Where do you prefer to live?
USA UK Canada ????????
Answer for POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Sarah is a site investigator for a large construction firm. She is considering Miguel, a former geology student with experience as an intern at an architecture firm, for an assistant site investigation position. Which of the following is most relevant to her decision?
Answer: A. whether his geology studies exposed him to principles of geotechnical engineering
Explanation:
The options include:
a. whether his geology studies exposed him to principles of geotechnical engineering
b. the size of the geology program he attended
c. the size of the architecture firm
d. whether the architecture firm was intending to offer Miguel a full-time position
Since Miguel, is a former geology student with experience as an intern at an architecture firm, and Sarah is considering him for an assistant site investigation position, the option that will be relevant for her to make a decision is to know whether his geology studies exposed him to principles of geotechnical engineering.
Geotechnical engineering, is a branch of engineering that makes use of principles of rock mechanics to solve engineering challenges. Since Sarah needs him for an assistant site investigation position, he'll need to investigate souls, rocks and evaluate them.
In low-speed external water flow over a bluff object, vortices are shed from the object. The vortex shedding produced by a particular object is to be studied in a water tunnel at a 1/4 scale model (model 1/4 the size of the prototype). After a dimensional study, it is found that the Pi terms of this phenomenon are the Reynolds number and the Strouhal number:Re = pVd/muSt = fd/Vwhere f is the frequency of the vortex shedding, V the velocity of the flow, d the characteristic length of the object, and p and mu the density and viscosity of the flow. 1. If the prototype speed is 7 m/s, determine the water velocity in the tunnel for the model tests. 2. If the model tests of part 1 produced a model shedding frequency of 200 Hz, determine the expected vortex shedding frequency on the prototype.
Answer:
1) the water velocity in the tunnel for the model tests is 28 m/s
2) the expected vortex shedding frequency on the prototype is 12.5 Hz
Explanation:
Given the data in the question;
1)
using the Reynolds number relation for prototype and model,
PpVpdp/mu(Prototype) = PmVmdm/mu(for model)
but we know that, density and viscosity in prototype and model will remain same so;
Vp × dp = Vm × dm
vm = Vp × dp/dm
we substitute
vm = 7 m/s × 4
vm = 28 m/s
Therefore, the water velocity in the tunnel for the model tests is 28 m/s
2)
we make use of the Strouhal number relation as given in the question;
fp × dp/Vp = fm × dm/Vm
frequency of the prototype will be;
fp = fm × dm/Vm / dp/Vp
we substitute
fp = 200 × 7 / ( 4 × 28 )
fp = 1400 / 112
fp = 12.5 Hz
Therefore, the expected vortex shedding frequency on the prototype is 12.5 Hz
Rainfall rates for successive 20-min period of a 140min storm are 1.5, 1.5, 6.0, 4.0, 1.0, 0.8, and 3.2 in/hr, totaling 6.0in. Determine the rainfall excess during successive 20-min periods by the NRCS method. The soil in the basin belongs to group A. It is an agriculture row crop land with contoured pattern in good hydrologic condition. The soil is in average condition before the storm (moisture condition II).
Can someone tell me what car, year, and model this is please
Answer:
Explanation:
2019 nissan altima 2.5 SV
have a good day /night
may i please have a branlliest
is a baby duck swimming in a lake a learned behavior
Answer:
True because some ducks can't swim but have to learn it
Also discuss how bandwidth is affected by increasing the number of signal elements.
Identify the true statements about the lumped system analysis.
A. The entire body temperature remains essentially uniform at all times during a heat transfer process.
B. The temperature of lumped system bodies can be taken to be a function of time only.
C. The Biot number is less than or equal to 0.1.
D. The Biot number is greater than or equal to 1.
Water vapor at 5 bar, 3208C enters a turbine operating at steady state with a volumetric flow rate of 0.65 m3/s and expands adiabatically to an exit state of 1 bar, 1608C. Kinetic and potential energy effects are negligible. Determine for the turbine (a) the power developed, in kW, (b) the rate of entropy production, in kW/K, and (c) the isentropic turbine efficiency.
Answer:
A) 371.28 kW
b) 0.1547 Kw/K
c) 85%
Explanation:
pressure (p1) = 5 bar
exit pressure ( p2 ) = 1 bar
Initial Temperature ( T1 ) = 320°C
Final temp ( T2 ) = 160°C
Volume ( V ) = 0.65 m^3/s
A) Calculate power developed ( kW )
P = m( h1 - h2 ) = 1.2 ( 3105.6 - 2796.2 ) = 371.28 kW
B) Calculate the rate of entropy production
Δs = m ( S2 - S1 ) = 1.2 ( 7.6597 - 7.5308 ) = 0.1547 Kw/K
c) Calculate the isentropic turbine efficiency
For an isentropic condition : S2s = S1
therefore at state , value of h2 at isentropic condition
attached below is the remaining part of the solution
Note : values of [ h1, h2, s1, s2 , v1 and m ] are gotten from the steam tables at state 1 and state 2
Name some technical skills that are suitable for school leavers .
Answer:
Welding, carpentry, masonry, construction worker, barber
Explanation:
Suppose we are given three boxes, Box A contains 20 light bulbs, of which 10 are defective, Box B contains 15 light bulbs, of which 7 are defective and Box C contains 10 light bulbs, of which 5 are defective. We select a box at random and then draw a light bulb from that box at random. (a) What is the probability that the bulb is defective? (b) What is the probability that the bulb is good?
Answer:
0.49
0.51
Explanation:
Probability that bulb is defective :
Let :
b1 = box 1 ; b2 = box 2 ; b3 = box 3
d = defective
P(defective bulb) = (p(b1) * (d|b1)) + (p(b2) * p(d|b2)) + (p(b3) * p(d|b3))
P(defective bulb) = (1/3 * 10/20) + (1/3 * 7/15) + (1/3 * 5/10))
P(defective bulb) = 10/60 + 7/45 + 5/30
P(defective bulb) = 1/6 + 7/45 + 1/6 = 0.4888
= 0.49
P(bulb is good) = 1 - P(defective bulb) = 1 - 0.49 = 0.51
1. A wastewater treatment plant (WWTP) releases effluent into a stream with mean depth 2 m and mean velocity 0.75 m/s. The BOD concentration at the WWTP is 15 mg/L, and the oxygen deficit is negligible. The deoxygenation rate in the stream is 0.8 d-1 and the reaeration rate is 1.2 d-1. a) Calculate the BOD concentration and DO deficit at a point 20 km downstream from the WWTP. (10 pts) b) What assumptions are inherent in these predictions (give at least two)
Answer:
A) BOD = 6.51 mg/l , DO = 2.46 mg/l
B) BOD of stream is negligible and DO of stream is at saturation level
Explanation:
Mean depth = 2 m
Mean velocity = 0.75 m/s
Bod concentration at WWTP = 15 mg/L
deoxygenation rate = 0.8 d-1
reaeration rate = 1.2 d-l
a) Calculate the BOD concentration and DO deficit
at 20 km
tc = (20 * 10^3) / (0.75 * 3600 * 24 )
= 0.309 days
[tex]BOD_{t}[/tex] = lo ( 1 - 10^- 0.8 * 0.309 )
= 15 ( 1 - 10^ - 0.2472 )
= 15 ( 0.434 ) = 6.51 mg/l
DO = ( Kd * lo / Kr ) * 10^ -Kd*tc
= ( 0.8 * 6.51 / 1.2 ) * 10 ^ - 0.8 * 0.309
= 4.34 * 10^-0.2472 = 2.46 mg/l
B) The assumptions are : BOD of stream is negligible and DO of stream is at saturation level
Air is a....
O Solid
O Liquid
O Gas
O Plasma
Answer:
Air is a gas
Explanation:
i think. beavuse it cant be a liqued or a solid. i dont think a plasma. i would answer gas