 2000 (3600)kW
= 0.55 kW
Thus
The ideal coefficient of performance = T_low/(T_high - T_low)
= (0+273)/(20-0)
= 13.65
So,
P ideal = 6000/13.65 = 439.6 kJ/hr
= 439.6/(60*60) kW
= 0.122 kW
(B)For the actual cost we have the following:
Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour
= 0.044*24 $/day
= 1.056 $/day
For the theoretical cost we have the following:
Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour
= 0.00976*24 $/day
= 0.234 $/day
Suppose a student carrying a flu virus returns to an isolated college campus of 9000 students. Determine a differential equation governing the number of students x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it. (Usek > 0for the constant of proportionality and x forx(t).)
Answer:
dx/dt = kx(9000-x) where k > 0
Explanation:
Number of students in the campus, n = 9000
Number of students who have contracted the flu = x(t) = x
Number of students who have bot yet contracted the flu = 9000 - x
Number of Interactions between those that have contracted the flu and those that are yet to contract it = x(9000 - x)
The rate of spread of the disease = dx/dt
Note: the rate at which the disease spread is proportional to the number of interactions between those that have contracted the flu and those that have not contracted it.
[tex]\frac{dx}{dt} \alpha [x(9000 -x)]\\[/tex]
Introducing a constant of proportionality, k:
dx/dt = kx(9000-x) where k > 0
For the following peak or rms values of some important sine waves, calculate the corresponding other value:
(a) 117 V rms, a household-power voltage in North America
(b) 33.9 V peak, a somewhat common peak voltage in rectifier circuits
(c) 220 V rms, a household-power voltage in parts of Europe
(d) 220 kV rms, a high-voltage transmission-line voltage in North America
Answer:
A) V_peak ≈ 165 V
B) V_rms ≈ 24 V
C) V_peak ≈ 311 V
D) V_peak ≈ 311 KV
Explanation:
Formula for RMS value is given as;
V_rms = V_peak/√2
Formula for peak value is given as;
V_peak = V_rms x √2
A) At RMS value of 117 V, peak value would be;
V_peak = 117 x √2
V_peak = 165.46 V
V_peak ≈ 165 V
B) At peak value of 33.9 V, RMS value would be;
V_rms = 33.9/√2
V_rms = 23.97 V
V_rms ≈ 24 V
C) At RMS value of 220 V, peak value is;
V_peak = 220 × √2
V_peak = 311.13 V
V_peak ≈ 311 V
D) At RMS value of 220 KV, peak value is;
V_peak = 220 × √2
V_peak = 311.13 KV
V_peak ≈ 311 KV
If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is the actual measure resistance of the circuit?
Answer:
The answer is 2.25 kΩ
Explanation:
Solution
Given that:
The resistance reading on a DMM'S meter face = 22.5 ohms
The range selector switch = R * 100 range,
We now have to find the actual measure resistance of the circuit which is given below:
The actual measured resistance of the circuit is=R * 100
= 22.5 * 100
=2.25 kΩ
Hence the measured resistance of the circuit is 2.25 kΩ
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Answer:
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Explanation:
what is the difference between erratic error and zero error
The negative mark is balanced by a positive mark on the set key scale while the jaws are closed.
It is common practice to shut the jaws or faces of the system before taking some reading to guarantee a zero reading. If not, please take care of the read. This read is referred to as "zero defect."
There are two forms of zero error:
zero-mistake positive; and
Non-null mistake.
----------------------------
Hope this helps!
Brainliest would be great!
----------------------------
With all care,
07x12!
which of the following tells the computer wha to do
operating system
the ROM
the motherboard
the monitor
Air at 80 °F is to flow through a 72 ft diameter pipe at an average velocity of 34 ft/s . What diameter pipe should be used to move water at 60 °F and average velocity of 71 ft/s if Reynolds number similarity is enforced? The kinematic viscosity of air at 80 °F is 1.69E-4 ft^2/s and the kinematic viscosity of water at 60 °F is 1.21E-5 ft^2/s. Round your answer (in ft) to TWO decimal places.
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity [tex]v_{air}[/tex] of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity [tex]v_{water}[/tex] of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
[tex]D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}[/tex]
[tex]D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft[/tex]
[tex]D_{water} =[/tex] 2.4686 ft
[tex]D_{water} =[/tex] 2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
2.4686 ft
2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft
How old are you? answer this question plz lol I will mark someone as brainliest
Answer:
100000000000000000000000
13- Convert the following numbers to the indicated bases. List all intermediate steps.
a- (36459080)10 to octal
b- (20960032010 to hexadecimal
c- (2423233303003040)s to base
25 36459080/8= 4557385 0/8 209600320/16=13100020 + 0/16 (2423233303003040)5 (36459080)10 =( 18 (209600320)10=( 1)16 (2423233303003040)5=( )125
Answer:
Following are the conversion to this question:
Explanation:
In point (a):
[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]
[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]
In point (b):
[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]
In point (c):
[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]
[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]
Symbols of Base 25 are as follows:
[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]
A gold vault has 3 locks with a key for each lock. Key A is owned by the
manager whilst Key B and C are in the custody of the senior bank teller
and the trainee bank teller respectively. In order to open the vault door at
least two people must insert their keys into the assigned locks at the same
time. The trainee bank teller can only open the vault when the bank
manager is present in the opening.
i) Determine the truth table for such a digital locking system (4 marks)
ii) Derive and minimize the SOP expression for the digital locking system
Answer:
i) Truth Table:
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) The minimized sum of products (SOP) expression is
O = AC + AB
Explanation:
We have three inputs A, B and C
Let O is the output.
We are given two conditions to open the vault door:
1. At least two people must insert their keys into the assigned locks at the same time.
2. The trainee bank teller (C) can only open the vault when the bank manager (A) is present in the opening.
i) Construct the Truth Table
A | B | C | O
0 | 0 | 0 | 0
0 | 0 | 1 | 0
0 | 1 | 0 | 0
0 | 1 | 1 | 0 (condition 2 not satisfied)
1 | 0 | 0 | 0
1 | 0 | 1 | 1 (both conditions satisfied)
1 | 1 | 0 | 1 (both conditions satisfied)
1 | 1 | 1 | 1 (both conditions satisfied)
ii) SOP Expression using Karnaugh-Map:
A 3 variable Karnaugh-map is attached.
The minimized sum of products (SOP) expression is
O = AC + AB
The orange pair corresponds to "AC" and the purple pair corresponds to "AB"
Bonus:
The above expression may be realized by using two AND gates and one OR gate.
Please refer to the attached logic circuit diagram.
Question 44
What should you do if you encounter a fishing boat while out in your vessel?
A
Make a large wake nearby.
B
Avoid making a large wake.
с
Pass on the side with the fishing lines.
D
Pass by close to the anglers.
Submit Answer
Answer:
The answer is B. Avoid making a large wake.
Explanation:
When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.
You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.
A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on a 30-kg carrier C until it impacts the end of the carrier.Knowing the impact between B and C is perfectly plastic determine (a) velocity of the bullet and B after the first impact, (b) the final velocity of the carrier
(Distance between C and B is 0.5 m)
Answer:
a.) 4.46 m/s
b.) 0.41 m/s
Explanation:
a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg
Velocity V = 450 m/s
From conservative of linear momentum,
Sum of momentum before impact = Sum of momentum after impact
0.03 × 450 = (0.03 + 3 ) × v₂
v₂ = 13.5/3.03 = 4.4554 m/s
Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately
(b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.
(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃
Where:
Mass of the carrier m₃ = 30 kg
Substitute all the parameters into the formula
3.03×4.4554 = (3.03 +30) × v₃
v₃ = 13.5 / 33.03 = 0.40872 m/s
Therefore the velocity of the carrier is 0.41 m/s approximately.
Air flows along a horizontal, curved streamline with a 20 foot radius with a speed of 100 ft/s. Determine the pressure gradient normal to the streamline.
Answer:
- 1.19 lb/ft^3
Explanation:
You are given the following information;
Radius r = 20 ft
Speed V = 100 ft/s
You should use Bernoulli equation pertaining to streamline. That is, normal to streamline.
The pressure gradient = dp/dn
Where air density rho = 0.00238 slugs per cubic foot.
Please find the attached files for the solution and diagram.
Many HVACR industry publications are published by
Answer:
HVACR Industry Trade Groups
Explanation: