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.
If 1 inch equals 10 feet, what would the measured distance be if the line scaled 1 1/2 inches?
A. 10 feet
b. 15 feet
c. 2 feet
d. 25 feet
Answer: B. 15 feet
Explanation:
1. The construction process begins with which of the following stages?
a) establishing the foundation
b) enclosing the building
c) site preparation
d) installing underground utilities
Answer:
c) site preparation
Explanation:
A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.
Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.
This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.
Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.
Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.
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]
Three 1.83 in. diameter bolts are used to connect the axial member to the support in a double shear connection. The ultimate shear strength of the bolts is 60 ksi, and a factor of safety of 3.9 is required with respect to fracture. Determine the allowable load P that can be applied to the axial member based on the shear strength of the bolts. Give your answer in kips. Enter a positive number.
Answer: the allowable load P is 242.7877 kips
Explanation:
Given that;
diameter bolts d = 1.83 in
ultimate shear strength of the bolts = 60 ksi
we know that
shear area = 2×(π/4)d²
= 2×(π/4)×(1.83)² = 5.2604 in²
so
p/3(5.2604) = 60000/3.9
p/15.7812 = 15384.6153
p = 15.7812 × 15384.6153
p = 242787.691 lb
p = 242.7877 kips
therefore the allowable load P is 242.7877 kips
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
On a Test please help
When towing a trailer, you should consider all of the following except
the minimum speed limit on roadways.
your personal protective equipment.
the gross weight of the load.
the increased stopping distance
Answer:
I'm not 100%sure but I think it's the first one
Compare and contrast the roles of agricultural and environmental scientists.
Answer:
HUHHHHHH BE SPECIFIC CHILE
Explanation:
ERM IRDK SORRY BOUT THAT
A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for ___ hours.
The amount of brine that must be fed to the evaporator is ___ kg/h.
Answer:
- the process should operate each day for 7.9286 hours
- the amount of brine that must be fed to the evaporator is 94.594 kg/hr
Explanation:
Given that;
concentration of brine = 26%
so water concentration will be 100% - 26% = 74%
for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)
so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr
Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr
Now for production of 195kg of pure dry salt, number of hours required will be
T = 195 / 24.5945 = 7.9286 hrs
Therefore the process should operate each day for 7.9286 hours.
Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg
Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr
Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr
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].
1) I love to swim. 2) A few years ago, my new year's resolution was to become a faster swimmer. 3) First, I started eating better to improve my overall health. 4) Then, I created a training program and started swimming five days a week. 5) I went to the pool at my local gym. 6) To measure my improvement, I tried to count my laps as I was swimming, but I always got distracted and lost track! 7) It made it very hard for me to know if I was getting faster. 8) This is a common experience for swimmers everywhere. 9) We need a wearable device to count laps, calories burned, and other real-time data. Summarey of the story
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