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Careers in Agricultural Science: Mastery Test
1
Select the correct answer.
A certain noodle brand was taken off supermarket shelves because it did not conform to standards set by US regulatory agencies. The internal
team investigating the matter concluded that Luke Brown failed in his duty. What do you think was Luke's position in the company?
OA.
food production supervisor
OB
food safety and quality assurance manager
C.
research and development technician
OD
food and drug inspector
Reset
Next
Answer:
B. food safety and quality assurance manager
Explanation:
PLATO
Test if a number grade is an A (greater than or equal to 90). If so, print "Great!". Hint: Grades may be decimals. Sample Run Enter a Number: 98.5 Sample Output Great!
Answer:
In Python:
grade = float(input("Enter a Number: "))
if grade >= 90:
print("Great!")
Explanation:
This prompts the user for grade
grade = float(input("Enter a Number: "))
This checks for input greater than or equal to 90
if grade >= 90:
If yes, this prints "Great"
print("Great!")
Factors such as brake shoe orientation, pin location, and direction of rotation determine whether a particular brake shoe is considered to be self-energizing or self-deenergizing. Sketch a brake drum where the left shoe is self-energizing and the right shoe is self-deenergizing. Which shoe will produce more braking torque, assuming equal actuation forces and identical brake shoes
Answer:
Self energizing brake shoe produces more Torque
Explanation:
Attached below is the sketches of the various cases
The cases are :
case 1 ; deenergizing
case 2 ; self energizing
case 3 ; produces high braking torque
For a brake to be self energizing when the Frictional torque and braking torque are in the same direction
summary from equation 3
when: b > uc the brake is controllable
b = uc It is self locking
b < uc The brake is uncontrollable
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.
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.
You would use a _____________ gauge to check the pressure in each tire. A technician should compare the tire-pressure reading with the tire pressure specified on the side of the driver's _____________ . When servicing tires follow the _____________ procedure outlined in the owner’s manual or online service information. Correct tire-inflation pressure is printed on a placard on the _____________. Why are tires rotated? Which tire rotation method is most often recommended? _____________ An easy way to remember effective tire rotation is, “Drive wheels straight, _____________ the nondrive wheels.” A _____________ is used to ensure that the proper torque is completed on a
Answer:
the first one is tire pressure gauge :)
Explanation:
A low-altitude meteorological research balloon, temperature sensor, and radio transmitter together weigh 2.5 lb. When inflated with helium, the balloon is spherical with a diameter of 4 ft. The volume of the transmitter can be neglected when compared to the balloon's size. The balloon is released from ground level and quickly reaches its terminal ascent velocity. Neglecting variations in the atmosphere's density, how long does it take the balloon to reach an altitude of 1000 ft?
Answer:
12 mins
Explanation:
The summation of the forces in vertical direction
= Fb - Fd - w = 0 ∴ Fd = Fb - w ----- ( 1 )
Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )
Pair = air density , Vballoon = volume of balloon
Vballoon = [tex]\frac{\pi D^3}{6}[/tex] , where D = 4 ∴ Vballoon = 33.51 ft^3
g = 32.2 ft/s^2
From property tables
Pair = 2.33 * 10^-3 slug/ft^3
μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s
Insert values into equation 2
Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 ) = 2.514 Ib
∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )
Assuming that flow is Laminar and RE < 1
Re = (Pair * vd) / μair -------- ( 3 )
where: Pair = 2.33 * 10^-3 slug/ft^3 , vd = ( 987 * 4 ) ft^2/s , μair = 3.8 * 10^-7 slug/ft.s
Insert values into equation 3
Re = 2.4 * 10^7 ( this means that the assumption above is wrong )
Hence we will use drag force law
Assume Cd = 0.5
Express Fd using the relation below
Fd = 1/2* Cd * Pair * AV^2
therefore V = 1.39 ft/s
Recalculate Reynold's number using v = 1.39 ft/s
Re = 34091
from the figure Cd ≈ 0.5 at Re = 34091
Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )
t = h/v
= 1000 / 1.39 = 719 seconds ≈ 12 mins
What can be used to measure the alcohol content in gasoline? A. Graduated cylinder B. Electronic tester C. Scan tool D. Either a graduated cylinder or an electronic tester
Answer:
GRADUATED CYLINDER
Explanation:
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
FOR BRAINLIST HELP PLEASE IS A DCP
A- Causes of the 13t Amendment
B- Reasons for Women's Suffrage
C- Reasons for the Freedmen's Bureau
D- Causes of the Plantation System
Answer:
C
Explanation: Freedmens Bureau provided resources for southerners and newly freed slaves
What 2 things can you never eat for breakfast?
i know the answer but lets see if you do
Answer:
you can't eat your school computer or a pencil.
is a baby duck swimming in a lake a learned behavior
Answer:
True because some ducks can't swim but have to learn it
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).
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
According to the article, what is one reason why commercial carmakers aim to develop driverless technology?
Incomplete question. However, I inferred you referring to the online article "How automakers can survive the self-driving era."
Explanation:
According to the article, as a result of the perceived demand for autonomous cars in the next few years, this has led to a heightened desire among commercial carmakers to develop driverless technology.
For example, carmakers such as Audi, Toyota have stated projections about the commercial availability of driverless cars.
A hydraulic cylinder pushes a heavy tool during the outward stroke, placing a compressive load of 400Ib in the piston rod. During the return stroke, the rod pulls on the tool with a force of 1500Ib. Calculate the resulting design factor for the 0.6 in-diameter rod when subjected to this pattern of forces for many cycles. Discuss your result and justify if the design factor is appropriate, too low or too high. The material is SAE 4130 WQT 1300 steel. Assume wrought steel, machined, and 99% reliability.
Nitrogen (N2) enters an insulated compressor operating at steady state at 1 bar, 378C with a mass flow rate of 1000 kg/h and exits at 10 bar. Kinetic and potential energy effects are negligible. The nitrogen can be modeled as an ideal gas with k 5 1.391. (a) Determine the minimum theoretical power input required, in kW, and the corresponding exit temperature, in 8C. (b) If the exit temperature is 3978C, determine the power input, in kW, and the isentropic compressor efficiency.
Answer:
A)
i) 592.2 k
ii) - 80 kw
B)
i) 105.86 kw
ii) 78%
Explanation:
Note : Nitrogen is modelled as an ideal gas hence R - value = 0.287
A) Determine the minimum theoretical power input required and exit temp
i) Exit temperature :
[tex]\frac{T_{2s} }{T_{1} } = (\frac{P2}{P1} )^{\frac{k-1}{k} }[/tex]
∴ [tex]T_{2s}[/tex] = ( 37 + 273 ) * [tex](\frac{10}{1} )^{\frac{1.391-1}{1.391} }[/tex] = 592.2 k
ii) Theoretical power input :
W = [tex]\frac{-n}{n-1} mR(T_{2} - T_{1} )[/tex]
where : n = 1.391 , m = 1000/3600 , T2= 592.2 , T1 = 310 , R = 0.287
W = - 80 kW ( i.e. power supplied to the system )
B) Determine power input and Isentropic compressor efficiency
Given Temperature = 3978C
i) power input to compressor
W = m* [tex]\frac{1}{M}[/tex] ( h2 - h1 )
h2 = 19685 kJ/ kmol ( value gotten from Nitrogen table at temp = 670k )
h1 = 9014 kj/kmol ( value gotten from Nitrogen table at temp = 310 k )
m = 1000/3600 , M = 28
input values into equation above
W = 105.86 kw
ii) compressor efficiency
П = ideal work output / actual work output
= ( h2s - h1 ) / ( h2 - h1 ) = ( T2s - T1 ) / ( T2 - T1 )
= ( 592.2 - 310 ) / ( 670 - 310 )
= 0.784 ≈ 78%
Name some technical skills that are suitable for school leavers .
Answer:
Welding, carpentry, masonry, construction worker, barber
Explanation:
A 0.06-m3 rigid tank initially contains refrigerant- 134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant- 134a at 1.2 MPa and 36°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has
entered the tank and (b) the amount of heat transfer.
Answer:
a) 0.50613
b) 22.639 kJ
Explanation:
From table A-11 , we will make use of the properties of Refrigerant R-13a at 24°C
first step : calculate the volume of R-13a ( values gotten from table A-11 )
V = m1 * v1 = 5 * 0.0008261 = 0.00413 m^3
next : calculate final specific volume ( v2 )
v2 = V / m2 = 0.00413 / 0.25 ≈ 0.01652 m^3/kg
a) Calculate the mass of refrigerant that entered the tank
v2 = Vf + x2 * Vfg
v2 = Vf + [ x2 * ( Vg - Vf ) ] ----- ( 1 )
where: Vf = 0.0008261 m^3/kg, V2 = 0.01652 m^3/kg , Vg = 0.031834 m^3/kg ( insert values into equation 1 above )
x2 = ( 0.01652 - 0.0008261 ) / 0.031834
= 0.50613 ( mass of refrigerant that entered tank )
b) Calculate the amount of heat transfer
Final specific internal energy = u2 = Uf + ( x2 + Ufg ) ----- ( 2 )
uf = 84.44 kj/kg , x2 = 0.50613 , Ufg = 158.65 Kj/kg
therefore U2 = 164.737 Kj/kg
The mass balance ( me ) = m1 - m2 --- ( 3 )
energy balance( Qin ) = ( m2 * u2 ) - ( m1 * u1 ) + ( m1 - m2 ) * he
therefore Qin = 41.184 - 422.2 + 403.655 = 22.639 kJ
Also discuss how bandwidth is affected by increasing the number of signal elements.
Question:
Why is my step-sis stuck in the washing machine every day?
———————————————————————————————-
Please, answer quick! I really need help figuring this out.
Alicia had to get over her fear of heights in order to become comfortable maintaining the generators in wind turbines. professional certification exam verifies the knowledge of the testee, whereas the CMP verifies green activity of the certified professional. A professional certification exam verifies the current qualifications of the testee, whereas the CMP ensures the certified professional continues to exhibit those qualifications. A professional certification exam verifies the qualifications of the testee, whereas the CMP measures the actual green commitment of the certified professional. A professional certification exam verifies the current knowledge of the testee, whereas the CMP ensures the certified professional continues to add to that knowledge.
Answer:a
Explanation:
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
FOR BRAINLIST ITS A DCP
The Battle of Sabine Pass
The Battle of Galveston
The Battle of Palmito Ranch
The Battle of Vicksburg
Answer:
I think The Battle of Sabine Pass
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
In this lab, we assumed that the flip-flops did not contribute to the timing constraints of the circuit. Unfortunately, this is not the case. As you saw when you simulated the D flip-flop, the sampling action does not happen instantaneously. In fact, a flip-flop will become unstable if the inputs do not remain stable for a certain amount of time prior to the rising-edge event (setup time) and a certain amount of time after the rising-edge event (hold time). Assume a setup and hold time of 2ns and 1ns, respectively. What would the theoretical maximum clock rate for the synchronous adder be in this scenario
The Aluminum Electrical Conductor Handbook lists a dc resistance of 0.01558 ohm per 1000 ft at 208C and a 60-Hz resistance of 0.0956 ohm per mile at 508C for the all-aluminum Marigold conductor, which has 61 strands and whose size is 1113 kcmil. Assuming an increase in resistance of 2% for spiraling, calculate and verify the dc resistance. Then calculate the dc resistance at 508C, and determine the percentage increase due to skin effect.
Answer:
a) 0.01558 Ω per 1000 feet
b) 0.0923 Ω per mile
c) 3.57%
Explanation:
a) Calculate and verify the DC resistance
Dc resistance = R = р [tex]\frac{l}{A}[/tex]
for aluminum at 20°C
р = 17 Ωcmil/ft
hence R = 17 * 1000 / ( 113000 ) = 0.01527 Ω per 1000 feet
there when there is an increase in resistance of 2% spiraling
R = 1.02 * 0.01527 = 0.01558 Ω per 1000 feet
b) Calculate the DC resistance at 50°C
R2 = R1 ( [tex]\frac{T+t2}{T+ t1}[/tex] )
where ; R1 = 0.01558 , T = 228 , t2 = 50, t1 = 20 ( input values into equation above )
hence R2 = ( 0.01746 / 0.189 ) Ω per mile = 0.0923 Ω per mile
c ) Determine the percentage increase due to skin effect
AC resistance = 0.0956 ohm per mile
Hence; Increase in skin effect
= ( 0.0956 -0.0923 ) / 0.0923
= 0.0357 ≈ 3.57%
Where do you prefer to live?
USA UK Canada ????????
Answer for POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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
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