Answer:
metal hub and the wire tension spokes?
Answer: metal hub and the wire tension spokes is correct
Lake Superior, the largest of the Great Lakes of North America, is also the world's largest freshwater lake by surface area (~ 82,100 km2), and the third largest freshwater lake by volume (~ 12,100 km3). The monthly average precipitation in 2018 in the lake area was 2.69 inches. According to the data provided by USGS water stations, the total flow rate in all the incoming streams was found to be 356,747 ft3/s (cubic feet per second). At the same time, the lake discharges 717,258 ft3/s to its surrounding water bodies. The monthly average evaporation was 18.7 mm. The groundwater inflow was 783.33 km3 less than the groundwater outflow in this year. (Hints: (1) The average monthly data can be used to calculate the annual data; (2) Volume = Area * Height; (3) Pay special attention to the units, convert all the units to be consistent first.)
(1) With the information above, please write the water budget for Lake Superior (10 points);
(2) Please estimate the change of storage (in m3) in the year of 2018 (20 points);
(3) Please estimate the increase or decrease of water level (in mm) in the year of 2018 (10 points).
Answer:
1) V_win = 7.05 10¹⁰ m³ , 2)V_ lost = 1.8471 10¹⁰ m³, 3) h = 633.4 mm
Explanation:
In this exercise we will start by reducing all units to the SI system
h1 = 2.69 inch (2.54 10-2 m / 1 inch) = 6.8326 10⁻² m
Ф1 = 356.747 ft3 / s (1 m3 / 35.3147 ft3) = 10.1019 m³ / s
Ф2 = 717,258 ft3 / s = 20,310 m³ / s
h2 = 18.7 mm (1 m / 1000 mm) = 18.7 10⁻³ m
V_subterránea = - 783.33 10³ m,
Now let's answer the questions
1) they ask us the amount of water that reaches the lake in 2018
volume of rainwater is
V₁ = A. h t
V₁ = 82 100 106 * 6.8326 10-2 * 12
V₁ = 6.7315 10 10 m3
stream water volume in a year
V₂ = Ф₁ t
t = 1 year (365 days / year) (24 h / 1 day) (3600 s / 1h) = 3.1536 10⁷ s
V₂ = 10.1019 3.1536 10⁷
v₂ = 31,857 10⁷ m³
The total water volume is
V_win = V₁ + V₂
V_win = 6.7315 10¹⁰ + 31.857 10⁷
V_win = 7.05 10¹⁰ m³
2) let's find the amount of water lost from the lake
volume of water to surrounding bodies
V₃ = Ф₂ t
V₃ = 20.310 3.1536 10⁷
V₃ = 6.40496 10 8 m3
Volume of evaporated water
V₄ = A h₂ t
V₄ = 82 100 10⁶ 18.1 10⁻³ 12
V⁴ = 1,783 10¹⁰ m³
groundwater volume
V⁵ = 7.83.33 10³ m³
The volume of water lost is
V_lost = V₃ + V₄ + V₅
V_ lost = 6.40496 10⁸ + 1.783 10¹⁰ + 783.33 10³
V_ lost = 1.8471 10¹⁰ m³
3) the change in the height of the lake
the change in volume is
ΔV = V_ won - V_ lost
ΔV = 7.05 10¹⁰ - 1.8471 10¹⁰
ΔV = 5.20 10¹⁰ m³
the volume is
v = A h
h = V / A
h = 5.20 10¹⁰/82100 10⁶
h = 6.334 10⁻¹ m
h = 633.4 mm
A 75-kg piano is hosted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What is the potential energy of the piano?
Answer:
14,700 J
Explanation:
PE = Mgh = (75 kg)(9.8 m/s²)(20 m) = 14,700 J
A parallel helical gearset consists of a 19-tooth pinion driving a 57-tooth gear. The pinion has a left-hand helix angle of 30°, a normal pressure angle of 20°, and a normal module of 2.5 mm.
Find:_______.
(a) The normal, transverse, and axial circular pitches
(b) The transverse diametral pitch and the transverse pressure angle
(c) The addendum, dedendum, and pitch diameter of each gear
Answer:
a)
normal circular pitch = 7.8539 mm
transverse circular pitch = 9.0689 mm
axial circular pitches = 15.7077
b)
transverse diametral pitch is 0.3464 teeth/mm
transverse pressure angle is 22.8°
c)
Addendum = 2.5 mm
dedendum = 3.125 mm
pinion diameter = 54.8482 mm and Gear diameter = 164.5448 mm
Explanation:
Given that;
module m = 2.5 mm
Number of teeth on Gear nG = 57 TEETH
Number of teeth on Pinion nP = 19 TEETH
Helix angle W = 30°
Normal Pressure angle β = 20°
finding the circular pitch
Pc = πm
we substitute
Pc = π * 2.5 mm = 7.8539 mm
now the diametral pitch p = π / Pc
= π / 7.8539
= 0.4 teeth/mm
a)
So the normal circular pitch
Pn = π / P
Pn = π / 0.4
Pn = 7.8539 mm
the transverse circular pitch
Pt = Pn / cosW
Pt = 7.8539 / cos30°
Pt = 9.0689 mm
for axial circular pitches
Px = Pt / tanW
Px = 9.0689 / tan30°
Px = 15.7077
b)
The transverse diametral pitch and the transverse pressure angle.
The transverse diametral pitch Pt = PcosW
= 0.4 * cos30°
= 0.3464 teeth/mm
transverse diametral pitch is 0.3464 teeth/mm
transverse pressure angle β1 = tan^-1 ( tan βn / cos W)
= tan^-1 ( tan20° / cos 30°)
= tan^-1 ( 0.42027 )
β1 = 22.8°
transverse pressure angle is 22.8°
c)
The addendum, dedendum, and pitch diameter of each gear
Now from table standard Tooth proportions for Helical Gears;
Addendum a = 1/p
= 1 / 0.4
= 2.5 mm
dedendum b = 1.25 / p
= 1.25 / 0.4
= 3.125 mm
now pinion diameter dP = Np / PcosW
= 19 / 0.4 (cos30°)
= 54.8482 mm
Gear diameter dG = nG / pcosW
= 57 / 0.4 (cos30°)
= 164.5448 mm
You can use this to listen to light from the big bang.
A. a microwave
C.
B. a cell phone
C. a radio
D. a tv
A test of a driver’s perception/reaction time was conducted on a special test track in a rural area. The friction factor f = 0.6. The driver’s initial speed is 55 MPH. The track is on level ground.
a) When a driver is not texting on his smart phone, a stop can be made just in time to avoid hitting an unexpected object that is first visible 520 feet ahead.
b) When a driver is texting on his smart phone, but with all other conditions exactly the same, the driver is distracted and fails to stop and hits the object at a speed of 35 MPH. Determine the driver’s perception/reaction time for both situations, for (a) and for (b).
Answer:
a) tD = 4.36 sec
b) tD = 5.2 sec
Explanation:
a)
V = 0
work-done = change in kinetic energy
f.m.g(d) = 1/2m(55^2 - v^2)
divided both sides by m
1/2(55^2 - v^2) = f.g(d)
1/2(55 * 1.467)^2 ft^2/sec^2 = 0.6 * (9.81 * 3.28) * d
3255.03 = 19.306d
d = 3255.03 / 19.306
d = 168.6 ft
now D = 520 - 168.6 = 351.4
therefore reaction/perception time = tD
tD = 351.4 / ( 55 * 1.467)
tD = 4.36 sec
b)
Also here, V = 35 mph
so
1/2(55^2 - 35^2)*(1.467)^2 ft^2/sec^2) = 0.6 * (9.81 * 3.28) * d
1936.88 = 19.306d
d = 1936.88 / 19.306
d = 100.325ft
also D = 520ft - 100.325ft = 419.675
so tD = 419.675 / ( 55 * 1.467)
tD = 5.2 sec
Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in each of the 11 units and then condensed and combined to obtain a product stream of fresh water. The brine leaving each evaporator but the 11this fed to the next evaporator. The brine leaving the 11th evaporator contains 5.00 wt% salt. It is desired to produce 2 x 104 L/h of fresh water. What is the mass flow rate of concentrated brine out of the process?
______kg/h Save for Later
Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Explanation:
F, W and B are the fresh feed, brine and total water obtained
w = 2 x 10^4 L/h
we know that
F = W + B
we substitute
F = 2 x 10^4 + B
F = 20000 + B .................EQUA 1
solute
0.035F = 0.05B
B = 0.035F/0.05
B = 0.7F
now we substitute value of B in equation 1
F = 20000 + 0.7F
0.3F = 20000
F = 20000/0.3
F = 66666.67 kg/hr
B = 0.7F
B = 0.7 * F
B = 0.7 * 66666.67
B = 46,666.669 kg/hr
the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft.
A) Determine the pressure at this elevation, based on the equation
p = pa(1 - βz/T0)g/Rβ
β = .00650 K/m = absolute pressure at z =0 R= 286.9 J/kg*k.
B) If the air is assumed to have a constant specific weight of 0.07647 lb/ft3 what would the pressure be at this altitude?
C) If the air is assumed to have a constant temperature of 59˚F what would the pressure be at this elevation? For all three cases assume standard atmospheric conditions at sea level.
Properties of U.S Standard Atmosphere at Sea Level
Property SI Units BG Units
Temperature, T 288.15 K(15 oC) 518.67 oR(59.00 oF)
Pressure, P 101.33 kPa(abs) 2116.2 Ib/ft2(abs)
(14.696 Ib/in.2(abs))
Density 1.225 kg/m3 0.00237 slugs/ft3
Specific weight 12.014 N/m3 0.07647 Ib/ft3
Viscosity, mu 1.789 * 10-5 N.s/m2 3.737 * 10-7 Ib.s/ft2
Answer:
A. 1236.49 lb/ft²
B. 1037.21 lb/ft²
C 2269.7 lb/ft²
Explanation:
P = pa(1-βz/Ta)^g/Rβ
a.)
Pa [1-(0.00357)(14110)/518.67]^32.2/1716*0.00357
= 2116.2(1-0.09712)^5.26
= 2116.2(0.5843)
= 1236.49566 lb/ft²
b.)
Pressure at higher elevation
P = Pa - yz
= 2116.2-(0.07647)(14110)
= 1037.2083 lb/ft³
C.
Pressure at higher elevation if air is isothermal
= (Pa)*exp[-(32.2)(14110)/(1716)(519)]
= 2116.2*exp[-454342/890604]
= 1269.7 lb/ft²
A new car design must carry a family of four, include safety features, get more than 25 miles per gallon on the highway, and weigh less than 1.5 tons. which is one of the criteria that should be part of the design to meet the proposal request?
Answer: weigh less than 1.5 tons
Explanation:
Answer:The answer is A, seats for five people.
Explanation: I took the test and it is the only option that fits the criteria listed above.
6. Dr. Li boils water using a kettle with a 1.5 kW Nichrome (80% Ni and 20% Cr) heating element (resister heater). The diameter and length of this heating element is 8 mm and 20 cm, respectively. When exposed to liquid water, the convection heat transfer between water and heating element is 800 W m2K . Please find: (25 Points) (1) The surface temperature and maximum temperature of the heating element when water is boiling in the kettle in Dr. Li’s office, ESB 751. (2) The surface temperature and maximum temperature when all liquid water has been evaporated and the heating element is exposed completely to the superheated vapor. You can assume the convection heat transfer between superheated water vapor and heating element is 24 W m2K . You can assume the temperature of the superheated water vapor is 100 °C. (3) Please justify if the heating element can survive in case (2) without melting.
Assuming that:
The heat generation is uniform throughout the heating element, Nichrome wire.
The cross-sections are insulated and heat transfer is taking place only in the radial direction.
All the heat generated is conducted, there is energy storage.
Now, from the properties of Nichrome:
The melting temperature of Nichrome, [tex]T_m=1400 ^{\circ}C[/tex]
Thermal conductivity, [tex]K=11.3 W/m^{\circ}C[/tex]
Given that:
The power generated by the heating element, [tex]Q_g=1.5kW=1500W[/tex]
Diameter, [tex]D=8mm=8\times10^{-3}m[/tex]
So, radius, [tex]R=4\times10^{-3}m[/tex]
Length, [tex]L=20cm=0.2m[/tex]
The volume of the Nichrome wire,
[tex]V=\pi R^2L=\pi\times (4\times10^{-3})^2\times0.2=1.005\times10^{-5}m^3[/tex]
Heat generation rate per unit volume,
[tex]q_g=\frac{Q_g}{V}=\frac{1500}{1.005\times10^{-5}}=149.2\times10^{6}W/m^3[/tex]
With the assumptions made above, this is the case of heat transfer in one direction.
Let [tex]T[/tex] be the temperature at the radius [tex]r[/tex].
Now, the heat generated within the cylinder of radius [tex]r[/tex] is conducted in a radially outward direction. i.e
[tex]q_g(\pi r^2)L=-K(2\pi r L)\frac{dT}{dr}[/tex] [where K is the thermal conductivity oof Nichrome]
[tex]\Rightarrow -\frac{q_g}{2K}rdr=dT[/tex]
[tex]\Rightarrow T=-\frac{q_g}{4K}r^2+ C_0[/tex] , where [tex]C_0[/tex] is constant.
Let the surface temperature is [tex]T_s[/tex], i.e at [tex]r=R, T=T_s[/tex].
Putting this boundary condition to get [tex]C_0[/tex], we have
[tex]T=T_s+\frac{q_g}{4K}(R^2-r^2)[/tex]
This is the temperature profile within the Nichrome wire, which is maximum at [tex]r=0[/tex].
So, the maximum temperature,
[tex]T_{max}=T_s+\frac{q_g}{4K}R^2\;\cdots(i)[/tex]
(1) The water is boiling, to the temperature of the water is, [tex]T_b=100^{\circ}C[/tex].
The total heat generated within the heating element is convected to the water from the surface. i.e
[tex]Q_g=h_w(2\pi RL)(T_s-T_b)[/tex]
where, [tex]h_w=800W/m^2K=800W/m^2^{\circ}C[/tex] is the convective heat transfer constant (Given).
[tex]\Rightarrow 1500=800\times (2\pi\times4\times10^{-3}\times 0.2(T_s-100)[/tex]
[tex]\Rightarrow T_s=100+373=473^{\circ}C[/tex]
So, the surface temperature is [tex]473^{\circ}C[/tex].
From equation (i), the maximum temperature is at the center of the wire which is
[tex]T_{max}=473+\frac{149.2\times10^{6}}{4\times11.3}(4\times10^{-3})^2[/tex]
[tex]\Rightarrow T_{max}=473+53=526^{\circ}C[/tex]
(2) In this case, the temperature of the superheated water vapor, [tex]T_b = 100^{\circ}C[/tex] (Given)
The heat transfer coefficient between the superheated water vapor and the heating surface is, [tex]h_v=24W/mK=24W/m^{\circ}C[/tex].
Similarly, [tex]Q_g=h_v(2\pi RL)(T_s-T_b)[/tex]
[tex]\Rightarrow 1500=24\times (2\pi\times4\times10^{-3}\times 0.2(T_s-100)[/tex]
[tex]\Rightarrow T_s=100+12434=12534^{\circ}C[/tex]
So, the surface temperature is [tex]12534^{\circ}C[/tex].
From equation (i), the maximum temperature is at the center of the wire which is
[tex]T_{max}=12534+\frac{149.2\times10^{6}}{4\times11.3}(4\times10^{-3})^2[/tex]
[tex]\Rightarrow T_{max}=12534+53=12587^{\circ}C[/tex]
(3) The melting temperature of Chromium is [tex]1400 ^{\circ}C[/tex].
So, the 1st case when the heating element is surrounded by water is safe as the maximum temperature within the element is below the melting temperature.
But, it the 2nd case, the heating element will melt out as the maximum temperature is much higher than the melting temperature of the element.
Technician A says you should place the air ratchet setting to
clockwise to loosen a fastener. Technician B says a fastener
must be set to the proper torque after using an air ratchet. Who
is correct?
a team of engineer's is designing a new Rover to explore the surface of Mars which statement describes the clearest constraint that applies to the solution
Answer:it must operate at temperatures below 0’C
Explanation:
Answer:
it must operate at temperatures below 0 degrees Celsius
Explanation:
5. What are the 3 basic types of electrical circuits?
OA. Simple, AC, and AC direct
OB. Series, Parallel, and DC indirect
OC. AC, DC, and DC direct
OD. Series, Parallel, and Series/Parallel
Answer:
D. Series, Parallel, and Series/Parallel
Explanation:
Categorized according to topology, circuits are ...
Series, Parallel, and Series/Parallel
__
Circuits can also be categorized according to the nature of the voltages and currents found in them. These categories would include AC, DC, and some mixture of AC and DC.
The terms "simple" and "direct" and "indirect" don't seem to have any defined meaning in this context.
Answer:
D. Series, Parallel, and Series/Parallel
Explanation:
on todays class
5.5 A scraper with a 275 hp diesel engine will be used to excavate and haul earth for a highway project. An evaluation of the job-site conditions indicates the scraper will operate 40 min./hr. For this project it is anticipated that the total cycle time will be 20 min. for a round trip. Previous job records show the scraper operated at full power for the 1.5 min. required to fill the bowl of the scraper and at 80% of the rated hp for the balance of the cycle time. Calculate the gallons per hour for fuel consumption of the scraper.
Answer: Fuel consumption per hour of the scrapper is 6 gal/hr
Explanation:
Given that;
Rated power = 275 hP
The scraper will work = 40min per hr
anticipated total cycle time = 20min
Previous Scrapper operated full power for 1.5min at 80% hP
First we find the Engine factor;
filling the bucket = 1.5/20 * 100% = 0.075
rest of cycle = (20-1.5)/20 * 80% = 0.74
so total engine factor = 0.075 + 0.74 = 0.815
now Time factor will be 40/60 = 0.6666 =
therefore operating factor = 0.6666 * 0.815 = 0.543
Now we know that for a diesel engine, a range fuel consumption = 0.04 gal/hP
so
Fuel consumption per hour of the scrapper is;
= 0.543 * 275 * 0.04
= 5.97 = 6 gal/hr
might give brainliest
Complete the following sentence.
_______ is a communication system that is composed of images that represent words.
Answer:
Blissymbols
Explanation:
Blissymbols is a constructed language with hundreds of basic symbols that represent words.
Answer:
Blissymbols
Explanation:
The evaporator section of a refrigeration unit consists of thin-walled, 10-mm-diameter tubes through which refrigerant passes at a temperature of −18°C. Air is cooled as it fows over the tubes, maintaining a surface convection coeffcient of 100 W/m2 ⋅ K, and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of −3°C, what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator’s defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range 0 ≤δ ≤ 4 mm. Frost may be assumed to have a thermal conductivity of 0.4 W/m ⋅ K. (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which T[infinity] = 20°C and natural convection maintains a convection coeffcient of 2 W/m2 ⋅ K, how long will it take for the frost to melt? The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of fusion of 334 kJ/kg
List five areas that increased energy prices impact.
Answer:
Supply, demand, global markets, imports and exports, and government Regulation.
Explanation:
There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W). Monitor a cellular phone call and observe the type of telephone and the speed of the user. The probability model for this experiment has the following information:
P[F]=0.5, P[HF]=0.2, P[MW]=0.1.
What is the sample space of the experiment?
Find the following probabilities P[W], P[MF], and P[H].
Answer:
A) P(W) = 0.5
B) P(MF) = 0.3
C) P(H) = 0.6
Explanation:
We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.
Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).
Thus, the sample space is combination of types and classification we are given and it is written as;
S = {HF, HW, MF, MW}
A) Now, phones can either be fast(F) or slow(W). Thus, we can write;
P(F) + P(W) = 1
We are given P(F) = 0.5
Thus;
0.5 + P(W) = 1
P(W) = 1 - 0.5
P(W) = 0.5
B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;
P(H ∩ F) + P(M ∩ F) = P(F)
We are given P[F] = 0.5 and P[HF] = 0.2.
P(H ∩ F) is same as P[HF]
Also, P(M ∩ F) is same as P(MF)
Thus;
0.2 + P(MF) = 0.5
P(MF) = 0.5 - 0.2
P(MF) = 0.3
C) Similarly, mobile Phone calls can either be fast or slow. It means the sample space partition is {F, W}
Thus;
P(M) = P(MW) + P(MF)
P(M) = 0.1 + 0.3
P(M) = 0.4
Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;
P(H) + P(M) = 1
P(H) + 0.4 = 1
P(H) = 1 - 0.4
P(H) = 0.6
A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and the return activated sludge concentration is 8000 mg/L. The biomass concentration is 3500 mg/L. The combined design flow rate of the primary and secondary clarifiers is 2.5 MGD. After primary treatment, the wastewater has an influent BOD concentration of 200 mg/L and an influent suspended solids concentration of 200 mg/L. Two secondary clarifiers, each 28 ft in diameter, are then used. After secondary treatment, the effluent BOD concentration is 15 mg/L, and the effluent suspended solids concentration is 20 mg/L. The volume of sludge produced is 0.5 MGD. What is most nearly the solids residence time
Answer:
11 hours approximately
Explanation:
We are to calculate mean cell residence time mcrt
= Mass of solid in reactor/mass of solid wasted in a day
Q = Qe + We
Q = 2.5
Qw = 0.5
Qe = 2.5 - 0.5
= 2 MGD
10⁶/svi
= 10⁶/125
= 8000
X = 3500
Xe = 20mg/
1MGD = 0.1337million
Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337
= 262500000/[4000000000+40000000} x 0.1337
= 262500000/574800000
= 0.45668 days
= 0.45668 x 24 hours
= 10.9603 hours
Approximately 11 hours
PLEASE HELP!!!! You are going to create a transcript for a persuasive presentation to convince your client that your design improvement is the best possible solution for their needs. Take your client through the different design options you considered for the product design improvement. Explain why you discarded the designs that were not the best possible solution. Finally, explain why your final product design covers each of the requirements and specifications you used to create your needs statement.
A lake with a surface area of 525 acres was monitored over a period of time. During onemonth period the inflow was 30 cfs (ie. ft3 /sec), the outflow was 27 cfs, and a 1.5 in seepage loss was measured. During the same month, the total precipitation was 4.25 inches. Evaporation loss was estimated as 6 inches. Estimate the storage change for this lake during the month.
Answer:
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] cubic feet during the month.
Explanation:
We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:
[tex]\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}[/tex]
Where [tex]\Delta V_{storage}[/tex] is the monthly storage change of the lake, measured in cubic feet.
Monthly inflow
[tex]V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)[/tex]
[tex]V_{inflow} = 77.76\times 10^{6}\,ft^{3}[/tex]
Monthly outflow
[tex]V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)[/tex]
[tex]V_{outflow} = 66.98\times 10^{6}\,ft^{3}[/tex]
Seepage losses
[tex]V_{seepage} = s_{seepage}\cdot A_{lake}[/tex]
Where:
[tex]s_{seepage}[/tex] - Seepage length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{seepage} = 1.5\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{seepage} = 2.86\times 10^{6}\,ft^{3}[/tex]
Evaporation losses
[tex]V_{evaporation} = s_{evaporation}\cdot A_{lake}[/tex]
Where:
[tex]s_{evaporation}[/tex] - Evaporation length loss, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{evaporation} = 6\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{evaporation} = 11.44\times 10^{6}\,ft^{3}[/tex]
Precipitation
[tex]V_{precipitation} = s_{precipitation}\cdot A_{lake}[/tex]
Where:
[tex]s_{precipitation}[/tex] - Precipitation length gain, measured in feet.
[tex]A_{lake}[/tex] - Surface area of the lake, measured in square feet.
If we know that [tex]s_{precipitation} = 4.25\,in[/tex] and [tex]A_{lake} = 525\,acres[/tex], then:
[tex]V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex]
[tex]V_{precipitation} = 8.10\times 10^{6}\,ft^{3}[/tex]
Finally, we estimate the storage change of the lake during the month:
[tex]\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}[/tex]
[tex]\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}[/tex]
The storage of the lake has increased in [tex]4.58\times 10^{6}[/tex] cubic feet during the month.
The volume of water gained and the loss of water through flow,
seepage, precipitation and evaporation gives the storage change.
Response:
The storage change for the lake in a month is 1,582,823.123 ft.³How can the given information be used to calculate the storage change?Given parameters:
Area of the lake = 525 acres
Inflow = 30 ft.³/s
Outflow = 27 ft.³/s
Seepage loss = 1.5 in. = 0.125 ft.
Total precipitation = 4.25 inches
Evaporator loss = 6 inches
Number of seconds in a month is found as follows;
[tex]30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds[/tex]
Number of seconds in a month = 2592000 s.
Volume change due to flow, [tex]V_{fl}[/tex] = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³
1 acre = 43560 ft.²
Therefore;
525 acres = 525 × 43560 ft.² = 2.2869 × 10⁷ ft.²
Volume of water in seepage loss, [tex]V_s[/tex] = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³
Volume gained due to precipitation, [tex]V_p[/tex] = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³
Volume evaporation loss, [tex]V_e[/tex] = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³
[tex]Storage \, change \, \Delta V = \mathbf{V_{fl} - V_s + V_p - V_e}[/tex]Which gives;
ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123
The storage change, ΔV = 1,582,823.123 ft.³Learn more about water resources and hydrology here:
https://brainly.com/question/10555667
ion 2 23
Ohm's law tells us that it takes 1 volt to push 1 amp through how many ohms?
A. 422
B. 292
O c. 10
D. 32
When starting your vehicle, what does it mean when the ABS indicator light 1 point
on instrument panel turns on for a few seconds before turning off? *
There is a problem with your braking system, do not drive
The ABS bulb has burned out, replace the bulb
O The vehicle safety check indicates the ABS is functioning normally
O The ABS system is not working properly but it's safe to drive
Answer:
C. The vehicle safety check indicates the ABS is functioning normally.
Explanation:
ABS, an antilock braking system, is a safe and secure slip-resistant braking system in cars and air-crafts. An ABS is there to prevent wheel-locking while using brakes in vehicles.
When one ignites a car, the ABS indicator will light up briefly as a part of safety check. The ABS indicator light comes on for a few seconds before it turns off again. This indicates that the ABS system is functioning normally. But, if the ABS indicator light remains after turning on the ignition, this indicates that there is a problem in the system.
In the given scenario, the ABS indicator is functioning properly, thus the correct answer is the third option (C).
When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizing this, estimate the total chronic daily intake of toluene from exposure to a city water supply that contains toluene at a concentration equal to the drinking water standard of 1 mg/L over a period of 10 years. Assume the exposed individual is an adult female that is exposed to the chemical via drinking water and inhaling gaseous toluene released while she showers. Use the given parameters to calculate the CDI for water consumption. For inhalation, assume the woman takes a 15-minute shower every day. Assume the average air concentration of toluene during the shower is 1 μg/m3 and that she breathes at the adult rate of 20 m3/day.
Answer:
The following are the solution to this question:
Explanation:
The Formula for calculating CDI:
[tex]\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}[/tex]
[tex]_{where} \\ CDI = \text{Chronic daily Intake rate} (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }[/tex]
calculating the value of AT:
[tex]= 365 \frac{days}{year} \times 70 \ year \\\\ = 25550 \ days[/tex]
calculating the value of Intake based drinking:
[tex]C = 1 \ \frac{mg}{L}[/tex]
[tex]CR = 2 \frac{L}{day}[/tex] Considering that adult females eat 2 L of water a day,
EF = 350 [tex]\frac{days}{year}[/tex] for drink
calculating the CDI value:
[tex]\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times 25550)}\\\\[/tex]
[tex]= \frac{(2 \times 3500)}{(70 \times 25550)}\\\\ = \frac{(7000)}{(70 \times 25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}[/tex]
Centered on inhalation, intake:
[tex]C = \frac{1 \mu g} { m^3} \ \ \ or \ \ \ \ 0.001 \ \ \frac{mg}{m^3}\\\\CR = 20 \frac{m^3}{day}\\\\EF = 15 \frac{min}{day} \ \ or\ \ 5475 \frac{min}{yr} \ \ \ or \ \ 3.80 \frac{days}{year}\\[/tex]
calculating the value of CDI:
[tex]\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}[/tex]
[tex]= \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}[/tex]
An engine manufacturer wants to develop a gasoline engine that produces 300 hP at 6000 rpm. They can achieve 10 bar BMEP at peak power with a naturally aspirated engine, and 20bar BMEP with a turbocharged engine. Most modern light-duty engines have displacements of ~0.5L per cylinder. Determine the displacement and minimum number of cylinders (rounded to nearest whole number) required to meet these specs for: a. A naturally aspirated engine. b. A turbocharged engine.
Answer:
a). 5 cylinders
b). 3 cylinders
Explanation:
To develop a gasoline engine that produces
Power, P = 300 hp
= 300 x 746 = 223710 watt
at speed, N = 600 rpm = 100 rps
and stroke volume [tex]$V_s=0.5$[/tex] L
= [tex]$0.5 \times 10^{-3}\ m^3$[/tex]
a). A naturally aspirated engine
BMEP at peak power = 10 bar
= [tex]$10 \times 10^5\ N/m^2$[/tex]
Suction volume, V = [tex]$V_s \times N$[/tex]
= [tex]$0.5 \times 10^{-3} \times 100 = 0.05\ m^3/s$[/tex]
Power produced in one engine cylinder , p = BMEP x V
= [tex]$10 \times 10^5 \times 0.05$[/tex]
= 50000 watts
No. of cylinders required, n = [tex]$\frac{P}{p}$[/tex]
= [tex]$\frac{223710}{50000}$[/tex]
= 4.47
So number of cylinders ≈ 5 nos.
b). A turbo charged engine
BMEP = 20 bar = [tex]$20 \times 10^5\ N/m^2$[/tex]
and Volume V = [tex]$0.05\ m^3 /s$[/tex]
Power produced in one engine cylinder, p = BMEP x V
= [tex]$20\times 10^5 \times 0.05$[/tex]
= 100000 watts
Therefore, number of cylinders, n = [tex]$\frac{P}{p}$[/tex]
= [tex]$\frac{223710}{100000}$[/tex]
= 2.23
So number of cylinders ≈ 3 nos.
Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of the window, calculate the heat loss rate through L = 12-mm-thick polycarbonate, soda lime glass, and aerogel windows, respectively. The thermal conductivities of the aerogel and polycarbonate are kag = 0.014 W/m ⋅ K and kpc = 0.21 W/m ⋅ K, respectively.
Evaluate the thermal conductivity of the soda lime glass at 300 K. If the aircraft has 130 windows and the cost to heat the cabin air is $1/kW ⋅ h, compare the costs associated with the heat loss through the windows for an 8-hour intercontinental fight.
Answer:
HEAT LOST
polycarbonate = 252 W
soda lime glass = 1680 W
aerogel = 16.8 W
COST associated with heat loss
polycarbonate = $ 262.08
soda lime glass = $ 1,747.2
aerogel = $ 17.472
The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
Explanation:
Given that;
surface area for each window = 0.4m * 0.4m = 0.16m^2
DeltaT = 90°C, L = 12mm = 0.012m
thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER
so at 300K
KsL = 1.4 W/mK
Kag = 0.014 W/mK
Kpc = 0.21 W/mK
Now HEAT LOSS
for polycarbonate;
Qpc = -KA dt/dx
NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)
so Qag = (0.21 * 0.16 * 90) / 0.012
= 252 W
for soda lime glass;
Qsl = (1.4 * 0.16 * 90) / 0.012
= 1680 W
for aerogel
Qaq = (0.014 * 0.16 * 90) / 0.012
= 16.8 W
Now for COST associated with heat lost
for polycarbonate;
cost = Qpc * 130 * 8 * 1/1000
= 252 * 130 * 8 * 1/1000
= $ 262.08
for soda lime glass;
cost = 1680 * 130 * 8 * 1/1000
= $ 1,747.2
for aerogel
cost = 16.8 * 130 * 8 * 1/1000
= $ 17.472
Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel
When we drive our car at 90 feet per second [ft/s], we measure an aerodynamic force (called drag) of 59 pounds-force [lbf] that opposes the motion of the car. How much horsepower [hp] is required to overcome this drag?
Answer:
P = 9.65 hp
Explanation:
Given that,
Speed of a car, v = 90 ft/s
Force that opposes the motion of the car, F = 59 pounds-force
90 ft/s = 27.432 m/s
1 pound-force = 4.448 N
59 pounds-force = 262.445 N
Power required is given by :
[tex]P=F\times v\\\\P=262.445\ N\times 27.432\ m/s\\\\P=7199.391\ W[/tex]
Since, 1 hp = 745.7
7199.391 W = 9.65 hp
So, 9.65 hp of power is required to overcome the drag.
You often experiment with everyday materials around you, trying to create new and unique items. You enjoy designing and creating things that improve your daily tasks. What characteristic of a STEM student are you displaying?
A. technologically-literate
B. inventor
C. innovator
D. problem-solver
answer:D. problem-solver
The answer is:
B) Inventor
Singularity is an important property of a square matrix. This is also known as degenerate. What is the value of the determinant of a singular matrix?
Answer:
For a Singular matrix, the determinant must be equivalent to 0.
Explanation:
A matrix is a rectangular array in which elements are arranged in rows and columns.
Each square matrix has a determinant. The determinant is a numerical idea that has a fundamental function in finding the arrangement just as investigation of direct conditions. For a Singular matrix, the determinant must be equivalent to 0.
Gear friction reduces power and engineers never use more gears than are need it.
A) True
B) False
Answer:
i personally think it is false
Explanation:
i think this because gear friction reduces next to no power
A group of scientists studied the environmental impact of internal combustion engines burning hydrocarbon fuels. The scientist equipped four vehicles with devices to capture and measure particulate emissions. One vehicle burned diesel fuel, one burned ordinary gasoline, one burned a gasoline/ethanol mixture and one burned natural gas. The four vehicles have equal masses and carried identical cargo. The scientists drove each vehicle 400 km, recording the volume of fuel burns in the quantity of particulate emissions generated. What is the independent variable in this experiment?
Answer: Combustion of Hydrocarbons
Explanation:
The Independent variable in an experiment is the one whose effect on the dependent variable is being measured. The independent variable therefore is controlled to see the effect it will have in the experiment.
In this experiment, the scientists combusted different types of hydrocarbons (diesel, gasoline, natural gas and a gasoline/ethanol mixture) as they aimed to find out the effect that this burning would have on the environment thereby making the combustion of hydrocarbons the independent variable.
Answer:
A. Type of Fuel
Explanation: The quantity of particulate Matter (PM) primarily depends upon the type of fuel used. Fine carbonaceous particles are mainly responsible for PM emissions. Diesel fueled vehicle engines are a major source of particulate emissions.
