Answer: A. Makes a commodity available for sale or exchange
Explanation: hope it helps ^w^
Which is currently the most supported theory about the future of the universe?
a.The universe could remain exactly as it is today.
b.The universe may continue to expand.
c.Stars could burn out, causing the universe to become dark and cold.
d.Gravity could pull galaxies back together, causing a reverse of the big bang.
Answer:
b. The universe may continue to expand.
Explanation:
The most supported theory about the future of the universe is that, it may continue to expand up to such point that it will become too cold to allow any forms of life to pull through. This is popularly known as the "Big Freeze."
Such acceleration of expansion is caused by the so-called, "dark energy." This also explains why the acceleration of expansion is not constant. This thereby causes the distance between galaxies to grow farther apart.
Answer:
b
Explanation:
edge 2021
200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.
Answer:
a) the tension in the cord T is 1200 N
b)
reaction at C along x-axis Cy is 400 N
reaction at C along y-axis Cy is 1200 N
reaction at C along z-axis Cz is 0 N
reaction at D along X-axis Dx is 1600 N
reaction at D along y-axis Dy is -480 N
Explanation:
a)
to find the tension in the cord, we say;
∑ Mz = 0
720(200) - T(120) = 0
144000 - 120T = 0
T = 144000/120
T = 1200 N
the tension in the cord T is 1200 N
b)
the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.
we make use of the moment law of equilibrium at point D about y-axis.
∑ MDy = 0
Cx(120) - T(40) = 0
we substitute value of T
Cx(120) - 1200(40) = 0
Cx 120 = 48000
Cx = 48000/120
Cx = 400 N
reaction at C along x-axis Cy is 400 N
Next we also apply the moment law of equilibrium at point D about x-axis.
∑MD z = 0
-Cy(120) + 720(80 + 120) = 0
-Cy(120) = - 144000
Cy = -144000 / - 120
Cy = 1200 N
reaction at C along y-axis Cy is 1200 N
we also apply the force law of equilibrium along z direction
∑Fz = 0
Cz = 0 N
reaction at C along z-axis Cz is 0 N
we also apply the force law of equilibrium along x direction
∑Fx = 0
Cx + Dx + T = 0
Substitute 400N for Cx and 1200 N for Dx
so
400 + Dx + 1200 = 0
Dx = 1600 N
therefore reaction at D along X-axis Dx is 1600 N
we also apply the force law of equilibrium along y direction
∑ Fy = 0
Cy + Dy - 720 = 0
we substitute Cy = 1200 N
so
1200 + Dy - 720 = 0
Dy = - 480 N
therefore reaction at D along y-axis Dy is -480 N
Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 900 cfs. Precipitation over the lake is 32 inches/year and evaporation over the lake is 55 inches/year; the area of the lake is 47,000 ac. If an average flow of 300 cfs must be released from the dam for the benefit of fish and downstream water users, calculate the amount of water that can be withdrawn from the lake to provide water supply for the Triangle area. Assume any other source/sink of water (such as groundwater), is negligible.
Answer:
The lake can withdraw a maximum of [tex]1.464\times 10^{10}[/tex] cubic feet per year to provide water supply for the Triangle area.
Explanation:
The maximum amount of water that can be withdrawn from the lake is represented by the following formula:
[tex]V = V_{in}+V_{p}-V_{e}-V_{out}[/tex] (Eq. 1)
Where:
[tex]V[/tex] - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.
[tex]V_{in}[/tex] - Inflow amount of water, measured in cubic feet per year.
[tex]V_{out}[/tex] - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.
[tex]V_{p}[/tex] - Amount of water due to precipitation, measured in cubic feet per year.
[tex]V_{e}[/tex] - Amount of evaporated water, measured in cubic feet per year.
Then, we can expand this expression as follows:
[tex]V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t[/tex]
[tex]V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}[/tex] (Eq. 2)
Where:
[tex]f_{in}[/tex] - Average watershed inflow, measured in cubic feet per second.
[tex]f_{out}[/tex] - Average flow to be released, measured in cubic feet per second.
[tex]\Delta t[/tex] - Yearly time, measured in seconds per year.
[tex]h_{p}[/tex] - Change in lake height due to precipitation, measured in feet per year.
[tex]h_{e}[/tex] - Change in lake height due to evaporation, measured in feet per year.
[tex]A_{l}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]f_{in} = 900\,\frac{ft^{3}}{s}[/tex], [tex]f_{out} = 300\,\frac{ft^{3}}{s}[/tex], [tex]\Delta t = 31,536,000\,\frac{second}{yr}[/tex], [tex]h_{p} = 32\,\frac{in}{yr}[/tex], [tex]h_{e} = 55\,\frac{in}{yr}[/tex] and [tex]A_{l} = 47,000\,acres[/tex], the available amount of water for supply purposes in the Triangle area is:
[tex]V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex][tex]V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}[/tex]
The lake can withdraw a maximum of [tex]1.464\times 10^{10}[/tex] cubic feet per year to provide water supply for the Triangle area.
A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a step-down transformer at the load end of a 2400-volt feeder whose series impedance is (1.0 + j2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the highvoltage (primary) side. The transformer is delivering rated load at a 0.8 power factor lagging and at a rated secondary voltage. Neglecting the transformer exciting current.
Determine:
a. The voltage at the transformer primary terminals
b. The voltage at the sending end of the feeder.
c. The real and reactive power delivered to the sending end of the feeder.
Drag each tile to the correct box. Not all tiles will be used.
Adam wants to become a certified professional engineer. What are the steps that he will have to follow?
Answer:
I think it is the 2,3,5 and 1 ones
How much would you spend on gasoline each year if you drove 10,000 miles over the year and your vehicle achieves 15 miles per gallon with gasoline prices at $4.00 a gallon? Now substitute your vehicle with a hybrid-electric automobile that achieves 60 miles per gallon. Calculate the yearly cost for fuel with this vehicle
Answer:
a. [tex]Total\ Cost = \$2667[/tex]
b. [tex]Total\ Cost = \$667[/tex]
Explanation:
(a)
Given
[tex]Distance = 10000\ miles[/tex]
[tex]Rate = 15\ miles/gallon[/tex]
[tex]Price = \$4.00/gallon[/tex]
Required
Determine the total amount spent in a year
First, we need to determine the number of gallons used in a year;
[tex]Total\ Gallons = Distance/Rate[/tex]
[tex]Total\ Gallons = 10000miles /\frac{15miles}{gallon}[/tex]
[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{15\ miles}[/tex]
[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{15}[/tex]
[tex]Total\ Gallons = \frac{10000\ gallons}{15}[/tex]
[tex]Total\ Gallons = \frac{10000}{15}\ gallons[/tex]
Next, is to determine the total cost:
[tex]Total\ Cost = Total\ Gallons * Price[/tex]
[tex]Total\ Cost = \frac{10000}{15}\ gallon * \frac{\$4}{gallon}[/tex]
[tex]Total\ Cost = \frac{10000}{15} * \$4[/tex]
[tex]Total\ Cost = \frac{10000 * \$4}{15}[/tex]
[tex]Total\ Cost = \frac{\$40000}{15}[/tex]
[tex]Total\ Cost = $2666.66666667[/tex]
[tex]Total\ Cost = \$2667[/tex] (approximated)
(b)
Given
[tex]Rate = 60\ miles/gallon[/tex]
Required
Determine the total amount spent in a year
First, we need to determine the number of gallons used in a year;
[tex]Total\ Gallons = Distance/Rate[/tex]
[tex]Total\ Gallons = 10000miles /\frac{60\ miles}{gallon}[/tex]
[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{60\ miles}[/tex]
[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{60}[/tex]
[tex]Total\ Gallons = \frac{10000\ gallons}{60}[/tex]
[tex]Total\ Gallons = \frac{10000}{60}\ gallons[/tex]
Next, is to determine the total cost:
[tex]Total\ Cost = Total\ Gallons * Price[/tex]
[tex]Total\ Cost = \frac{10000}{60}\ gallon * \frac{\$4}{gallon}[/tex]
[tex]Total\ Cost = \frac{10000}{60} * \$4[/tex]
[tex]Total\ Cost = \frac{10000 * \$4}{60}[/tex]
[tex]Total\ Cost = \frac{\$40000}{60}[/tex]
[tex]Total\ Cost = \$666.666666667[/tex]
[tex]Total\ Cost = \$667[/tex] (approximated)
Driving next to another vehicle can sometimes take away the option to _____.
A. steer around a hazard
B. improve your mileage
C. conduct your scan
D. manage your tire wear
Answer: A. steer around a hazard
Explanation: say there was a vehicle next to you on your left and there was a pile of rocks it would prevent you from steer from going straight but instead have you steer around the rocks and cross into another lane that’s what it said in the lesson
Driving next to another vehicle can sometimes take away the option to A. steer around a hazard
What are Traffic Rules?This refers to the rules and regulations that guide road users and pedestrians to ensure their safety.
Hence, we can see that when driving on the road next to another vehicle, the option to steer around a hazard is taken away, and hence, it is not advisable to do this.
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1. Elevations are isometric drawings.
True or false
What is usually the strongest part of a unibody
A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3 /s. A factory dumps 4.3 m^3 /s of waste with 100 mg/L of the same non-conservative pollutant into the lake. The pollutant has a first order decay coefficient K of 0.10/day. Assuming the lake is well mixed, find the steady-state concentration of pollutant in the lake.
Answer:
12.84 mg/L
Explanation:
We are given;
Volume of lake; V = 1.1 x 10^(6) m³
decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s
Factory rate: Q_f = 4.3 m³/s
Factory concentration: C_f = 100 mg/L
Stream rate: Q_s = 34 m³/s
Stream Concentration: C_s = 2.3 mg/L
Now, to find the steady state concentration of pollutant in the lake, we will use the formula;
(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)
Where C_L is the steady state concentration of pollutant in the lake.
Thus, making C_L the subject, we have;
C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)
Plugging in the relevant values gives;
C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))
C_L = 12.84 mg/L
The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm (0.01 inn) thick. If the extruder head and polymer extrudate velocities are both 40mm/s, estimate the production time for the generation of a 38-mm (1.5 in., edge length) solid cube. Assume that there is a 15- second delay after deposition of each layer as the extruder head is moved over a wire brush for cleaning.
Answer:
The time taken will be "1 hour 51 min". The further explanation is given below.
Explanation:
The given values are:
Number of required layers:
= [tex]\frac{38}{0.25}[/tex]
= [tex]152 \ layers[/tex]
Diameter (d):
= 1.25 mm
Velocity (v):
= 40 mm/s
Now,
The area of one layer will be:
= [tex]38\times 38 \ mm^2[/tex]
= [tex]1444 \ mm^2[/tex]
The area covered every \second will be:
= [tex]d\times v[/tex]
= [tex]1.25\times 40[/tex]
= [tex]50 \ mm^2[/tex]
The time required to deposit one layer will be:
= [tex]\frac{1444}{50}[/tex]
= [tex]28.88 \ sec[/tex]
The time required for one layer will be:
= [tex]15 \ sec[/tex]
∴ Total times required for one layer will be:
= [tex]15+28.88[/tex]
= [tex]43.88 \ sec[/tex]
So,
Number of layers = 152
Therefore,
Total time will be:
= [tex]152\times 43.88[/tex]
= [tex]6669.76 \ sec[/tex]
= [tex]1 \ hour \ 51 \ min[/tex]
Given values are:
Number of layers required = [tex]\frac{38}{0.25}[/tex] = [tex]152 \ layers[/tex]Diameter (d) = [tex]1.75 \ mm[/tex]Velocity (v) = [tex]40 \ m/s[/tex]Now,
Area of one layer will be:
= [tex]38\times 38[/tex]
= [tex]1444 \ mm^2[/tex]
Area covered per second:
= [tex]d\times v[/tex]
= [tex]1.25\times 40[/tex]
= [tex]50 \ mm^2[/tex]
The time required to deposit 1 layer:= [tex]\frac{1444}{50}[/tex]
= [tex]28.88 \ sec[/tex]
The time required for one layer:= [tex]15 \ sec[/tex]
∴ Total time for 1 layer will be:
= [tex]15+28.88[/tex]
= [tex]43.88 \ sec[/tex]
hence,
The total time required:
= [tex]152\times 43.88[/tex]
= [tex]6669.76 \ sec[/tex]
= [tex]1 \ hour \ 51 \ minutes[/tex]
Thus the above response is right.
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When joining two pieces of wood for an exterior application,
should be used. *
Answer:
A tongue- and -groove joint
Explanation:
In this method, you first cut a a groove on one piece of the wood and then cut a matching tongue on the other piece. The cuts can be made using a dado head on a table saw. A glue is applied last in a line all around, thus it is vital to cut the tongue a bit smaller than the groove to allow gluing.
What should wheel bearing seals be checked for
Answer:
drugs
Explanation:
which statement about life on earth is true ?
Answer:
Humans have been on Earth for a very short amount of time.
Explanation:
edgeunity 2021
Answer:
Humans have been on Earth for a very short amount of time<3
D.
Explanation:
What components should you inspect if the crankshaft end play is out of specifications?
Answer:
gdyc ddxtfvytg4dgtfxdwcftcd3rcby
If the crankshaft end play is out of specifications, check for:
Thrust Bearings
Main Bearings
Crankshaft
Crankshaft Thrust Washers
Engine Block
To understand the crankshaft when it is out of specifications, check for:
Thrust Bearings: Check the condition of the thrust bearings, which are located at the front and/or rear of the engine block. Excessive wear or damage to the thrust bearings can contribute to increased crankshaft end play.
Main Bearings: Inspect the main bearings, which support the crankshaft within the engine block. Worn or damaged main bearings can cause excessive movement of the crankshaft.
Crankshaft: Check the crankshaft itself for any signs of damage, such as scoring or bending. A damaged crankshaft may not sit properly within the bearings, leading to increased end play.
Crankshaft Thrust Washers: Some engines use thrust washers to control the end play of the crankshaft. Inspect these washers for wear, damage, or improper installation.
Engine Block: Check the engine block for any signs of damage or distortion that may affect the alignment and support of the crankshaft.
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What is the full form of AWM
Answer:
its a gun
Explanation:
The Accuracy International AWM (Arctic Warfare Magnum or AI-Arctic Warfare Magnum) is a bolt-action sniper rifle manufactured by Accuracy International designed for magnum rifle cartridges. ...
Effective firing range: 1,100 m (1,203 yd) (.300 ...
Manufacturer: Accuracy International
In service: 1996–present
Draw an ERD for each of the following situations. (If you believe that you need to make additional assumptions, clearly state them for each situation.) Draw the same situation using the tool you have been told to use in the course. a. A company has a number of employees. The attributes of EMPLOYEE include Employee ID (identifier), Name, Address, and Birthdate. The company also has several projects. Attributes of PROJECT include Project ID (identifier), Project Name, and Start Date. Each employee may be assigned to one or more projects or may not be assigned to a project. A project must have at least one employee assigned and may have any number of employees assigned. An employee's billing rate may vary by project, and the company wishes to record the applicable billing rate (Billing Rate) for each employee when assigned to a particular project. Do the attribute names in this description follow the guidelines for naming attributes? If not, suggest better names. Do you have any associative entities on your ERD? If so, what are the identifiers for those associative entities? Does your ERD allow a project to be created before it has any employees assigned to it? Explain. How would you change your ERD if the Billing Rate could change in the middle of a project? b. A laboratory has several chemists who work on one or more projects. Chemists also may use certain kinds of equipment on each project. Attributes of CHEMIST include Employee ID (identifier), Name, and Phone No. Attributes of PROJECT include Project ID (identifier) and Start Date. Attributes of EQUIPMENT include Serial No and Cost The organization wishes to record Assign Date—that is, the date when a given equipment item was assigned to a particular chemist working on a specified project A chemist must be assigned to at least one project and one equipment item. A given equipment item need not be assigned, and a given project need not be assigned either a chemist or an equipment item. Provide good definitions for all of the relationships in this situation. c. A college course may have one or more scheduled sections or may not have a scheduled section. Attributes of COURSE include Course ID, Course Name, and Units. Attributes of SECTION include Section Number and Semester ID. Semester ID is composed of two parts: Semester and Year. Section Number is an integer (such as 1 or 2) that distinguishes one section from another for the same course but does not uniquely identify a section. How did you model SECTION? Why did you choose this way versus alternative ways to model SECTION? d. A hospital has a large number of registered physicians. Attributes of PHYSICIAN include Physician ID (the identifier) and Specialty. Patients are admitted to the hospital by physicians. Attributes of PATIENT include Patient ID (the identifier) and Patient Name. Any patient who is admitted must have exactly one admitting physician. A physician may optionally admit any number of patients. Once admitted, a given patient must be treated by at least one physician. A particular physician may treat any number of patients, or may not treat any patient& Whenever a patient is treated by a physician, the hospital wishes to record the details of the treatment (Treatment Detail). Components of Treatment Detail include Date, Time, and Results. Did you draw more than one relationship between physician and patient? Why or why not? Did you include hncnithi ac an antitv type? Why or why not?
Answer:
it wqas red
Explanation:
red
_____is the ability of a system to grow as the volume of users increases.
For others taking this class that have not found the answer
Answer:
Scalability
Explanation:
just took a test and it was right.
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 5008C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa. Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.
This question is incomplete, the complete question is;
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.
Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.
Answer:
- the quality of the steam exiting the second stage of the turbine is 0.9329
- the thermal efficiency is 36.05%
Explanation:
get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .
Specific enthalpy h1= 3192.3 kJ/kg
Specific entropy s1 = 5.9566 kJ/kg.K
Process 1 to 2s is isentropic expansion process in the turbine
S1 = S2s
get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K
h2s = 2822.2 kJ/kg
get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)
0.78 = (3192.3 - h2) / (3192.3 - 2822.2)
h2 = 2903.6 kJ/kg
get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C
h3 = 3422.2 kJ/kg
s3 = 6.8803 kJ/kg.K
Process 3 to 4s is isentropic expansion process in the turbine
S3 = S4s
get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K
h4s = 2118.8 kJ/kg
get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)
0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )
h4 = 2405.5 kJ/kg
get the properties at pressure, p5 = 6 kPa
h5 = hf
= 151.53 kJ/kg
v5 = Vf
= 0.0010064 m³/kg
get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at
p6 = p1 = 28 MPa
np = v5( p6 - p5) / (h6 - h5)
0.82 = ((0.0010064)( 28000 - 6)) / (h6 - 151.53)
h6 = 185.89 kJ/kg
Now to find the quality of the steam at the exit of the second stage of the turbine
At stat4, p4 = 6kPa
h4f = 151.53 kJ/kg
h4fg = 2415.9 kJ/kg
h4 = h4f + x4h4fg
2405.5 = 151.53 + (x4 (2415.9))
x4 = 0.9329
the quality of the steam exiting the second stage of the turbine is 0.9329
Also to find the efficiency of the power plant, we use the following equation;
n = Wnet / Qin
= (Wt1 + Wt2 - Wp) / (Q61 + Q23)
= [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]
[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]
= 0.3605
n = 36.05%
therefore the thermal efficiency is 36.05%
The next day at SLS found everyone in technical support busy restoring computer systems to their former state and installing new virus and worm control software. Amy found herself learning how to install desktop computer operating systems and applications as SLS made a heroic effort to recover from the attack of the previous day.
Required:
Do you think this event was caused by an insider or outsider?
Answer:
Yes, it was caused by an insider or outsider.
Explanation:
Yes, I believe that this event was caused by an insider or outsider. This is because, if a USB removable flash drive is attached to all the office computers by an insider and now an outsider sends a mail to the company’s General group mail; if the mail has a link that contains some infected worms and virus; once anybody clicks on the link in the mail then, all the computers will be affected by the virus.
Therefore, this event may be caused by the insider who clicked the infected link or by the outsider sent the infected link in the mail.
Oxygen initially is at 35°F and 16.8 lbf/in2. It fills a closed, rigid 5 ft3 tank filled with a paddle wheel. During the process, a paddle wheel provides 4 Btu of energy transfer by work to the gas. During the process, the gas temperature increases up to 90°F. Assuming ideal gas behavior and ignoring K.E and P.E effects, determine the mass of the oxygen in the tank, the amount of heat transfer in Btu, and the final pressure of the tank.
Answer:
m = 228 gm
ΔQ = 4.35 Btu
P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa
Explanation:
FOR MASS OF OXYGEN:
We will use ideal gas equation for initial conditions:
P₁V₁ = nRT₁
P₁V₁ = mRT₁/M
where,
P₁ = Initial Pressure = (16.8 lbf/in²)(6894.76 Pa/1 lbf/in²) = 1.15 x 10⁵ Pa
V₁ = Initial Volume = (5 ft³)(0.028316 m³/1 ft³) = 0.1415 m³
m = mass of oxygen = ?
R = Universal Gas Constant = 8.314 J/mol.k
T₁ = Initial Temperature = 35°F = 274.67 k
M = Molecular Mass of Oxygen = 32 gm
Therefore,
(1.15 x 10⁵ Pa)(0.1415 m³) = m(8.314 J/mol.k)(274.67 k)/(32 gm)
m = (1.15 x 10⁵ Pa)(0.1415 m³)(32 gm)/(8.314 J/mol.k)(274.67 k)
m = 228 gm
FOR AMOUNT OF HEAT TRANDFER:
Using First Law of Thermodynamics:
ΔQ = ΔU +W
Since, this is a constant volume process. Therefore, the work done in this process must be equal to zero:
ΔQ = ΔU + 0
ΔQ = ΔU
and,
ΔU = m Cv ΔT
Therefore,
ΔQ = m Cv ΔT
where,
ΔQ = Heat Transfer Amount = ?
m = mass of oxygen = 228 g = 0.228 kg
Cv = Molar Specific Heat of Oxygen at Constant Volume = 0.659 KJ/kg.k
ΔT = Change in temperature = 305.22 k - 274.67 k = 30.55 k
Therefore,
ΔQ = (0.228 kg)(0.659 KJ/kg.k)(30.55 k)
ΔQ = (4.6 KJ)(0.9478 Btu/1 KJ)
ΔQ = 4.35 Btu
FOR FINAL PRESSURE IN TANK:
Using equation of state:
P₁V₁/T₁ = P₂V₂/T₂
Here, the volume will be constant due to rigid tank. Hence,
V₁ = V₂ = V
Therefore,
P₁V/T₁ = P₂V/T₂
P₁/T₁ = P₂/T₂
(16.8 lbf/in²)/(35°F) = P₂/(90°F)
P₂ = (16.8 lbf/in²)(90°F)/(35°F)
P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa
list five properties of most transition metals
Turning operations that require heavy material removal typically use what setting on the
engine lathe?
1. Why does condensed water drip from the air-conditioning system?
Answer:
Your answer is: Evaporator, the condensed water drip comes from something called the evaporator. When warm air touches the cold coil ( or the cold pipe) the water vapor condensed while the cold coil absorbs.
Explanation:
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A house is maintained at a comfortable temperature by means of an electrical resistor heater during winter. The heater is operated at a constant current (I) under an applied voltage of 110 V. For simplicity, let us assume that the heater operates at a steady state, with 100% of conversion efficiency from the electrical energy to the internal energy of air in the house.
Required:
Find if the instructor pays $8.80/day for heating, with electricity cost $0.09/kWh.
b. Calculate the heat produced by such a heater hourly.
Answer:
$8.89
Explanation:
$8.80/a day for heat and $0.09kwh for electricity
Given a 250Ω strain gage with a gage factor of 1 which is mounted to a metal bar 0.6m long. The bar is stretched under a tension force and the resistance changes to 251.4 Ω. How much was the bar stretched? _ _mm (Answer in mm to 2 decimal places) What is the length of the bar after it is stretched?
Answer:
the bar was stretched by [tex]\mathbf{\Delta L = 3.36 \ mm}[/tex]
the length of the after it was stretched is [tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]
Explanation:
From the information given:
The strain gauge resistance R = 250 Ω
The gauge factor = 1
The original length L = 0.6 m = 600 mm
After the bar is being stretched under tension force;
the new resistance [tex]R_{new} = 251.42[/tex]
The gauge factor [tex]G = \dfrac{\Delta R/R}{\Delta L /L }[/tex]
where;
[tex]\Delta R = R_{new} - R[/tex] and [tex]\Delta L = L_{new} - L[/tex]
ΔR = 251.4 - 250
ΔR = 1.4 Ω
[tex]\Delta L = L_{new} - L[/tex]
[tex]L_{new} = L + L (\dfrac{\Delta R/R}{G})[/tex]
[tex]L_{new} = 0.6 + 0.6 (\dfrac{\Delta 1.4/250}{1})[/tex]
[tex]L_{new} = 0.60336 \ m[/tex]
[tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]
Thus, the length of the after it was stretched is [tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]
Thus, the bar was stretched by [tex]\Delta L = L_{new} - L[/tex]
[tex]\Delta L = (603.36 - 600) \ mm[/tex]
[tex]\mathbf{\Delta L = 3.36 \ mm}[/tex]
The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought to me from the Mission Valley site. I measured the mass of the sample to be 600 grams. I then measured the mass of the sample after placing it in the oven for 24 hrs. I found this mass to be 200 grams. Can you help me determine the water content?" Determine the water content.
Answer:
The water of the saturated clayed soil is 66.67 %.
Explanation:
Given;
mass of saturated clayed soil, [tex]M_s[/tex] = 600 g
mass of dry soil sample, [tex]M_d[/tex] = 200 g
mass of water content, [tex]M_w[/tex] = [tex]M_s[/tex] - [tex]M_d[/tex] = 600 g - 200 g = 400 g
The water content is determined as;
[tex]M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%[/tex]
Therefore, the water of the saturated clayed soil is 66.67 %.
anything that is made to meet a need or desire is a
A 200 Ω transmission line is to be matched to a computer terminal with ZL = (50 − j25) Ω by inserting an appropriate reactance in parallel with the line. If f = 800 MHz and r = 4, determine the location nearest to the load at which inserting:
A) A capacitor can achieve the required matching, and the value of the capacitor. PROBLEMS 131 B) An inductor can achieve the required matching, and the value of the inductor.
Answer:
The answer is below
Explanation:
a) To solve this problem, we are going to use the smith chart. After entering the value of Zo = 200 ohm and ZL = (50 − j25) Ω, this give us [tex]z_L\ and\ y_L[/tex], the intersection between [tex]y_L[/tex] and the SWR line gives:
[tex]y(d)=1.026-j1.54\\\\We\ are\ using\ the\ imaginary\ part\ to calculate\ the\ capacitance,hence:\\\\wC=1.54*Y_o\\\\C=\frac{1.54}{Z_o*w}=\frac{1.54}{200*2\pi*800*10^6}=1.53*10^{-12}\\ \\C=1.53*10^{-12}F[/tex]
b) Also, we get:
[tex]y(d)=1.0-j1.52\\\\We\ are\ using\ the\ imaginary\ part\ to\ calculate\ the\ inductance,hence:\\\\1/wL=1.52*Y_o\\\\L=\frac{Z_o}{1.52*w}=\frac{200}{1.52*2\pi*800*10^6}=2.6*10^{-8}\\ \\L=2.6*10^{-8}H[/tex]
Which AWS specification covers electrodes used for welding low-alloy steels?
Answer:
American Welding Society filler metal specification A5. 5-96
Explanation:
You prepare low alloy electrodes by adding the correct alloying elements to the electrode coating. This specification covers the following;
requirements for size and lengths of requirements for marking and manufacturingrequirements for chemical composition and mechanical properties of weld metalmetal soundness test and moisture testsusability test for electrodes