Answer:
0.4 gallons per second
Explanation:
A function shows the relationship between an independent variable and a dependent variable.
The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.
The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).
Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.
Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second
Technician A says that a way to prevent galvanic corrosion is to duplicate the original installation method. Technician B says that a way to prevent galvanic corrosion is to reuse coated bolts. Who is right
Answer:
Technician A
Explanation:
Galvanic corrosion is not on only one metal alone but caused when two metals are interacting. Thus, Duplicating the original installation method is a better option because re-using a coated bolt doesn't prevent galvanic corrosion because both materials must be coated and not just the bolt and in technician B's case he is coating just the bolt. Thus, technician B's method will not achieve prevention of galvanic corrosion but technician A's method will achieve it.
Which system provides an easier way for people to communicate with a computer than a graphical user interface (GUI)?
a) data mining
b) transaction-processing
c) natural language processing
d) data abstraction
Answer:
C.
Explanation:
Natural language processing or NLP can be defined as an artificial intelligence that aids computers to understand, interpret, and manipulate human language. NLP is subfiled of many disciplines, for instance, artifical intelligence, linguistics, computer science, etc. NLP helps computers to interact with humans in their own language.
So, the easier way through which people can communicate with computers apart from GUI is NLP. Thus option C is correct.
An ECG has a scalar magnitude of 1 mV on lead II and a scalar magnitude of 0.5 mV on lead III. Calculate the scalar magnitude on lead I.
Answer: the scalar magnitude on lead I is 0.5 mV
Explanation:
Given that;
scalar magnitude on lead II = 1 mV
scalar magnitude on lead III = 0.5 mV
the scalar magnitude on lead I = ?
we know that;
Lead I Voltage = LA - RA -----------let this be equation 1
where LA is left arm electrode and RA is right am electrode
Also
Lead II = LL - RA
where LL is the left leg of electrode
we substitute
1 mV = LL - RA ---------------------let this be equation 2
Again
Lead III = LL - LA
we substitute
0.5 mV = LL - LA ------------------let this be equation 3
now subtract equation 3 and 2
1 mV - 0.5 mv = LL - RA - (LL - LA)
0.5 mV = LL - RA - LL + LA
0.5 mV = -RA + LA
0.5 mV = LA - RA
now taking a look at our equation 1 ( Lead I Voltage = LA - RA )
hence, Lead I Voltage = LA - RA = 0.5 mV
Therefore the scalar magnitude on lead I is 0.5 mV
A site is underlain by a soil that has a unit weight of 118 lb/ft3. From laboratory shear strength tests that closely simulated the field conditions, the total stress parameters were measured to be C total = 250 lb/ft2 and φ total = 29°. Estimate the shear strength on a horizontal plane at a depth of 12 ft below the ground surface at this site in lbs/ sq ft
Answer: the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
Explanation:
Given that;
Weight of soil r = 118 lb/ft³
stress parameter C = 250 lb/ft²
φ total = 29°
depth Z = 12 ft
The shear strength on a horizontal plane at a depth of 12ft
ζ = C + δtanφ
where δ = normal stress
normal stress δ = r × z = 118 × 12 = 1416
so
ζ = C + δtanφ
ζ = 250 + 1416(tan29°)
ζ = 250 + 1416(tan29°)
ζ = 250 + 784.9016
ζ = 1034.9015 lb/ft²
Therefore the shear strength at a depth of 12 ft is 1034.9015 lb/ft²
If the length of the arm were to increase, how would the required torque change? a. Required torque would increase b. Required torque would decrease c. Required torque would remain the same d. It is impossible to calculate required torque with the values given.
A _________ is interesting only if the statistics computed from transactions covered by the rule are different than those computed from transactions not covered by the rule.
Answer: Quantitative association rule.
Explanation:
The quantitative rules of association apply to the basic type of rules of association which exists as X and Y, with X and Y consisting of a collection of numerical and/or categorical attributes. Unlike general association laws, where both the left and right sides of the law should have categorical (nominal or discrete) attributes, a numerical attribute must be included in at least one attribute of the quantitative association rule (left or right).
Given that the skin depth of graphite at 100 (MHz) is 0.16 (mm), determine (a) the conductivity of graphite, and (b) the distance that a 1 (GHz) wave travels in graphite such that its field intensity is reduced by 30 (dB).
Answer:
the answer is below
Explanation:
a) The conductivity of graphite (σ) is calculated using the formula:
[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]
where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012
Substituting:
[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]
b) f = 1 GHz = 10⁹ Hz.
[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]
