Answer:
Because it helps to speed up recovery and required recovery efficiency When recovering refrigerant from a household refrigerator who's compressor does not run
What is the purpose of an engineer?
Answer:
Their purpose is to invent, design, analyze, build and test machines, complex systems, structures, gadgets and materials to fulfill functional objectives and requirements while considering the limitations imposed by practicality, regulation, safety and cost.
Answer:
To make things
Explanation:
Tech a says that only a certain few people will find jobs in the automotive industry. Tech b says the automotive industry offers numerous career choices . Who's correct?
Answer: Tech B
Explanation:
From the question above, tech B saying the automotive industry offers numerous career choices is the correct answer. Pursuing a career in automotive industry is challenging and rewarding as one works with both vehicles and technology.
Examples of career choices in the automotive industry are car detailer, tire technician, car rental agent, auto mecahnic, vehicle inspector, auto electrician, auto engineer etc.
The resultant force is directed along the positive x axis and has a magnitude of 1330 N.
Determine the magnitude of F_A. Express your answer to three significant figures and include the appropriate units. Determine the direction theta of F_A. Express your answer using three significant figures.
Answer:
the magnitude of F_A is 752 N
the direction theta of F_A is 57.9°
Explanations:
Given that,
Resultant force = 1330 N in x direction
∑Fx = R
from the diagram of the question which i uploaded along with this answer
FB = 800 N
FAsin∅ + FBcos30 = 1330 N
FAsin∅ = 1330 - (800 × cos30)
FA = 637.18 / sin∅
Now ∑Fx = 0
FAcos∅ - FBsin30 = 0
we substitute for FA
(637.18 / sin∅)cos∅ = 800 × sin30
637.18 / 800 × sin30 = sin∅/cos∅
and we know that { sin∅/cos∅ = tan∅)
so tan∅ = 1.59295
∅ = 57.88° ≈ 57.9°
THEREFORE FROM THE EQUATION
FA = 637.18 / sin∅
we substitute ∅
so FA = 637.18 / sin57.88
FA = 752 N
The amount of time an activity can be delayed and yet not delay the project is termed:_________
A. Total slack.
B. Free slack.
C. Critical float.
D. Float pad.
E. Slip pad.
The correct answer is A. Total slack
Explanation:
Projects include multiple tasks or activities; moreover, each activity and the project itself have a specific due date. Besides this, activities can be delayed and this might affect or not the due date of the project. Indeed, some activities can have extra time without any effect on the date of the project, which is known as total slack or total float (delay in activities that does not delay the project.) Additionally, total slack is possible if activities or tasks are flexible, for example, if each task in a project should take 1 day and one of the tasks takes one day and a half, this can be compensated if another task takes less time. According to this, the correct answer is A.
A spherical ball of solid, nonporous naphthalene, a "mothball," is suspended in still air. The naphthalene ball slowly sublimes, releasing the naphthalene vapor into the surrounding air by the molecular diffusion-limited process. Estimate the time required to reduce the diameter from 2.0 to 0.50 cm when the surrounding air is at 347 K and 1.0 atm. Naphthalene has a molecular weight of 128 g/mol, a solid density of 1.145 g/cm3, and diffusivity in the air of 8:19 10␣6 m2=s, and exerts a vapor pressure of 5.0 Torr (666 Pa) at 347 K.
Answer:
The time required to reduce the diameter of solid spherical ball from 2 cm to 0.50 cm = 61.60 hours
Explanation:
Given that:
The bulk gas temperature T = 347 K
The pressure = 1.0 atm
Molecular weight of naphthalene = 128 g/mole
density of the solid = 1.145 g/cm³
diffusivity in the air of coefficient = 8.19 × 10 ⁻⁶ m²/s
vapor pressure = 5.0 Torr (666 Pa)
The shape of the ball in this given study is spherical. So, the first step we need to carry out is to state the general differential equation for the mass transfer in the system of the spherical coordinates. After that , we will make use of simplified differential expression form of Fick's Flux. From there we'll determine the surface concentration (c) of the species and finally determine the material balance equation on the naphthalene which will be used in solving the time required to reduce the diameter from 2.0 cm to 0.50cm.
THE DETAILED EXPLANATION OF WHAT WE JUST SAID CAN BE SEEN IN THE ATTACHED IMAGES BELOW.
Which of the three predominant equations for friction loss in fluid flow can be used only for water and is primarily used for open-channel (gravity) flow?
A. Darcy-Weisbach
B. Manning
C. Hazen-Williams
D. All of the above
Answer:
B. Manning
Explanation:
The Manning formula is an empirical formula for velocity estimation of a liquid flowing in a open channel, only used for water. The Manning formula is:
[tex]V = \frac{k}{n} \cdot R_{h}^{2/3}\cdot S^{1/2}[/tex]
Where:
[tex]V[/tex] - Cross sectional average velocity, measured in meters per second.
[tex]k[/tex] - Conversion factor, dimensionless. ([tex]k = 1[/tex] for SI units)
[tex]n[/tex] - Gauckler-Manning constant, measured in [tex]\frac{s}{m^{1/3}}[/tex].
[tex]R_{h}[/tex] - Hydraulic radius, measured in meters.
[tex]S[/tex] - Slope of the hydraulic grade line, dimensionless.
Both Darcy-Weisbach and Hazen-Williams formulas are used only for fluid flowing in pipes.
In a nutshell, the correct answer is B.
The input resistance at the base of a biased transistor depends mainly on
(a) bDC
(b) RB
(c) RE
(d) bDC and RE
Answer:
D
Explanation:
Exceeding critical mach may result in the onset of compressibility effects such as:______.
Answer:
Sound barrier.
Explanation:
Sound barrier is a sudden increase in drag and other effects when an aircraft travels faster than the speed of sound. Other undesirable effects are experienced in the transonic stage, such as relative air movement creating disruptive shock waves and turbulence. One of the adverse effect of this sound barrier in early plane designs was that at this speed, the weight of the engine required to power the aircraft would be too large for the aircraft to carry. Modern planes have designs that now combat most of these undesirable effects of the sound barrier.
Where should the Q point of an amplifier be on a DC load line to have the LARGEST linear output?
Answer:
At the middle of the DC load
Explanation:
For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region
which of the following devices are used in networking? check all that apply?
switches, hubs,TCP or routers
Answer:
Hub , switch port , transmission media , hope I helped
Explanation:
Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb which acts up to the left, in the same direction as from B towards A. Take θ = 30 ∘.
Answer:
design angle ∅ = 4.9968 ≈ 5⁰
Explanation:
First calculate the force Fac :
Fac = [tex]\sqrt{400^2 + 650^2 - 2(400)(650)cos30}[/tex]
= [tex]\sqrt{160000 + 422500 - 80210}[/tex]
= 708.72 Ib
using the sine law to determine the design angle
[tex]\frac{sin}{400} = \frac{sin 30}{Fac}[/tex]
hence ∅ = [tex]sin^{-1} (\frac{sin 30 *400}{708.72} )[/tex]
= [tex]sin^{-1} 0.0871[/tex] = 4.9968 ≈ 5⁰
The design angle of the struts system given is;
ϕ = 38.26°
Parallelogram law of vector additionThe diagram showing the struts system is missing and so i have attached it.
Now, from the forces given to us acting on the system attached, I have drawn a force diagram in form of a parallelogram showing the direction of the forces action. The second attached image shows this force diagram called diagram a.From the diagram, we can see that it follows parallelogram law of vector addition.
Now, I have narrowed down the force system to the upper part of the strut as depicted in the third image attached.
From the third image attached, we can use the concept of parallelogram law of vector addition and the law of cosine to get the force on the strut AC which is;F_AC = √(400² + 600² - 2(400 × 600)cos 30)
F_AC = 322.97 lb
Now, to get the design angle, we will solve as;(sinϕ)/(sin 30) = 400/322.97
Where; ϕ is the design angle.
Thus;
(sinϕ) = (400/322.97) × sin 30
(sinϕ) = 1.2385 × 0.5
(sinϕ) = 0.61925
ϕ = sin^(-1) 0.61925
ϕ = 38.26°
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a mercury barometer located in a lab has a heigh tof 750 mm. what is the atmosphereic pressur ein kPa
Answer:
The atmospheric pressure from a mercury barometer equals to 99.992 kilopascals.
Explanation:
760 milimeters of mercury column is equal to 101.325 kilopascals. The atmospheric pressure is found by simple rule of three:
[tex]p_{atm} = \frac{750\,mm\,Hg\times 101.325\,kPa}{760\,mm\,Hg}[/tex]
[tex]p_{atm} = 99.992\,kPa[/tex]
The atmospheric pressure from a mercury barometer equals to 99.992 kilopascals.
You are driving on a roadway with multiple lanes of travel in the same direction, and are approaching an emergency vehicle parked along the roadway. You must:a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.b. Yield to the emergency vehicle.c. If you are in the lane closest to the emergency vehicle, immediately come to a complete stop and ask if anyhelp is needed.d. All of the above.
Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.
Explanation:
Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.
The output current of an amplifier is 190 mA and the input current is 32 mA. The current gain of the amplifier is:
Answer:
-5.9375 (or 5.9375 depending)
Explanation:
The current gain of the amplifier is simply the output current over the input current negated.
-190 mA / 32 mA == -5.9375
So the current gain for the amplifier was -5.9375
The answer a program may be looking for may not be negative, but based on the derivation, the gain should be negative for amplifiers.
Cheers.
A data bus can be visualized as a multilane highway
A. and each component is located at an intersection where it will turn or go straight
B. with each component having an individual address
C. and each component located in a curve where the data will slow down
D. with each component acting as a traffic light, stop-go
Answer:
B. with each component having an individual address
Explanation:
Data bus is a system within a computer or device, consisting of a connector or set of wires, that provides transportation for data. Data bus needs an address unique to each component in order to deliver the right data to the right place. Every memory location has a unique binary address. A microprocessor architecture is mainly composed of two main buses: The data bus and the address bus.
If a transistor is operating with 5 mA of Collector Current, 100uA of Base Current and VCE = 10 V. What is the power dissipation?
Answer:
well its 31.4226804124
Explanation:
A circular hoop sits in a stream of water, oriented perpendicular to the current. If the area of the hoop is doubled, the flux (volume of water per unit time) through it:___________
Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2
When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is known as spiral gearing.
a) true
b) false
Answer:
False
Explanation:
The spiral gearing is the application that is commonly used in any vehicle where the drive from the shaft will turn perpendicular to the drive of the wheel.
The two parallel and coplanar shafts are connected by gears having helical teeth perpendicular to the axis of the shaft.
Spiral gearing application is appropriate for any machine or vehicle with a demand of great velocity and large torque power.
When two parallel and coplanar shafts are connected by gears having teeth parallel to the axis of the shaft, the arrangement is known as spiral gearing is therefore false.
Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determine the final temperature of argon.
Answer:
181 °C
Explanation:
Initial pressure [tex]P_{1}[/tex] = 100 kPa
Initial temperature [tex]T_{1}[/tex] = 30 °C = 30 + 273 K = 303 K
Final pressure [tex]P_{2}[/tex] = 1200 kPa
Final temperature [tex]T_{2}[/tex] = ?
n = 1.2
For a polytropic process, we use the relationship
([tex]T_{2}[/tex]/[tex]T_{1}[/tex] ) = ([tex]P_{2}[/tex]/[tex]P_{1}[/tex])^γ
where γ = (n-1)/n
γ = (1.2-1)/1.2 = 0.1667
substituting into the equation, we have
([tex]T_{2}[/tex]/303) = (1200/100)^0.1667
[tex]T_{2}[/tex]/303 = 12^0.1667
[tex]T_{2}[/tex]/303 = 1.513
[tex]T_{2}[/tex] = 300 x 1.513 = 453.9 K
==> 453.9 - 273 = 180.9 ≅ 181 °C
The final temperature of Argon in polytropic process is T = 181°C
Given data:
The initial pressure [tex]P_{1}=100kPa[/tex]
The final pressure [tex]P_{2}=1200kPa[/tex]
The initial temperature [tex]T_{1}=303K[/tex]
The value of n = 1.2
For a polytropic process:
[tex]\frac{T_{2}}{T_{1}} =\frac{P_{1}}{P_{2}}^{\beta }[/tex]
The value of [tex]\beta = \frac{(n-1)}{n}[/tex]
So, [tex]\beta =0.1667[/tex]
Substituting the values in the equation:
[tex]\frac{T_{2}}{303} =\frac{1200}{100}^{0.1667 }[/tex]
So, the value of [tex]T_{2}=453.9K[/tex]
The temperature in degree Celsius is [tex]T_{2}=181C[/tex].
Hence, the final temperature is T = 181°C.
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What is the approximate probability that no people in a group of seven have the same birthday?
(A) 0.056
(B) 0.43
(C) 0.92
(D) 0.94
Answer: (D) 0.94
P(7 distinct birthdays) = 363/365 = 0.9438 (0.94)
The probability that no people in a group of seven have the same birthday is; D: 0.94
How to find the probability?The probability that no people in a group of seven have the same birthday is;
P(no two people in a group of seven) = P(1) * P(2) * P(3) * P(4) * P(5) * P(6) *P(7)
⇒ 365/365 * 364/365 * 363/365 * 362/365 * 361/365 * 360/365 * 359/365 = 0.9438 ≈ 0.94
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Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
a. 2.17000000 A-2.258571429 A
b. 2.20000000 A-2.29000000 A
c. 2.211445 A-2.30000000 A
d. 2.20144927 A-2.29130435 A
e. 2.00000000 A-2.30000000 A
Answer:
the only one that meets the requirements is option C .
Explanation:
The tolerance of a quantity is the maximum limit of variation allowed for that quantity.
To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula
x_average = ∑ [tex]x_{i}[/tex] / n
The tolerance or error is the current value over the mean value per 100
Δx₁ = x₁ / x_average
tolerance = | 100 -Δx₁ 100 |
bars indicate absolute value
let's look for these values for each case
a)
x_average = (2.1700000+ 2.258571429) / 2
x_average = 2.2142857145
fluctuation for x₁
Δx₁ = 2.17000 / 2.2142857145
Tolerance = 100 - 97.999999991
Tolerance = 2.000000001%
fluctuation x₂
Δx₂ = 2.258571429 / 2.2142857145
Δx2 = 1.02
tolerance = 100 - 102.000000009
tolerance 2.000000001%
b)
x_average = (2.2 + 2.29) / 2
x_average = 2,245
fluctuation x₁
Δx₁ = 2.2 / 2.245
Δx₁ = 0.9799554
tolerance = 100 - 97,999
Tolerance = 2.00446%
fluctuation x₂
Δx₂ = 2.29 / 2.245
Δx₂ = 1.0200445
Tolerance = 2.00445%
c)
x_average = (2.211445 +2.3) / 2
x_average = 2.2557225
Δx₁ = 2.211445 / 2.2557225 = 0.9803710
tolerance = 100 - 98.0371
tolerance = 1.96%
Δx₂ = 2.3 / 2.2557225 = 1.024624
tolerance = 100 -101.962896
tolerance = 1.96%
d)
x_average = (2.20144927 + 2.29130435) / 2
x_average = 2.24637681
Δx₁ = 2.20144927 / 2.24637681 = 0.98000043
tolerance = 100 - 98.000043
tolerance = 2.000002%
Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017
tolerance = 2.0000002%
e)
x_average = (2 +2,3) / 2
x_average = 2.15
Δx₁ = 2 / 2.15 = 0.93023
tolerance = 100 -93.023
tolerance = 6.98%
Δx₂ = 2.3 / 2.15 = 1.0698
tolerance = 6.97%
Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .
Based on the information given, the correct option will be C. 2.211445 A-2.30000000 A.
It should be noted that the tolerance range simply means the maximum limit of variation that can be allowed for a particular quantity.
It can be noted that the range that depicts the 2% tolerance range to the full 9 digits provided is 2.211445 A-2.30000000 A. It illustrates a high precision regarding the question.
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Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.
Answer:
Explanation:
Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.
Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.
Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease.
Answer:
Option D - the moisture content at turbine exit will decrease
Explanation:
In an ideal rankine system, the phenomenon of superheating occurs at a state where the vapor state of the fluid is heated above its saturation temperature and the phase of the fluid is changed from the vapor phase to the gaseous phase.
Now, a vapour phase has two different substances at room temperature, whereas a gas phase consists of just a single substance at a defined thermodynamic range, at standard room temperature.
At the turbine exit, since it's just a single substance in gaseous phase, it means it will have less moisture content.
Thus, the correct answer is;the moisture content at turbine exit will decrease
In a four-stroke engine, the piston rises in the cylinder, which triggers the _______ stroke.
intake
compression
exhaust
power
Answer: It is power stroke
Steam enters an adiabatic turbine with a mass flow rate of 6 kg/s with a specific enthalpy of h = 3248.4 kJ/kg). It exits the turbine at a pressure of 15 kPa and a quality of 0.9. The power generated by this turbine, in kW, is
Answer:
5318.346
Explanation:
From the question we have
M = 6kg/s
P = 15kpa
h1 = 3248.4kj
x = 0.9
H2 is unknown
From the steam table
hf = 225.85
hg = 2599.36
H2 = HF + x(hg -hf)
h2= 225.85+0.9(2599.36 - 225.85)
= 225.85 + 0.9(2373.51)
= 225.85 + 2136.159
= 2362.009
To find The power generated by this turbine:
We insert all the parameters into the equation below:
m(h1 - h2)
= 6(3248.4 - 2362.009)
= 6 x 886.391
= 5318.346KW
what is
entrained
the difference between Air-
and Air- entrapped,
2
Answer:
sometimes small air bubbles are intentionally incorporated (entrained) into the mix using admixtures; other times larger bubbles are entrapped during mixing. When the bubbles are smaller than 0.04 inch, the air is called entrained; larger, and it's called entrapped.
Explanation:
In a high-quality coaxial cable, the power drops by a factor of 10 approximately every 5 km. If the original signal power is 0.25 W (=2.5 x 10-1), how far will a signal be transmitted before the power is attenuated to 25 μW? As part of your answer, include a Table showing the signal power vs. distance in 5 km intervals. If optical fibre is used instead of the coaxial cable, briefly explain how you would expect the above calculated distance value to change. You are not required to include another Table.
Answer:
20 km for 40 dB loss80 km for 40 dB loss, or 10 dB loss for 20 kmExplanation:
Here's your table of (distance, power level):
(0 km, 250 mW), (5 km, 25 mW), (10 km, 2.5 mW),
(15 km, 250 μW), (20 km, 25 μW)
The signal can be transmitted 20 km before being attenuated to 25 μW.
__
Reportedly, the loss in fiber optic cable is about 0.5 dB/km. This compares to 10 dB/5 km = 2 dB/km for the coaxial cable. Loss in dB/km is a factor of 4 less for fiber optic cable, so the distance for the same loss would be multiplied by 4.
5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025
Answer:
LMC wall thickness= 5.05
Explanation:
Given:
Minimum inner diameter = 25 - 0.05 = 24.95
Maximum outer diameter = 35 + 0.05 = 35.05
Find:
LMC wall thickness
Computation:
LMC wall thickness = (maximum outer diameter - minimum inner diameter) / 2
LMC wall thickness = (35.05 - 24.95) / 2
LMC wall thickness= 5.05
A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the appropriate units.Part BSpecify his mass in kilograms.Express your answer to three significant figures and include the appropriate units.Part CSpecify his weight in newtons.Express your answer to three significant figures and include the appropriate units.Part DIf the man is on the moon, where the acceleration due to gravity is gm = 5.30 ft/s2, determine his weight in pounds.Express your answer to three significant figures and include the appropriate units.Part EDetermine his mass in kilograms.Express your answer to three significant figures and include the appropriate units.
Answer:
a) 4.51 lbf-s^2/ft
b) 65.8 kg
c) 645 N
d) 23.8 lb
e) 65.8 kg
Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound = [tex]x[/tex] slug
[tex]x[/tex] = 145/32.174 = 4.51 lbf-s^2/ft
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds = [tex]x[/tex] kg
[tex]x[/tex] = 145/2.205 = 65.8 kg
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = 645 N
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
[tex]x[/tex] m/s^2 = 5.30 ft/s^2
[tex]x[/tex] = 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = 23.8 lb
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = 65.8 kg
The term route of entry on an sds refers to the way a ___ enters the body
The correct answer is A) Chemical
Explanation:
The abbreviations SDS means Safety Data Sheet, which refers to a very complete document about chemicals and safety related to these, including their properties and potential hazards, as well as, procedures or steps to avoid hazards and accidents.
This means the main focus of this document is chemicals; moreover, in this, there is information about the route of entry, which refers to the way chemical can enter the body and includes through inhalation, ingestion, contact with skin, among others. Also, depending on the chemical some routes of entry represent a major hazard. According to this, the correct answer is A.
Answer: Chemical!!!
Explanation: