Answer:
I got iron
Explanation:
on my plato test
who is the strongest avenger i say hulk but who knows at this point
Answer:
or is the strongest evenger she hulk
Explanation:
?????????
Answer:
Thor!
Explanation:
In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.
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
Thermodynamics fill in the blanks The swimming pool at the local YMCA holds roughly 749511.5 L (749511.5 kg) of water and is kept at a temperature of 80.6 °F year round using a natural gas heater. If you were to completely drain the pool and refill the pool with 50°F water, (blank) GJ (giga-Joules) of energy are required to to heat the water back to 80.6 °F. Note: The specific heat capacity of water is 4182 J/kg ⋅°C. The cost of natural gas per GJ is $2.844. It costs $ (blank) to heat the pool (to the nearest dollar).
Answer:
[tex]95.914\ \text{GJ}[/tex]
[tex]\$272.78[/tex]
Explanation:
m = Mass of water = 749511.5 kg
c = Specific heat of water = 4182 J/kg ⋅°C
[tex]\Delta T[/tex] = Change in temperature = [tex]80.6-50=30.6^{\circ}\text{F}[/tex]
Cost of 1 GJ of energy = $2.844
Heat required is given by
[tex]Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}[/tex]
Amount of heat required to heat the water is [tex]95.914\ \text{GJ}[/tex].
Cost of heating the water is
[tex]95.914\times 2.844=\$272.78[/tex]
Cost of heating the water to the required temperature is [tex]\$272.78[/tex].
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 garden hose fills a 2-gallon bucket in 5 seconds. The number of gallons, g (y), is proportional to the number of seconds, t (x), that the water is running. What is the constant of proportionality?
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
Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structure, (2) an atomic radius of 0.154 nm, and (3) a saturation flux density of 0.83 tesla. Determine the number of Bohr magnetons per atom for this material. A: What is the volume, in m3, for this unit cell?B: How many atoms are there per m3 in this material?C: What is the number of Bohr magentons per atom for this material?
Answer:
a. 2.9218x10^(-29) m^3
b. 7.1230x10^(28) atoms/m^3
c. 2.0812 BM/atom
Explanation:
Atomic radius r = 0.154 nm
saturation flux density Bs = 0.83 tesla
- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3
= (2 x 0.154x10^-9)^3
= 2.9218x10^-29 m^3
- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;
Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)
= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)
= 2.4251x10^-29 / 1.1652x10^-29
= 2.0812 BM/atom
- Number of atoms per Vc, N = nb / Vc
= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms
A blown fuse or tripped circuit breaker shows
Answer:
Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.
Explanation:
It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.
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]
Compare and contrast the roles of agricultural and environmental scientists.
Answer:
HUHHHHHH BE SPECIFIC CHILE
Explanation:
ERM IRDK SORRY BOUT THAT
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 work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1.
(a) Determine the normal times in TMUs for these motion elements.
(b) What is the total time for this work element in sec
Answer:
a)
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b) 3.1 secs
Explanation:
a) Determine the normal times in TMUs for these motion elements
1) R16C ; Tn = 17 TMU
2) G4A ; Tn = 7.3 TMU
3) M10B5 ; Tn = 15.1 TMU
4) RL1 ; Tn = 2 TMU
5) R14B ; Tn = 14.4 TMU
6) G1B ; Tn = 3.5 TMU
7) M8C3 ; Tn = 14.7 TMU
8) P1NSE ; Tn = 10.4 TMU
9) RL1 ; Tn = 2 TMU
b ) Determine the total time for this work element in seconds
first we have to determine the total TMU = ∑ TMU = 86.4 TMU
note ; 1 TMU = 0.036 seconds
hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds
A rectangular bar has a edge crack at the bottom and is subjected to a pure bending moment. The crack length is a = 1 mm. The height of the bar is b = 12.5 cm. Knowing that the failure strength of the material is Sigma = 1,400 MPa, what is the fracture toughness of the material, K_ic.
Answer:
The answer is "[tex]\bold{87.3906 \ MPa \sqrt{m}}[/tex]".
Explanation:
Given value:
[tex]\sigma = 1400 \ MPa \ \ \ \ \ \ \ \ where \ \sigma = failure \ strength\\\\a = 1 \ mm = 1 \times 10^{-3} \ m \ \ \ \ \ \ \ \ \ \ where\ a = crack\ length\\\\b= 12.5 \ cm = 125 \ mm = 0.125 \ m\\\\[/tex]
[tex]\to \alpha = \frac{a}{b} =\frac{1}{125} = 8 \times 10^{-3}\\\\[/tex]
[tex]k_{b} = \frac{1.12 + \alpha (2.62 \alpha -1.59)}{1-0.7 \alpha}\\[/tex]
[tex]= \frac{1.12 + (8\times 10^{-3}(2.62(8\times 10^{-3}) -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(0.02096 -1.59))}{1-(0.7 \times 8\times 10^{-3})}\\\\= \frac{1.12 + (8\times 10^{-3}(-1.56904))}{1-(0.0056)}\\\\= \frac{1.12 + (-0.01255232)}{0.9944}\\\\= \frac{-1.10744768}{0.9944}\\\\= -1.11368431\\\\[/tex]
[tex]k_{ic} = \sigma \sqrt{\pi a} \ y_b[/tex]
[tex]=1400 \times \sqrt{\pi \times 1 \times 10^{-3} } \times -1.11368431\\\\=1400 \times 0.00177200451 \times -1.11368431\\\\=87.3906 \ MPa \sqrt{m}[/tex]
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.
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).
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.
