Answer:11. False 12. True 13. True 14. True 15. True 16. True 17. False 18. False 19. True 20. False
Explanation:
True/False Indicate whether the statement is true or false. 11. Electronic leak detectors can only be used to detect CFCs. 12. Disposable cylinders are designed to be used for refrigerant recovery. 13. Refrigerant must be recovered from a leaking system before repairs can be performed. 14. A gauge that reads absolute pressure should be adjusted depending on the elevation. 15. The pressure of a perfect vacuum is 29.92 in. Hg. 16. Both valves need to be closed on a gauge manifold in order to evacuate a system. 17. A decrease in temperature can cause the pressure in a cylinder to increase. 18. Service valves are front seated for normal system operation. 19. The compound gauge measures pressure both above and below atmospheric pressure. 20. Fluorescent leak detectors require the use of a halogen light.
The statements are determined whether true or false for the refrigerants and Disposable cylinders.
Given data:
False. Electronic leak detectors are capable of detecting a wide range of refrigerants, not just CFCs. They can detect various types of refrigerants such as HFCs, HCFCs, and others.
True. Disposable cylinders are designed to be used once for specific purposes, such as refrigerant recovery. They are typically not refillable and are intended to be discarded after use.
True. It is necessary to recover refrigerant from a leaking system before performing repairs. This ensures that the refrigerant is not released into the atmosphere, which is environmentally harmful. Proper recovery helps to comply with regulations and minimize refrigerant loss.
False. A gauge that reads absolute pressure does not require adjustment based on elevation. Absolute pressure readings are already referenced to a complete vacuum and are not affected by atmospheric pressure changes at different elevations.
False. The pressure of a perfect vacuum is 0 in. Hg. A perfect vacuum implies the absence of any gas or pressure.
True. When evacuating a system using a gauge manifold, it is essential to close both valves to isolate the system and create a proper vacuum.
False. According to the ideal gas law, a decrease in temperature typically results in a decrease in pressure, assuming the volume and the amount of gas remain constant.
False. Service valves, also known as shutoff valves or isolation valves, are typically back seated (closed) for normal system operation to prevent refrigerant flow. Front seating refers to opening the valve fully.
True. A compound gauge is designed to measure pressure both above and below atmospheric pressure. It usually has two scales or sections, one for positive pressure (above atmospheric) and one for negative pressure (below atmospheric).
False. Fluorescent leak detectors do not require the use of a halogen light. They utilize fluorescent dyes or additives that react with refrigerants to emit a fluorescent glow when illuminated with UV (ultraviolet) light. The UV light source can be separate from the detector itself.
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A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed elastically in tension with a force of 49100 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34 respectively, Determine the following:(a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. The entry field with incorrect answer now contains modified data.(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.
Answer:
0.6727
-0.02017
Explanation:
diameter = 15.7
l = 178
E =elastic modulus = 67.1 Gpa
poisson ratio = 0.34
p = force = 49100N
first we calculate the area of the cross section
[tex]A=\frac{\pi }{4} d^{2}[/tex]
[tex]A=\frac{\pi }{4} (15.7)^{2} \\A = \frac{774.683}{4} \\[/tex]
A = 193.6mm²
1. Change in directon of the applied stress
[tex]= \frac{pl}{AE}[/tex]
= 49100*178/193.6*67.1*10³
= [tex]=\frac{8739800}{12990560}[/tex]
δl = 0.6727 mm
2. change in diameter of the specimen
equation for poisson distribution =
m = -(δd/d) / (δl/l)
0.34 = (δd/15.7) / (0.6727/178)
0.34 = (-δd * 178) / 15.7 * 0.6727
0.34 = -178δd / 10.56139
we cross multiply
10.56139*0.34 =-178δd
3.5908726 = -178δd
δd = 3.5908726/-178
δd = -0.02017 mm
the change in dimeter has a negative sign. it decreases
Tech A says that LED brake lights illuminate faster than incandescent bulbs. Tech B says that LED brake lights have
more visibility and last longer. Who is correct?
Answer:
Both
Explanation:
who was part of dempwolf his firm when he first started
Explanation:
Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.
Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligible medium resistance. The cake solids (dry basis) per volume of filtrate was 20 g/liter. It is desired to operate a larger rotary vacuum filter (diameter 8 m and length 12 m) at a vacuum pressure of 80 kPA with a cake formation time of 20 s and a cycle time of 60 s. Determine the filtration rate in volumes/hr expected for the rotary vacuum filter.
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( [tex]\frac{60}{20}[/tex] ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
If the hypotenuse of a right triangle is 12 and an acute angle is 37 degrees find leg a and leg b lengths
scrity añao devid codicie
What current works best when the operator
encounters magnetic arc blow?
•DCEP
•ACEN
•CC
•AC
Answer:
AC
Explanation:
One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.
Current works best when the operator encounters magnetic arc blow is AC
Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.
This is caused as a result of the following;
- if the material being welded has residual magnetism at an intolerable level
- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).
Since it is caused by DC(Direct Current) which means constant polarity , it means the opposite will be better which is AC(alternating current) because it means that electricity direction will be switching to and fro and as such the polarity will also be revered in response to this back and forth switch manner.Thus, Current works best when the operator encounters magnetic arc blow is AC
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An example of a power consuming device would be a(n) ____ while an example of a non-power consuming device would be a(n) ____.
An example of a power consuming device would be a(n) light bulb, a computer device while an example of a non-power consuming device would be a(n) switch or button.
What is power consumption?The quantity of energy utilized per unit of time is known as power consumption. Power use is a significant factor in digital systems.
Power consumption is a limiting factor for the battery life of portable devices like laptop computers and cell phones.
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Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 399 MPa causes the specimen to plastically elongate to a length of 54.4 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.7 mm to a length of 57.8 mm.
Answer:
The answer is "583.042533 MPa".
Explanation:
Solve the following for the real state strain 1:
[tex]\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}[/tex]
Solve the following for the real stress and pressure for the stable.[tex]\sigma_{r1}=K(\varepsilon_{r1})^{n}[/tex]
[tex]K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}[/tex]
Solve the following for the true state stress and stress2.
[tex]\sigma_{r2}=K(\varepsilon_{r2})^n[/tex]
[tex]=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa[/tex]
An isolated pretimed signalized intersection has an approach with a traffic flow rate of 750 veh/h and a saturation flow rate of 3200 veh/h. This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay a) 4,6 s b) 30.2 s c) 34.8 s d) 35.0 s
Answer: This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27). A) 34.8 s.
Describe how to contribute to
zero/low carbon work outcomes
within the built environment.
Answer:
day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes
Explanation:
If the sum of the two numbers is 4 and the sum of their squares minus three times their product is 76,find the number
Answer:
-2 and 6
Explanation:
Let "x" and "y" be 2 numbers.
The sum of the two numbers is 4. The mathematical expression is:
x + y = 4
y = 4 - x [1]
The sum of their squares minus three times their product is 76. The mathematical expression is:
x² + y² - 3 x y = 76 [2]
If we substitute [1] in [2], we get:
x² + (4 - x)² - 3 x (4 - x) = 76
x² + 16 - 8 x + x² - 12 x + 3 x² = 76
5 x² - 20 x - 60 = 0
We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.
If we replace these x values in [1], we get:
y₁ = 4 - x₁ = 4 - 6 = -2
y₂ = 4 - x₂ = 4 - (-2) = 6
As a consequence, one of the numbers is 6 and the other is -2.
The pressure less than atmospheric pressure is known as:
Suction pressure
Negative gauge pressure
Vacuum pressure
All of the above
Answer:
answer is option (d) all of the above
The driver of the truck has an acceleration of 0.4g as the truck passes over the top A of the hump in the road at constant speed. The radius of curvature of the road at the top of the hump is 98 m, and the center of mass G of the driver (considered a particle) is 2 m above the road. Calculate the speed v of the truck.
Answer:
19.81 m/s
Explanation:
The total acceleration of the truck (a) is due to the centripetal acceleration and as a result of the linear acceleration. Therefore the total acceleration (a) is given by:
[tex]a^2=a_n^2+a_t^2\\\\where\ a_n=centripetal\ acceleration=\frac{v^2}{r},a_t=linear \ acceleration\\\\But\ since\ the \ speed\ is \ constant, the \ linear \ acceleration(a_t)\ would\ be\ 0.\\\\a^2=a_n^2+a_t^2\\\\a^2=a_n^2\\\\a=a_n=\frac{v^2}{r} \\\\v^2=ar\\\\v=\sqrt{ar} \\\\a=0.4g=0.4*9.81,r=98\ m+2\ m=100\ m\\\\v=\sqrt{0.4*9.81*100} \\\\v=19.81\ m/s[/tex]
A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16.5 s later. The descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude. Assume that g = 32.2 ft/s2.
Determine
(a) the speed v1 of the rocket at the end of powered flight,
(b) the maximum altitude reached by the rocket.
Answer:
[tex]u = 260.22m/s[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Explanation:
Given
[tex]S_0 = 89.6ft[/tex] --- Initial altitude
[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds
[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
[tex]S = ut + \frac{1}{2}at^2[/tex]
The final altitude after 16.5 seconds is represented as:
[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]
Substitute the following values:
[tex]S_0 = 89.6ft[/tex] [tex]S_{16.5} = 0ft[/tex] [tex]a = -g = -32.2ft/s^2[/tex] and [tex]t = 16.5[/tex]
So, we have:
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]
[tex]0 = 89.6 + 16.5u- 4383.225[/tex]
Collect Like Terms
[tex]16.5u = -89.6 +4383.225[/tex]
[tex]16.5u = 4293.625[/tex]
Make u the subject
[tex]u = \frac{4293.625}{16.5}[/tex]
[tex]u = 260.21969697[/tex]
[tex]u = 260.22m/s[/tex]
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
[tex]v=u + at[/tex]
At the maximum height:
[tex]v =0[/tex] --- The final velocity
[tex]u = 260.22m/s[/tex]
[tex]a = -g = -32.2ft/s^2[/tex]
So, we have:
[tex]0 = 260.22 - 32.2t[/tex]
Collect Like Terms
[tex]32.2t = 260.22[/tex]
Make t the subject
[tex]t = \frac{260.22}{ 32.2}[/tex]
[tex]t = 8.08s[/tex]
The maximum height is then calculated as:
[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]
This gives:
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]
[tex]S_{max} = 1141.0676[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Hence, the maximum height is 1141.07ft
Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per year. On the second site, the wind blows steadily at 10 m/s for 2000 hours per year. The density of air on the both sites is 1.25 kg/m3 . Assuming the wind power generation is negligible during other times.Calculate the maximum power of wind on each site per unit area, in kW/m2 .
Solution :
Given :
[tex]$V_1 = 7 \ m/s$[/tex]
Operation time, [tex]$T_1$[/tex] = 3000 hours per year
[tex]$V_2 = 10 \ m/s$[/tex]
Operation time, [tex]$T_2$[/tex] = 2000 hours per year
The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]
The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]
[tex]$= \dot{m} \times 0.5 V^2$[/tex]
The power generation is the time rate of the kinetic energy which can be calculated as follows:
Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]
Regarding that [tex]$\dot m \propto V$[/tex]. Then,
Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]
Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.
For the first site,
Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]
[tex]$P_1 = \text{const.} \times 343 \ W$[/tex]
For the second site,
Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]
[tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]
I have an AC waveform with the voltage peak value of 100 V. I'm trying to find the RMS value of the voltage
Answer:
98 x 100=9,00
Explanation:
With _____, only one criterion must evaluate true in order for a record to be selected and with _____, all criteria must be evaluate true in order for a record to be selected.
a. parameter criteria, double criteria
b. function criteria, IF criteria
c. simple criteria, complex criteria
d. OR criteria, AND criteria
Answer:
d
Explanation:
OR criteria, AND criteria
In an OR criteria, it doesn't need all the records to be true. Just one record is enough and all other criterion becomes true.
In an AND criteria, it's unlike the OR criteria and works in opposite. It needs every member of the record to be true to be able to adjudge the whole record as true.
And as such, we have
With OR criteria, only one criterion must evaluate true in order for a record to be selected and with AND criteria, all criteria must be evaluate true in order for a record to be selected.
During a shrinkage limit test, a 19.3 cm3 saturated clay sample with a mass of 37 g was placed in a porcelain dish and dried in the oven. The oven-dried sample had a mass of 28 g with a final volume of 16 cm3 . Determine the shrinkage limit and the shrinkage ratio.
Answer:
shrinkae limit = 20.35%
shrinkage ratio = 1.45
Explanation:
1. to get the shrinkage limit we would first calculate the moisture content w.
w = (37-28)/28
= 9/28
= 0.3214
then the formula for shrinkage limit is
[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]
w = 0.3214
V = 19.3
Vd = 16
Wd = 28
when we put these values into the formula:
[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]
= 20.35%
2. the shrinkage limit = Wd/V
= 28/19.3
= 1.45
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.
Answer:
Explanation:
From the given information:
The equation for applied stress can be expressed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]
where;
[tex]\phi[/tex] = angle between the applied stress [100] and [111]
To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system
Using the equation:
[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [111]
[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]
Thus;
[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]
[tex]\phi= 54.74^0[/tex]
To determine [tex]\lambda[/tex] for [tex][1 \overline 1 0][/tex]
where;
for [100]
[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]
for [tex][1 \overline 1 0][/tex]
[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]
Thus;
[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]
[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]
[tex]\phi= 45^0[/tex]
Thus, the magnitude of the applied stress can be computed as:
[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]
[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]
[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]
This is silence I couldent find the tab... 30 points plus marked brainliest if corrects!
The most recent evidence supporting the theory of plate tectonics would include
es )
A)
GPS monitoring of plate speeds and movements.
B)
the WWII discovery of paleomagnetic reversals.
Elimi
O
the 1963 mapping of the tectonic plate boundaries.
D
C-14 dating of marine fossils found in the Himalayas.
Find the derivative of x
Answer:
this is your answer. if mistake don't mind.
Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.
When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.
Pls help im giving 100 points for this i have this due in minutes
Answer:
u need to plan it out
Explanation:
u need to plan it out
Answer:
use the turn 1 degrees option and put a repeat loop on it
Explanation:
u can add sound in ur loop
What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them
An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.
Solution :
Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :
[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]
[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]
[tex]$h_T=46.3 \ W/m^2 .K$[/tex]
The total heat transfer coefficient will be :
[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]
[tex]$=296.3 \ W/m^2 .K$[/tex]
Now calculating the maximum volumetric heat generation :
[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]
[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]
[tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]
The heat generation inside the wire is given as :
[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]
Here, R is the resistance of the wire
V is the volume of the wire
∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]
[tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]
where, ρ is the resistivity.
[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]
[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]
= 28.96 A
Now considering the relation for the current flow through the finite potential difference.
[tex]$E=I_{max} \times R$[/tex]
[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]
[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]
[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]
= 2.983 m
Now calculating the power rating of the heater:
[tex]$P= E \times I_{max}$[/tex]
[tex]$P= 110 \times 28.96}$[/tex]
= 3185.6 W
= 3.1856 kW
Which of the following choices accurately contrasts a categorical syllogism with a conditional syllogism?
An argument constructed as a categorical syllogism uses deductive reasoning whereas an argument constructed as a conditional syllogism uses inductive reasoning.
A categorical syllogism contains two premise statements and one conclusion whereas a conditional syllogism contains one premise statement and one conclusion.
A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.
An argument constructed as a categorical syllogism is valid whereas an argument constructed as a conditional syllogism is invalid.
Answer:
The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.
Explanation:
As,
Categorical syllogisms follow an "If A is part of C, then B is part of C" logic.
Conditional syllogisms follow an "If A is true, then B is true" pattern of logic.
So,
The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.
I don’t know the answer to this question
Answer:
I dont know the answer either
Explanation:
Answer:
flux
Explanation:
what is the distance in term of wavelengh between successive minima in the standing wave ratio
Answer:
hi there
Explanation:
how skateboards works?
Answer
The skateboarder applies pressure to the trucks and gives/releases pressure on the levers. Second, the wheels and the axles are also examples of simple machines. They help the skater ride, spin, grind, and do a bunch of other radical movements on a skateboard.:
Explanation:
A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:
a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.
Answer:
hello some parts of your question is missing attached below is the missing part
answer :
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
Explanation:
Given data :
50-mm cube of graphite fiber reinforced polymer matrix
subjected to 125-KN force in direction 2,
direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3
A) Determine changes in the 50-mm dimensions
The changes are : 0.006mm compression in y-direction
0.002 mm expansion in x and z directions
B) the stress required are evenly distributed
attached below is the detailed solution
A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade
Solution :
Finding the cohesion of the soil(c) using the relation:
[tex]$c = \frac{q_u}{2}$[/tex]
Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;
[tex]$c = \frac{800}{2}$[/tex]
= 400 psf
∴ The cohesion value is greater than 0
So the use of the angle of internal friction is 0
Referring to the table relation between bearing capacity factors and angle of internal friction.
For the angle of inter friction [tex]$0^\circ$[/tex]
[tex]$N_c = 5.14$[/tex]
[tex]$N_q = 1.0$[/tex]
[tex]$N_r = 0$[/tex]
Therefore,
[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]
= 2386 psf
∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]
[tex]$=\frac{30}{B^2}$[/tex]
∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]
[tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]
∴ B = 0.2 ft
Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft
[tex]$=0.04 \ ft^2$[/tex]