Answer:
newspaper, radio, televison
Explanation:
had avid in 7th :)
Answer:
1. will help for colcollares.
2. gives you money for studies
3. helps you choose the perfect collage
In warm climates, a vapor barrier is placed on the exterior side of the insulation, and in cold climates it is installed on the interior side of the
insulation. Which of the following explains this placement of the barrier?
The barrier should always be placed on the side opposite from where the water condenses.
The barrier should always be placed on the side opposite where rain or snow hit.
The barrier should always be placed on the side where rain or snow hit.
The barrier should always be placed on the side where the water condenses
Answer: its c
Explanation:
if the output of a signal is 36% on and 64% off and repeats itself is it considered periodic
helppppp
Answer:
No, it is not a periodic Signal
Explanation:
No, it is not a periodic Signal
This signal is repeating itself with the fixed on and off values but the major point to note here is that is this signal repeating after a fixed length of time every time. No, such information is provided in the question and hence, this signal cannot be termed as periodic.
Which two technologies were combined to create product life cycle management (PLM) software?
CAD and a database
spreadsheets and graphics
a database and spreadsheets
CAD and spreadsheets
Answer:
CAD and a database
Explanation:
The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.
Help meeeeeeeee plzzzzz need explanation
the picture is blank for me what does it say i can comment the answer plz mark brainlyist
In the production of soybean oil, dried and flaked soybeans are brought in contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil.
a. Draw flow diagram of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract).
b. The soybeans contain 18.5 wt% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to 2.0 kg hexane per kg beans. The residual solids leaving the extraction unit contain 35.0 wt% hexane, all of the non-oil solids that entered with beans, and 1.0% of the oil that entered the beans. For a feed rate of 1000 kg/h of dried flaked soybeans, calculate mass flow rates of extract and residual solids and the composition of extract.
Answer: its c
Explanation:
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Answer:
Which option identifies the step of the implementation phase represented in the following scenario?
A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.
establishing a process and budget
using communication tools
building and assembling a team
setting up a change order process
Explanation:
#carryonml
Answer:
using communication tools
Explanation:
The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.
5. The pin support at A allows _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction (d) none of the above 6. The support at B does not allow _______. Select the one that applies. (a) displacement in the x direction (b) rotation about its central axis (c) displacement in the y direction
Answer: Diagram associated with your question is attached below
5) B
6) C
Explanation:
5) The pin support at A allows ; Rotation about its central axis
This is because pin supports does not allow the translation of its structural member in any direction i.e. y or x but only rotation about its axis
6) The support at B does not allow displacement in y direction
This is because roller support allows displacement only in the direction that they are situated and in this case it is the x - direction
In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B. the ratio of a machine's power to the force of its input Ос. the rate at which a machine performs work D. the rate at which a machine consumes energy E. the ratio of the work output of a machine to the work input
Answer:
E
Explanation:
I have a big brain and I just took the test and got it correct.
For a bronze alloy, the stress at which plastic deformation begins is 284 MPa and the modulus of elasticity is 106 GPa. (a) What is the maximum load that can be applied to a specimen having a cross-sectional area of 310 mm2 without plastic deformation? (b) If the original specimen length is 120 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Answer:
a) the maximum load is 88,040 N
b)
the maximum length to which the specimen may be stretched is 0.12032148 mm
Explanation:
Given the data in the question;
the stress at which plastic deformation begins σ = 284 MPa = 2.84 × 10⁸ Pa
modulus of elasticity E = 106 GPa = 1.06 × 10¹¹ Pa
a)
Area A = 310 mm² = 310 × 10⁻⁶ m ( without plastic deformation )
now, lets consider the equation relating to stress and cross sectional area.
σ = F / A₀
hence, maximum load F = σA₀
so we substitute
F = (2.84 × 10⁸) × (310 × 10⁻⁶)
F = 88,040 N
Therefore, the maximum load is 88,040 N
b)
Initial length specimen l₀ = 120 mm = 120 × 10⁻³ m
using engineering strain, ε = (l₁ - l₀)/l₀
Also from Hooke's law, σ = Eε
so from the equation above;
l₁ = l₀( ε + 1 )
l₁ = l₀( σ/E + 1 )
so we substitute
l₁ = (120 × 10⁻³)( (2.84 × 10⁸)/(1.06 × 10¹¹)) + 1 )
l₁ = (120 × 10⁻³) ( 1.002679 )
l₁ = 0.12032148 mm
Therefore, the maximum length to which the specimen may be stretched is 0.12032148 mm
When an arbitrary substance undergoes an ideal throttling process through a valve at steady state, (SELECT ALL THAT APPLY) inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet pressures will be equal. inlet and outlet mass flowrates will be equal. inlet and outlet specific enthalpies will be equal. inlet and outlet temperatures will be equal.
Answer:
15x
Explanation:
Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife
Answer: i got you its d
Explanation:had the smae question as you
A tiger cub has a pattern of stripes on it for that is similar to that of his parents where are the instructions stored that provide information for a tigers for a pattern
probably in it's chromosomes
1)Saturated steam at 1.20bar (absolute)is condensed on the outside ofahorizontal steel pipe with an inside and outside diameter of 0.620 inches and 0.750 inches, respectively. Cooling water enters the tubes at 60.0°F and leaves at 75.0°F at a velocity of 6.00ft/s. (HINT: You may assume laminar condensate flow.You many also assume that the mean bulk temperature of the cooling water is equal to the wall temperature on the outside of the pipe, T".You may also neglect the viscosity correction in your calculations.)a)What are the inside
Answer:
hi = 7026.8 W/m^2.k
Explanation:
Given data :
pressure of saturated steam = 1.2 bar
Horizontal steel pipe : inside diameter = 0.620 , outside diameter = 0.750 inches
temperature of water at entry = 60°F
temperature of water at exit = 75°F
velocity of water = 6 ft/s
Calculate the Inside convective heat transfer coefficient ( hi )
mean temperature ( Tm ) = 60 + 75 / 2 = 67.5°F ≈ 292.877 K ≈ 19.727°C
next : find the properties of water at this temperature ( 19.727°C )
thermal conductivity = 0.598 w/m.k
density = 1000 kg/m^3
specific heat ( Cp ) = 4.18 KJ/kg.k
viscosity = 0.001 pa.s
velocity of water = 6 ft/s ≈ 1.8288 m/s
∴ Re ( Reynolds number ) = 28712.16
and Prandtl number ( Pr ) = (4180 * 0.001) / 0.598 = 6.989
finally to determine the inside convective heat transfer coefficient we will apply the Dittos - Bolter equation
hi = 7026.8 w/m^2.k
attached below is the remaining solution
A cylindrical metal specimen having an original diameter of 10.33 mm and gauge length of 52.8 mm is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.38 mm, and the fractured gauge length is 73.9 mm. Calculate the ductility in terms of (a) percent reduction in area (percent RA), and (b) percent elongation (percent EL).
Answer
a) 62 percent
b) 40 percent
Explanation:
Original diameter ( d[tex]_{i}[/tex] ) = 10.33 mm
Original Gauge length ( L[tex]_{i}[/tex] ) = 52.8 mm
diameter at point of fracture ( d[tex]_{f}[/tex] ) = 6.38 mm
New gauge length ( L[tex]_{f}[/tex] ) = 73.9 mm
Calculate ductility in terms of
a) percent reduction in area
percentage reduction = [ (A[tex]_{i}[/tex] - A[tex]_{f}[/tex] ) / A[tex]_{i}[/tex] ] * 100
A[tex]_{i}[/tex] ( initial area ) = π /4 di^2
= π /4 * ( 10.33 )^2 = 83.81 mm^2
A[tex]_{f}[/tex] ( final area ) = π /4 df^2
= π /4 ( 6.38)^2 = 31.97 mm^2
hence : %reduction = ( 83.81 - 31.97 ) / 83.81
= 0.62 = 62 percent
b ) percent elongation
percentage elongation = ( L[tex]_{f}[/tex] - L[tex]_{i}[/tex] ) / L[tex]_{i}[/tex]
= ( 73.9 - 52.8 ) / 52.8 = 0.40 = 40 percent
A noisy transmission channel has a per-digit error probability p = 0.01.
(a) Calculate the probability of more than one error in 10 received digits?
Answer:
The appropriate answer is "0.0043".
Explanation:
The given values is:
Error probability,
p = 0.01
Received digits,
n = 10
and,
[tex]x\sim Binomial[/tex]
As we know,
⇒ [tex]P(x)=\binom{n}{x}p^xq^{n-x}[/tex]
Now,
⇒ [tex]P(x >1) =1- \left \{ P(x=0)+P(x=1) \right \}[/tex]
⇒ [tex]=1-\left \{\binom{10}{0}(0.01)^0(0.99)^{10-0}+\binom{10}{0}(0.01)^1(0.99)^{10-1} \right \}[/tex]
⇒ [tex]=1-0.9957[/tex]
⇒ [tex]=0.0043[/tex]
A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Poisson's ratio.Calculate the stress (in MPa) necessary to cause a 0.0105 mm reduction in diameter.
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
[tex]E^z[/tex] = - ( εˣ / v )
[tex]E^z[/tex] = - ( -0.00088 / 0.34 )
[tex]E^z[/tex] = - ( - 0.002588 )
[tex]E^z[/tex] = 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
The evaporator:
A. directs airflow to the condenser.
B. absorbs heat from the passenger compartment.
C. removes moisture from the refrigerant.
D. restricts refrigerant flow.
Answer:
Option B
Explanation:
An evaporator along with cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.
Hence, option B is correct
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
immediate possession at any time when operating a motor vehicle.
True
False
Answer:
true
Explanation:
the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once
A 10 cm thick slab (density 8530 kg/m3 , specific heat 380J/kg K, conductivity 110 W/m K) initially at 650C is being cooled by air at 15C (convection coefficient 220 W/m2 K) at left surface. The right surface is insulated. We want to estimate temperature in the slab. a. What will be the temperature of the slab after a very long time
Answer:
The temperature after a long time will return to 15°C
Explanation:
Determine the temperature of the slab after a very long time
First we calculate the heat flow for m^2 area normal to the surface
= q / A = 650°c - 15°C / ( 1 / h + L / K )
= 635°c / ( 1 / 220 + 0.1 / 110 ) = 116.416 kw/m^2
Total heat content in the slab is calculated as
= m* c * ΔT
= 8530 * A * 0.1 * 380 * ( 650 - 15 )
= 205828.9 kJ/m^2
The temperature will return to 15°C after a long time
what is the answer to life the universe and everything
(worth 95 points!)
Answer:
In the absence of dark energy, a flat universe expands forever but at a continually decelerating rate, with expansion asymptotically approaching zero; with dark energy, the expansion rate of the universe initially slows down, due to the effects of gravity, but eventually increases, and the ultimate fate of the universe ...
Explanation:
I think it goes on forever.
Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.
Answer:
The issues related to the privacy are:
1. Informational privacy
2. Discrimination factors
3. Biased grouping on the basis of Data mining
4. Lack of consent
5. Morally wrong
6. Illegal distribution of information risks
7. Possibility of threat to life
Let's look at some major concerns:
1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.
2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.
3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:
Barries of effective
communication?
Answer: barries
Explanation:
A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v[tex]_f[/tex] = 0.001057 m³/kg
v[tex]_g[/tex] = 1.0037 m³/kg
u[tex]_f[/tex] = 486.82 kJ/kg
u[tex]_g[/tex] 2524.5 kJ/kg
h[tex]_g[/tex] = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v[tex]_g[/tex]
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m[tex]_{out[/tex] = m₁ - m₂
m[tex]_{out[/tex] = 1.89414 - 0.003985
m[tex]_{out[/tex] = 1.890155 kg
so, Initial internal energy will be;
U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]
U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex] + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E[tex]_{in[/tex] - E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]
QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁
QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -16oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.
Answer:
mass flow rate = 0.0534 kg/sec
velocity at exit = 29.34 m/sec
Explanation:
From the information given:
Inlet:
Temperature [tex]T_1 = -16^0\ C[/tex]
Quality [tex]x_1 = 0.2[/tex]
Outlet:
Temperature [tex]T_2 = -16^0 C[/tex]
Quality [tex]x_2 = 1[/tex]
The following data were obtained at saturation properties of R134a at the temperature of -16° C
[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]
[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]
[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]
[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]
If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.
This question is incomplete, The missing image is uploaded along this answer below;
Answer:
the maximum stress in the cylinder is 3.23 ksi
Explanation:
Given the data in the question and the diagram below;
First we determine the initial Kinetic Energy;
T = [tex]\frac{1}{2}[/tex]mv²
we substitute
⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²
T = 34.16149 lb.ft
T = ( 34.16149 × 12 ) lb.in
T = 409.93788 lb.in
Now, the volume will be;
V = [tex]\frac{\pi }{4}[/tex]d²L
from the diagram; d = 0.5 ft and L = 1.5 ft
so we substitute
V = [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )
V = 508.938 in³
So by conservation of energy;
Initial energy per unit volume = Strain energy per volume
⇒ T/V = σ²/2E
from the image; E = 6.48(10⁶) kip
so we substitute
⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]
508.938σ² = 5,312,794,924.8
σ² = 10,438,982.5967
σ = √10,438,982.5967
σ = 3230.9414
σ = 3.2309 ksi ≈ 3.23 ksi { three significant figures }
Therefore, the maximum stress in the cylinder is 3.23 ksi
Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?
Answer:
PERT Chart and GANTT Chart
As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart. With the PERT chart, the sequence of tasks is clearly mapped out. Dependent tasks are carried out when other tasks that they depend on have been executed.
Explanation:
By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars. On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams. It displays all the project tasks in separate boxes. The boxes are then connected with arrows which clearly show the task dependencies.
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]
[tex]\rho = 7900 \ kg/m^3[/tex]
[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]
[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]
Here, we can't apply the lumped capacitance method, since Bi > 0.1
[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]
[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]
[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]
However, on a single rod, the energy extracted is:
[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]
Hence, for centerline temperature at 50 °C;
The surface temperature is:
[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]
Which of these parts reduces the amount of pollutants in the engine exhaust?
A. Transmission
B. Catalytic converter
C. Tailpipe
D. Muffler
Answer:
Option B
Explanation:
Catalytic convertor with in the exhaust pipes of any vehicle converts harmful gases (such as hydrocarbons (HC), carbon monoxide (CO) and nitrogen oxides (NOx)) produced by combustion of fossil fuels (petrol, diesel etc.) into less harmful/toxic pollutants by acting as a catalyst for redox reaction. A catalytic convertor releases some mild gases from the exhaust like less harmful CO2, nitrogen and water vapour (steam)
Hence, option B is correct
MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False
An asphalt binder is mixed with aggregate and compacted into a sample. The mass of the dry sample is 1173.5 g, the mass of the sample submerged and then surface-dried with a damp towel is 1175.5 g, and the mass of the sample completely submerged in water is 652.5 g. Find the bulk specific gravity of the compacted sample.
Answer:
[tex]\mathbf{G_m = 2.25}[/tex]
Explanation:
From the given information:
Let the weight of the mix in the air be = [tex]W_{ma}[/tex]
Let the weight of the mix in water be = [tex]W_{mw}[/tex]; &
the bulk specific gravity be = [tex]G_m[/tex]
SO;
[tex]W_{mw} = W_{ma} - v \delta _{w} --- (1)[/tex]
Also;
[tex]G_m = \dfrac{W_{mw}}{v \delta_w} --- (2)[/tex]
From (2), make[tex]v \delta_w[/tex] the subject:
[tex]v \delta_w = \dfrac{W_{ma}}{G_m}[/tex]
Now, equation (1) can be rewritten as:
[tex]W_{mw} = W_{ma} - \dfrac{W_{ma}}{G_m}[/tex]
[tex]G_m = \dfrac{W_{ma}}{W_{ma} - W_{mw}}[/tex]
Replacing the values;
[tex]G_m = \dfrac{1173.5}{1173.5 -652.5}[/tex]
[tex]G_m = \dfrac{1173.5}{521}[/tex]
[tex]\mathbf{G_m = 2.25}[/tex]
