To prevent pests from coming it!
A correctly designed and installed air curtain can be used to prevent insects, dust, and debris from entering a building
How to determine what it preventsAn air curtain, when correctly designed and installed, can prevent insects, dust, and debris from entering a building. It can also act as a thermal barrier, blocking the transfer of heat or cold, disperse odors, and contain smoke in the event of a fire.
A correctly designed and installed air curtain can be used to prevent or minimize several things.
It's important to note that the effectiveness of an air curtain depends on various factors, such as its design, installation, and maintenance. Proper sizing, airflow velocity, and positioning are crucial to achieving the desired results.
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List down 10 items made from plastic in your home. Describe why each item is made from plastic.
Answer:
each item is made from plastic are:
water bottlebucketjuggallonplastic bags as poly thin bagto cover wirehandle of knife,pressure cooker,etc plastic roofdustbinplastic plates ,cupsExplanation:
Items made from plastic in our home are
water bottlesjugbucketfood wrappersdisposable plastic cupshand sanitizer bottlesplastic grocery bagsstraws fruit basket Dustbinitems and their uses
Water bottles are used to drink waterfood wrappers are used to wrap the different types of food itemsplastic grocery bags are used to carry grocery itemshand sanitizer bottles are used to store sanitizerfruit basket is used to store and carry fruits .hope it is helpful to you
Voltage signals are generated on the output of a piezoelectric pressure transducer that is equipped with an amplifier.
a. True
b. False
Answer: True
Explanation: Piezoelectric pressure transducers are equipments that helps to generate voltage through the mechanical actions such as applied pressure. Piezoelectric transducer is made up of quartz crystals which are made of silicone and oxygen formed in a crystalline structure.
Piezoelectric transducer can also be described as a Piezoelectric sensor as it helps to detect electric voltage generated through applied pressure.
A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight of steel is 490 pcf. Compute:
a. the maximum tensile stress due to the weight of the wire
b. the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi.
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ[tex]_w[/tex] = W / A
we substitute
σ[tex]_w[/tex] = 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ[tex]_{all[/tex] - σ[tex]_w[/tex] )A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb
A series of end-milling cuts is currently used to produce an aluminum part that is an aircraft component. The purpose of the machining operation is to remove 95% of the part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle due to tool repositioning. The part has a length = 1.6 m, width=0 m, and mm. The operation uses a four-tooth indexable-insert end mill with mm at a cutting speed 600 m/min, chip load 0.15 mm/tooth, and average cross-sectional area of cut-240 mm. High-speed machining has to replace the conventional milling process. The same chip load and average area of cut will be used, but the cutting speed will be increased to 3600 m/min , and the time lost for tool repositioning will be reduced to 2.0 min.
Determine :
(a) The cycle time of the current milling operation and
(b) The cycle time of the proposed HSM operation.
(c) Is this part a good candidate for high-speed machining?
Answer:
what-
Explanation:
I dont understand
A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. The damping ratio is 0.4.
Required:
a. Find natural frequency (in Hz) of this instrument.
b. Determine the bandwidth of this instrument in Hz
c. How can you increase the bandwidth of this instrument?
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
a) Calculate Natural frequency
Wn = √k/m = [tex]\sqrt{12000 / 5*10^{-3} }[/tex]
= 1549.19 rad/s ≈ 246.56 Hz
b) Bandwidth of instrument
W / Wn = [tex]\sqrt{1-2(0.4)^2}[/tex]
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
A steel stamping is to be nickel plated for corrosion protection. The part is a rectangular flat plate that is 3.2 mm thick and whose face dimensions are 13 cm by 16 cm. The plating operation is carried out in an acid sulfate electrolyte, using a current of 25 amps for duration of 60 min.
Required:
Determine the average thickness of the plated metal resulting from this operation.
Answer:
the average thickness of the plated metal resulting from this operation is 0.0744 mm
Explanation:
Given the data in the question;
thickness = 3.2 mm and face dimensions are 13 cm by 16 cm; i.e 130 mm by 160 mm
Total surface area of the steel plate to be electroplated will be;
⇒ A = 2( ( 130 × 160 ) + ( 3.2 × 130 ) + ( 3.2 × 160 ) )
A = 2( 20800 + 416 + 512 )
A = 2 × 21728
A = 43456 mm²
Given that; Current I = 25 amps
time t = 60 min = 60 × 60 = 3600 sec
From table C below;
For Nickel ( Acid Sulfate ); Cathode efficiency n = 95% = 0.95
Panting Constant C = 3.42 × 10⁻² mm³/amp-s
Hence, Required Thickness of plated metal will will be;
⇒ ( C × I × t ) / ( A × n )
we substitute
⇒ ( (3.42 × 10⁻²) × 25 × 3600 ) / ( 43456 × 0.95 )
⇒ 3078 / 41283.2
⇒ 0.0744 mm
Therefore, the average thickness of the plated metal resulting from this operation is 0.0744 mm
A surface grinding operation is used to finish a flat plate that is 5.50 in wide and 12.500 in long. The starting thickness is 1.085 in. After grinding the surface, the thickness is 1.000 in. The grinding wheel had a starting diameter of 6.013 in and a width of 0.50 in. After the operation, the diameter of the grinding wheel is 5.997 in. Determine the grinding ratio in this operation.
Answer:
77.40
Explanation:
Initial width = 5.5 Inches
Initial length = 12.5 inches
Initial thickness = 1.085 inches
After grinding
Thickness of flat plate = 1 inch
Grinding wheel
starting diameter( di )= 6.013 inches
width of grinding wheel = 0.5 inch
After operation
Diameter of grinding wheel ( df ) = 5.997
Calculate the grinding ratio in this operation
First step : determine the volume of material removed from flat plate
= Length of flat plate * width of flat plate * change in thickness
= 12.5 * 5.5 * ( 1.085 - 1 )
= 5.8437 in^3
Volume of material from grinding wheel
= π / 4 * ( di^2 - df^2 ) * width
= π / 4 * ( 6.013^2 - 5.997^2 ) * 0.5
= 0.0755 in^3
Finally the Grinding ratio
= 5.8437 / 0.0755
= 77.4
The following data represent the time of production (in hours) for two different factories for the same product. Which factory has the best average time of production? Which factory will you select and why?
Factory A
14, 10, 13, 10, 13, 10, 7
Factory B
9, 10, 14, 14, 11, 10, 2
To inspect a 12,500 N car, it is raised with a hydraulic lift. If the radius of the small piston is 5.0 cm, and the radius of the large piston (the car is sitting on the large piston) is 50 cm, find the force that must be exerted on the small piston to lift the car. N (precision to the nearest whole number)
Answer:
The force will be "125 N".
Explanation:
The given values are:
[tex]F_1=12500 \ N[/tex]
[tex]R_1 = 50 \ cm[/tex]
[tex]R_2=5 \ cm[/tex]
As we know,
⇒ [tex]A=\pi(H)^2[/tex]
then,
⇒ [tex]A_2=\pi(5)^2[/tex]
⇒ [tex]A_1=\pi(50)^2[/tex]
Since,
The pressure on both the pistons are equal, then
⇒ [tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}[/tex]
or,
⇒ [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]
By substituting the values, we get
⇒ [tex]\frac{F_2}{12500} =\frac{\pi(5)^2}{\pi(50)^2}[/tex]
⇒ [tex]\frac{F_2}{12500} =\frac{\pi(25)}{\pi(2500)}[/tex]
⇒ [tex]F_2=\frac{25}{2500}\times 12500[/tex]
⇒ [tex]=0.01\times 12500[/tex]
⇒ [tex]=125 \ N[/tex]
The 5-lb cylinder is falling from A with a speed vA = 10 ft/sonto the platform. Determine the maximum displacementof the platform, caused by the collision. The spring has anunstretched length of 1.75 ft and is originally kept incompression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision.
Answer: hello some pictorial details related to your question is missing attached below is the missing detail
answer : 0.0735 ft
Explanation:
weight of cylinder = 5 IB
speed ( v ) = 10 ft/s
Calculate the maximum displacement of platform
Express the initial energy of the system as
potential energy of cylinder + potential energy of spring + Kinetic energy of cylinder
= mgh + 1/2 ky^2 + 1/2 mv^2
y = initial spring compression = 1.75 - 1 = 0.75
hence the initial energy can be expressed as
= 5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 ------------ ( 1 )
Next : determine the Final energy of the system
Final energy of the system = potential energy of the system
= 1/2 k ( x + 0.75 )^2 ----- ( 2 )
Equating equations ( 1 ) and ( 2 ) to determine the value of x
5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 = 1/2 k ( x + 0.75 )^2
solve for x
x ( max displace of platform ) = 0.0735 ft
1. ¿Cómo esta simbolizada la Cena en el Antiguo testamento?
Answer:
La Comunión o la Cena del Señor es el partir y comer el pan para simbolizar el cuerpo de Cristo partido por nosotros y beber vino para recordar la sangre que derramó por nuestros pecados.
Explanation:
A blizzard is a massive snowstorm. Definitions vary, but for our purposes, we will assume that a blizzard is characterized by both winds of 30 mph or higher and blowing snow that leads to visibility of 0.5 miles or less, sustained for at least 4 hours. Data from a storm one day has been stored in a file stormtrack.dat. There are 24 lines in the file, one for each hour of the day. Each line in the file has the wind speed and visibility at a location. Create a sample data file. Read this data from the file and determine whether blizzard conditions were met during this day or not.
Answer:
The weather conditions do not suggest a blizzard. It is explained below.
Explanation:
wind_speed=randi([12 56],24,1);
visibility=randi([1 10],24,1)/10;
storm_data=[wind_speed,visibility];
save stormtrack.dat storm_data -ascii
clear
load stormtrack.dat
fprintf('Below is the storm data from the file stormtrack.dat \n');
disp(stormtrack)
wind_speed=stormtrack(:,1);
visibility=stormtrack(:,2);
L=length(wind_speed);
count=0;
i=0;
while count<4 && i<L
i=i+1;
if wind_speed(i)>=30 && visibility(i)<=0.5
count=count+1;
else
count=0;
end
end
if count==4
fprintf('The weather conditions suggest a blizzard.\n')
else
fprintf('The weather conditions do not suggest a blizzard.\n')
end
Results:
Below is the storm data from the file stormtrack.dat
49.0000 0.1000
56.0000 0.4000
44.0000 0.6000
27.0000 0.7000
38.0000 0.5000
16.0000 0.9000
52.0000 0.8000
1 51.0000 1.0000
48.0000 0.6000
23.0000 0.4000
38.0000 0.2000
13.0000 0.7000
31.0000 0.8000
26.0000 0.5000
19.0000 0.1000
20.0000 0.3000
31.0000 0.2000
16.0000 0.3000
38.0000 0.5000
33.0000 0.6000
43.0000 0.5000
43.0000 0.9000
40.0000 0.6000
13.0000 1.0000
The weather conditions do not suggest a blizzard.
It is desired to produce a continuous and oriented carbon fiber-reinforced epoxy having a modulus of elasticity of at least 83 GPa in the direction of fiber alignment. The modulus of elasticity of the graphite fiber is 260 GPa, and the modulus of elasticity of the epoxy is 2.4 GPa. What is the minimum volume fraction of fibers required to meet this requirement. (this is a Chapter 16 problem) Answer: the minimum fiber volume fraction is:__________
Answer:
0.3129
Explanation:
Calculate The minimum fiber volume fraction using the relation below
E = x * E1 + ( 1 - x ) E2 ------ ( 1 )
given that :
E = 83 GPa , x = ?
E1 = 260 GPa , E2 = 2.4 GPA
Insert values back into equation 1
Then ; 80.6 = 257.6 x
∴ x ( volume fraction ) = 80.6 / 257.6 = 0.3129
Which federal agency issues the model food code?
The FDA! (food and drug administration)
The federal agency issues the model food code is FDA
How to determine the agencyThe U. S. Food and Drug Administration (FDA) publishes the Food Code, a model that assists food control jurisdictions at all levels of government by providing them with a scientifically sound technical and legal basis for regulating the retail and food service segment of the industry (restaurants and grocery stores.
The Food Code is a model for safeguarding public health and ensuring food is unadulterated and honestly presented when offered to the consumer. It represents FDA's best advice for a uniform system of provisions that address the safety and protection of food offered at retail and in food service.
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*Avoid waiting in long drive-thru lines, for example, at fast-food restaurants or banks. Park your car and go in. *When possible, use public transportation, walk, or ride a bike. *Get regular engine tune ups and car maintenance checks. *Join a carpool or vanpool to get to work or school. *Keep tires properly inflated and aligned. Conservation. The choices in the list above help to regulate important biogeochemical cycles as well as conserve important natural resources. Making these good choices impacts several cycles on Earth.
Required:
Which set of answers - biogeochemical cycle; natural resource; consequence - is not affected by the list above?
Answer:
When possible, use public transportation, walk, or ride a bike
Explanation:
Answer:
water; water; Oxygen enters the waterways and causes an imbalance of plant life to animal life.
Explanation:
3. In order to obtain your commercial driver's license (CDL) you must first:
A. obtain a commercial learner's permit (CLP).
B. have all restrictions removed from your personal driver's license.
C. be at least 15-years old.
D. both A and C.
Answer: is A
Explanation:
first thing first you cannot drive CMV with out a permit and To operate the CMV you need to be an adult
An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C, and V1 = 0.004 m3. The maximum cycle temperature is 1127°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature.
Answer:
i) Heat rejection = 1.0288 KJ
Network production = 1.212 kJ
ii) Thermal efficiency = 54.08%
mean effective pressure = 353.5 kPa
Explanation:
Given data :
Compression ratio ( r ) = 7
P1 ( initial pressure ) = 90 kPa,
T1 ( initial temperature ) = 27°C + 273 = 300 K
V1 = 0.004 m^3
Max cycle Temperature = 1127°C for each repetition of the cycle
i) Determine the heat rejection and net work production
considering that process 1-2 is an Isentropic compression
Find ; T2 = T1 ( r )^1.4-1
T2 = 300 ( 7 )^0.4 = 653.371 K
P2 = P1( r )^k
= 90 ( 7 )^1.4 = 1372.081 kPa
considering that process 3-4 is an Isentropic expansion
T4 = T3 / r^k-1
= 1400 / 7^(1.4 -1 ) = 642.82 K
Next ; Calculate the value of m
m = P1V1 / RT1 = 90(0.004) / 0.287 ( 300 )
m = 4.18 * 10^-3 kg
Finally :
amount of heat rejected = mCv ( T4 - T1 )
= 4.18 * 10^-3 ( 0.718 ) ( 642.82 - 300 )
Qout = 1.0288 KJ
amount of heat added ( Qin ) = mCv ( T3 - T2 )
= 4.18 * 10^-3 ( 0.718 ) ( 1400 - 653.371 )
hence Qin = 2.2408 kJ
Network production( Wnet) = Qin - Qout
= 2.2408 - 1.0288 ) KJ
= 1.212 kJ
ii) Determine the thermal efficiency and mean effective pressure of the cycle
Thermal efficiency = 1 - Qout / Qin
= 1 - ( 1.0288 / 2.2408 ) = 54.08%
mean effective pressure = Wnet / V1 ( 1 - 1/r )
= 1.212 / 0.004 ( 1 - 1/7 )
= 353.5 kPa
The Boeing 787 Dreamliner is billed to be 20% more fuel efficient than the comparable Boeing 767 and will fly at Mach 0.85. The midsize Boeing 767 has a range of 12,000 km, a fuel capacity of 90,000 L, and flies at Mach 0.80. For Boeing 787, assume the speed of sound is 700 mph and calculate the projected volumetric flow rate of fuel for each of the two Dreamliner engines in m3/s.
Answer:
the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
Explanation:
Given the data in the question;
First we determine the fuel economy of Boeing 767
Range = 12,000 km
fuel capacity = 90,000 L
so, fuel economy of Boeing 767 will be
η[tex]_f[/tex] = Range / fuel capacity
η[tex]_f[/tex] = ( 12,000 km / 90000 L ) ( 1000m / 1 km) ( 3.7854 L/gal × 264.2 gal/m² )
η[tex]_f[/tex] = 133,347.024 m/m³
Now, Boeing 787 is 20% more fuel efficient than Boeing 767
so fuel economy of Boeing 787 will be;
⇒ (1 - 20%) × fuel economy of Boeing 767
⇒ (1 - 0.2) × 133,347.024 m/m³
⇒ 0.8 × 133,347.024 m/m³
⇒ 106,677.6 m/m³
Hence, fuel economy of Boeing 787 dream line engine is
⇒ 106,677.6 m/m³ / 2 = 53,338.8 m/m³
Next, we find the velocity of Boeing 787
[tex]V_{787[/tex] = Mach number of 787 × speed of sound
given that; Mach number is 0.85 and speed of sound is 700 mph
we substitute
[tex]V_{787[/tex] = (0.85 × 700 mph) × ( 1 hr / 3600 s ) × ( 1609 m / 1 mile )
[tex]V_{787[/tex] = 265.9319 m/s
Now, to get the Volume flow rate for each dream liner engine { Boeing 787 };
Volumetric flow rate = velocity of flight / fuel economy
we substitute
= 265.9319 m/s / 53,338.8 m/m³
= 0.0049857 ≈ 0.005 m³/s
Therefore, the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
A structural element for a new bridge is designed for a constant load of 1000 psi. Its mean resistance is 1200 psi and the probability of failure is 1.9 X 10–3 . Find the element's safety factor as well as standard deviation, variance and coefficient of variation of its resistance. Clearly show units as appropriate.
Answer:
i) SF = 0.83
ii) 10 psi
iii) 3.16 psi
iv) 0.0083
Explanation:
Constant load = 1000 Psi
mean resistance = 1200 psi
Probability of failure = 1.9 * 10^-3
Determine
i) Safety factor
S.F = 1 / (mean load / load design ) = 1 / ( 1200 / 1000 ) = 0.83
ii) Standard deviation
1.9 X 10^-3 = e ( -1/2 ( μ / б) ^3 / (2.5 * 6 )
0.004756 = e^- 20000 / б^3
hence std = 10 psi
iii) Variance
= [tex]\sqrt{10}[/tex] = 3.16 psi
iv) coefficient of variation
Cv = std / mean resistance
= 10 / 1200 = 0.0083
The volume of the pyramid is 36 cubic cm, find the volume of the prism.
Answer:
Given :- the volume of the pyramid is 36 cubic cm , find the volume of the prism on same base and same height as pyramid .
Answer :-
we know that,
Volume of pyramid = (1/3) * Base area * height .
Volume of prism = Base area * height .
so,
→ Volume of pyramid = 36 cm³
→ (1/3) * Base area * height = 36
→ Base area * height = 36 * 3
→ Base area * height = 108 cm³.
then,
→ Volume of prism = Base area * height .
→ Volume of prism = 108 cm³ (Ans.)
Explanation:
Series-parallel circuits can be analyzed using the same electrical laws that apply to separate series or parallel circuits; you just have to apply the series laws to the series portion and the parallel laws to the parallel portion. True or False?
False
Series-parallel circuits have their own electrical laws purpose made for them
If A is the brother of B; B is the sister of c; and is
the father of D, how d is related to A?
Answer:
d
Explanation:
We know that daughter of one's brother or sister is Niece, If it is son Nephew. Here D's sex is not determined. A, B, and C are siblings. So, D is niece of A if D is female, D is nephew of A if D is male.
A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loaded tractor-trailer crosses over the bridge, however, the stress in the beam increases to 45 ksi. The beam is fabricated from steel with an ultimate tensile strength of 76 ksi, a yield strength of 50 ksi, and an endurance limit of 38 ksi. Find the safety factor for an infinite fatigue life:
a. if the effect of mean stress on fatigue strength is ignored
b. when the effect of mean stress on fatigue strength is considered.
Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
Given:
Stress due to dead weight = 14 KsiDue to loaded tractor = 45 KsiTensile strength of beam = 76 KsiYield strength = 50 KsiEndurance limit = 38 Ksi(a)
According to the question,
→ [tex]\beta = \frac{45-14}{2}[/tex]
[tex]= \frac{31}{2}[/tex]
[tex]= 15.5 \ Ksi[/tex]
hence,
The safety factor will be:
= [tex]\frac{Endurance \ limit}{\beta}[/tex]
= [tex]\frac{38}{15.5}[/tex]
= [tex]2.452[/tex]
(b)
→ [tex]\beta_2 = \frac{45+14}{2}[/tex]
[tex]= \frac{59}{2}[/tex]
[tex]= 29.5 \ Ksi[/tex]
Ration will be:
= [tex]\frac{\beta}{\beta_2}[/tex]
= [tex]\frac{15.5}{29.5}[/tex]
= [tex]0.5254[/tex]
By applying Goodman method,
→ [tex]Sa = \frac{0.5254\times 38\times 76}{0.5254\times 76+38}[/tex]
[tex]= 19.47 \ Ksi[/tex]
hence,
The safety factor will be:
= [tex]\frac{Sa}{\beta}[/tex]
= [tex]\frac{19.47}{15.5}[/tex]
= [tex]1.256[/tex]
Thus the above response is correct.
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A company sent prototypes of a newly designed hair dryer for trial use to 100 homes along with a satisfaction survey. A month later the company received 82 completed surveys. This process is part of which step of the engineering design process?
Answer:
The answer is "evaluating outcomes ".
Explanation:
The assessment of the results shows that maybe a program has achieved its goals. The assessment with results measures the results of the program and decides whether intended results were achieved. Its results allow you to achieve people by the measurable changes made through participating in your project by students enrolled. It is a form of assessment that examines if changes occur for students enrolled or if these adjustments are related to a program.
Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______
( ) Differentiating output
( ) Integrating output
( ) Impulse response is identical to unit step response
( ) Multiplying impulse response by t
( ) This is not possible. Unit Impulse response has nothing to do with unit step response
Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
When you see a school bus stop with its stop sign extended or its lights flashing, you must stop, except for cases when:
Answer:
Whenever you approach a school bus from any direction, which has stopped to pick up or let off passengers while operating its flashing red lights, you must stop your vehicle at least 25 feet from the school bus. The only time you do not have to stop is when you are on the other side of a divided highway. You must stay stopped until the bus has started again or the bus driver stops operating the flashing red lights.
6. Which of the following is considered a major disqualifying offense?
A. Following the vehicle ahead too closely
B. Leaving the scene of an accident
C. Texting while driving a CMV
D. Violating an out-of-service order
The option that is considered a major disqualifying offense is; B. Leaving the scene of an accident
What are major traffic offenses?The federal motor carrier safety administration has come up with different rules and regulations that can classify as a major disqualifying offense by drivers of motor vehicles such as;
Driving under the influence of alcohol or hard substanceLeaving the scene of an accidentUsing the vehicle to commit a felonyLooking at the given options, the only correct one is Option B
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2 Select the correct answer from each drop-down menu. Identify the jobs with the type of risks they include. Michael is a/n engineer with the risk of inhaling very small crystalline particles that can cause silicosis, pulmonary diseases, or lung cancer. Tommy is a/n engineer with the risk of exposure to poisonous pollutants in production units when a certain reaction does not follow h. textile Reset Next chemical automobile
Answer:
2nd option is correct
Explanation:
as Tommy is an engineer only exposed to poisonous pollutants while Michael is an engineer exposed to small crystalline particles that can cause lung cancer and other diseases
I Hope You Got Your Answer
What is the link between fossil fuel use and flowers blooming earlier in the spring season
Answer:
a warmer earth due to fossil fuel emissions has milder winters and earlier springs
Explanation:
fossil fuels cause global warming which would make the winters less cold and spring and summer hotter.
This determines whether two different String objects contain the same string.
a. the == operator
b. the = operator
c. the equals method
d. the stringCompare method
Answer: c. the equals method
Explanation:
There are more than one comparison that could be made about string objects. Hence, depending on the comparison we want to make, we could be in need of the == operator, equal() or the Stringcompare method. For the question posed above, we want to kno of two string objects containa the same string, hence we are concerned about comparing the content of each string. For this purpose, we make use of the equal(). In the case of making comparison about string reference, we use the == method while the StringCompare method comes in handy while making alphabetical or lexicographic comparison.