Answer:
a) The stresses due to bending moments is much more than the stresses from transverse shear.
c) The deformations due to bending moments is much more than the deformations from transverse shear.
Explanation:
Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.
7. The surface finish for the cylinder walls usually depends on the
O A. type of engine oil used.
O B. sharpness of the cylinder bore edges.
O C.type of piston rings used
O D. cylinder wall-to-piston clearance.
A rigid tank of volume of 0.06 m^3 initially contains a saturated mixture of liquid and vapor of H2O at a pressure of 15 bar and a quality of 0.2. The tank has a pressure-regulating venting valve that allows pressure to be constant. The tank is subsequently being heated until its content becomes a saturated vapor (of quality 1.0). During heating, the pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. You can neglecting the kinetic and potential energy effects.
Required:
a. Determine the total mass in the tank at the initial and final states, in kg.
b. Calculate the amount of heat (in kJ) transferred from the initial state to the final state.
Answer:
The total mass in the tank = 0.45524 kg
The amount of heat transferred = 3426.33 kJ
Explanation:
Given that:
The volume of the tank V = 0.06 m³
The pressure of the liquid and the vapor of H2O (p) = 15 bar
The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]
By applying the energy rate balance equation;
[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]
where;
[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]
Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]
If we integrate both sides; we have:
[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
We obtain the following data from the saturated water pressure tables, at p = 15 bar.
Since:
[tex]h_e =h_g[/tex]
Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]
[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]
[tex]v_g = 0.1318 \ m^3/kg[/tex]
Hence;
[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]
[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]
[tex]v_1 = 0.02728 \ m^3/kg[/tex]
Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar
[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]
Hence;
[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]
[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]
[tex]u_1 = 1193.428[/tex]
However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:
[tex]m_1 = \dfrac{V}{v_1}[/tex]
[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]
[tex]m_1 = 2.1994 \ kg[/tex]
From the question, given that the final quality; [tex]x_2 = 1[/tex]
[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]
[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]
[tex]v_2 = 0.1318 \ m^3/kg[/tex]
Also;
[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]
[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]
[tex]u_2 = 2594.5 \ kJ/kg[/tex]
Then the final mass can be calculated by using the formula:
[tex]m_2 = \dfrac{V}{v_2}[/tex]
[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]
[tex]m_2 = 0.45524 \ kg[/tex]
Thus; the total mass in the tank = 0.45524 kg
FInally; from the previous equation (1) above:
[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]
[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]
Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]
Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]
Q = 3426.33 kJ
Thus, the amount of heat transferred = 3426.33 kJ
A uniform edge load of 500 lb/in. and 350 lb/in. is applied to the polystyrene specimen. If it is originally square and has dimensions of a = 2 in., b = 2 in., and a thickness of t = 0.25 in., determine its new dimensions a, b, and t after the load is applied. Ep = 597(10)3 psi, vp = 0.25.
The image of the load applied to the polystyrene is missing, so i have attached it.
Answer:
a_new = 2.00302 in
b_new = 2.00552
Explanation:
From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.
Now, formula for stress is;
Stress(σ) = Force/Area
We are not given force and area but the load and plate thickness.
Thus, stress = load/thickness
We are given;
Load in x - direction = 500 lb/in.
Load in y - direction = 350 lb/in.
Thickness; t = 0.25 in
Thus;
σ_x = 500/0.25
σ_x = 2000 ksi
σ_y = 350/0.25
σ_y = 1400 ksi
From Hooke's law for 2 dimensions, strain is given by the formula;
ε_x = (1/E)(σ_x - vσ_y)
ε_y = (1/E)(σ_y - vσ_x)
We are given v_p = 0.25 and Ep = 597 × 10³ psi
Thus;
ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)
ε_x = 0.00276
ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)
ε_y = 0.00151
From elongation formula, we know that;
Startin is: ε = ΔL/L
Thus; ΔL = Lε
We are given a = 2 and b = 2
Thus;
ΔL_x = 2 × 0.00276
ΔL_x = 0.00552
ΔL_y = 2 × 0.00151
ΔL_y = 0.00302
New dimensions are;
a_new = 2 + 0.00302
a_new = 2.00302 in
b_new = 2 + 0.00552
b_new = 2.00552
Calculate the normal stress along the x-direction.
[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]
Calculate the normal stress along the y-direction.
[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]
Calculate the strain along the x-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]
[tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]
Calculate the strain along the y-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]
[tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]
Calculate the strain along the z-direction.
[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]
[tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]
Calculate the new dimensions.
[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]
[tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]
[tex]\to a' = a + \varepsilon_y a\\\\[/tex]
[tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]
[tex]\to c'= c + \varepsilon_x c\\\\[/tex]
[tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]
Find out more about the dimensions here:
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Which method of freezing preserves the quality and taste of food?
Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
Consider the following ways of handling deadlock:
(1) banker’s algorithm,
(2) detect deadlock and kill thread, releasing all resources,
(3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait,
(5) resource ordering, and
(6) detect deadlock and roll back thread’s actions.
b) Another criterion is efficiency; in other words, which requires the least processor overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient, assuming that deadlock is a very rare event. Comment on your ordering
Answer:
banker’s algorithm,
(2) detect deadlock and kill thread, releasing all resources,
(3) reserve all resources in advance,
(4) restart thread and release all resources if thread needs to wait,
(5) resource ordering, and
(6) detect deadlock and roll back thread’s actions.
b) Another criterion is efficiency; in other words, which requires the least processor overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient, assuming that deadlock is a very rare event. Comment on your ordering
Explanation:
Identify how the average friction and heat transfer coefficients are determined in flow over a flat plate.
A) They are determined by differentiating the local friction and heat transfer coefficients at the mid-length of the plate, and then multiplying them by the length of the plate.
B) They are determined by by integrating the local Reynolds number and Nusselt numbers over the entire plate.
C) They are determined by integrating the local friction and heat transfer coefficients over the entire plate, and then dividing them by the length of the plate.
D) They are determined by by differentiating the local Reynolds number and Nusselt numbers at the mid-length of the plate.
Answer:
C.
Explanation:
Let's have
Q = heat transfer surface
∆T = average temperature
F = area of the heat surface
Then the heat transfer coefficient = Q/∆T*F
In a flow over flat plate, the average friction and the heat transfer coefficient are determined by the integration of local friction and also great transfer coefficients over the plate entirely and then dividing by the plates length.
Therefore answer option C is the answer to this question.
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the distance can be measured using full tape measures, what is the maxim error per tape measure allowed?
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
Refrigerant-22 absorbs heat from a cooled space at 50°F as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at 10°F at a rate of 0.08 lbm/s with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine:
a. The rate of cooling provided, in Btu/h.
b. The rate of exergy destruction in the evaporator.
c. The second-law efficiency of the evaporator.
Take T0 = 77°F. The properties of R-22 at the inlet and exit of the evaporator are: h1 = 107.5 Btu/lbm, s1 = 0.2851 Btu/lbm·R, h2 = 172.1 Btu/ lbm, s^2 = 0.4225 Btu/lbm·R.
Answer:
a) the rate of cooling provided is 18604.8 Btu/h
b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) the second-law efficiency of the evaporator is 37.39%
Explanation:
Given that;
Temperature of sink TL = 50°F = 510 R
Temperature at evaporator inlet TI = 10°F = 470 R
mass flow rate m" = 0.08 lbm/s
quality of refrigerant at evaporator inlet x1 = 0.3
quality of refrigerant at evaporator exit x2 = 1.0
T₀ = 77°F = 537 R
h1 = 107.5 Btu/lbm
s1 = 0.2851 Btu/lbm·R,
h2 = 172.1 Btu/ lbm,
s2 = 0.4225 Btu/lbm·R.
a) rate of cooling provided, in Btu/h.
QL = m"( h2 - h1)
we substitute
QL = 0.08( 172.1 - 107.5
= 0.08 × 64.6
= 5.168 Btu/s
we convert to Btu/h
5.168 × 60 × 60
QL = 18604.8 Btu/h
Therefore the rate of cooling provided is 18604.8 Btu/h
b) The rate of exergy destruction in the evaporator
Entropy generation can be expressed as;
S_gen = m"(s2 - s1) - QL/TL
so we substitute
S_gen = 0.08( 0.4225 - 0.2851 ) - 5.168 / 510
= 0.010992 - 0.01013
S_gen = 0.00086 Btu/ibm.R
now the energy destroyed expressed as;
X_dest = T₀ × S_gen
so
X_dest = 537 × 0.00086
X_dest = 0.46 Btu/Ibm
Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) The second-law efficiency of the evaporator.
Energy expended is expressed as;
X_exp = m"(h1 - h2) - m"T₀(s1 - s2)
we substitute
= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )
= -5.168 - [ - 5.9027)
= -5.168 + 5.9027
= 0.7347 Btu/s
Now second law efficiency is expressed as;
nH = 1 - (X_dest / X_esp)
= 1 - ( 0.46 / 0.7347 )
= 1 - 0.6261
= 0.3739
nH = 37.39%
Therefore the second-law efficiency of the evaporator is 37.39%
Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the dam (in ft). Neglect friction. [h = 155 ft]
Answer:
155fts
Explanation:
We apply the bernoulli's equation to get the depth of water.
We have the following information
P1 = pressure at top water surface = 0
V1 = velocity at too water surface = 0
X1 = height of water surface = h
Hf = friction loss = 0
P2 = pressure at exit = 0
V2 = velocity at exit if penstock = 100ft/s
X2 = height of penstock = 0
g = acceleration due to gravity = 32.2ft/s²
Applying these values to the equation
0 + 0 + h = 0 + v2²/2g +0 + 0
= h = 100²/2x32.2
= 10000/64.4
= 155.28ft
= 155
please help i have no xlue
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.
help me asap I rely need help u will be my fav
Answer:
The Answer is C!
Explanation:
Which of the following is not one of the common classifications of product liability defects? A. Manufacture B. Materials C. Packaging D. Both "Materials" and "Packaging" E. Design
Answer:
D. Both "Materials" and "Packaging"
Explanation:
Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.
There are majorly three product defects. They are :
1. Manufacturing defect
2. Design defect
3. Marketing defect
Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate
Answer:
The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.
Explanation:
If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:
60 = 4,000
95 = X
((95 x 4,000) / 60)) = X
(380,000 / 60) = X
6,333.3 = X
Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.
The nuclear reactions resulting from thermal neutron absorption in boron and cadmium are 10B5 + 1 n0 ï 7Li3 + 4He2 113Cd48 + 1 n0 ï 114Cd48 + γ[5 MeV] The microscopic thermal absorption cross sections for B-10 and Cd-113 are 3841 b and 20,600 b respectively. Which of these two materials would be the more effective radiation shield? Explain
Solution :
The nuclear reaction for boron is given as :
[tex]$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$[/tex]
And the reaction for Cadmium is :
[tex]$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$[/tex]
We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.
A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 1200 C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N. What is the rate of heat transfer from both sides of the plate to the air?
This question is incomplete, the complete question is;
A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120°C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N.
What is the rate of heat transfer from both sides of the plate to the air?
Answer:
the rate of heat transfer from both sides of the plate to the air is 236.54 W
Explanation:
Given the data in the question,
first we calculate the Reynold's number for the flow
Re = pu∞d / Ц
Re = (1.12 × 40 × 0.2) / 1.983 × 10⁻⁵
Re = 451840
Now the Local skin friction coefficient is given as;
Cfx = T / ( 1/2pu∞²)
Cfx = (Fd/A) / ( 1/2pu∞²)
Cfx = (0.075/(2×0.2×0.2)) / ( 1/2 × 1.12 × 40²)
= 0.9375 / 896
= 0.0010463
Cfx = 1.0463 × 10⁻³
Apply Reynold's- cOLBURN analogy
Cfx/2 = StₓPr^2/3
so
1.0463 × 10⁻³ / 2 = (h/pu∞Cp) × ( 0.711)^2/3
5.2315 × 10⁻⁴ × 1.12 × 40 × 1.005 × 1000 = h(0.711)^2/3
h = 23.554 / 0.7966
h = 29.56 W/m².K
so
The heat transfer rate from both the sides of the plate will be;
Q = 2 × 29.56 × 0.2 × 0.2 × ( 120 - 20 )
Q = 236.54 W
Therefore the rate of heat transfer from both sides of the plate to the air is 236.54 W
A cylindrical specimen of Aluminium having a diameter of 12.8 mm and gauge length of 50.8 is pulled in tension. Use the data given below to:A) Plot the data as engineering stress versus engineering strain. B) Compute the modulus of elasticity. C) Determine the yield strength at a strain offset of 0.002. D) Determine the tensile strength of this alloy.E) What is the approximate ductility, in percent elongation?Load (N) Length0 50.8007330 50.85115100 50.90223100 50.95230400 51.00334400 51.05438400 51.30841300 51.81644800 52.83246200 53.84847300 54.86447500 55.88046100 56.89644800 57.65842600 58.42036400 59.182
Answer:
Hello the needed data given is not properly arranged attached below is the properly arranged data
Answer:
b) 62.5 * 10^3 MPa
c) ≈ 285 MPa
d) 370Mpa
e) 16%
Explanation:
Given Data:
cylindrical aluminum diameter = 12.8 mm
Gauge length = 50.8 mm
A) plot of engineering stress vs engineering strain
attached below
B ) calculate Modulus of elasticity
Modulus of elasticity = Δб / Δ ε
= ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa
C) Determine the yield strength
at strain offset = 0.002
hence yield strength ≈ 285 MPa
D) Determine tensile strength of the alloy
The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress vs strain ( complete plot )
E) Determine approximate ductility in percent elongation
ductility in percent elongation = plastic strain at fracture * 100
total strain = 0.165 , plastic strain = 0.16
therefore Ductility in percent elongation = 0.16 * 100 = 16%
In a p+-n Si junction, the n side has a donor concentration of 1016 cm^-3. If ni = 1010 cm^-3, relative dielectric constant Pr = 12, calculate the depletion width at a reverse bias of 100 V? What is the electric field at the mid-point of the depletion region on the n side?
Answer:
This graph shows linear
y = f(x) and y = g(x).
Find the solution to the equation f(x) - g(x) = 0
An inductor has an inductance of 0.025 H and a wire resistance of . How long will it take the current to reach its full Ohm’s law value?
Answer:
Time constant t = 0.0083 (Approx)
Explanation:
Given:
L = 0.025
Missing resistance r = 3Ω
Find:
Time constant t
Computation:
Time constant t = L / r
Time constant t = 0.025 / 3
Time constant t = 0.0083 (Approx)
Using the inductance - resistance relation to calculate the time constant , the time constant would be 0.08333
Given the Parameters :
Inductance in Henry = 0.025 H Resistance of wire in ohms, R = 3 ohmsThe time taken for current to reach the wire can be calculated thus :
Time constant = (Inductance, L) ÷ resistance, RTime constant = 0.025 ÷ 3 = 0.083333
Therefore, the time constant for current to reach is 0.08333
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While at a concert you notice five people in the crowd headed in the same direction. Your tendency to group them is due to? *
Answer:
common fate
Explanation:
The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.
Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.
Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.
I need help with simply science
Answer:
mountain ranges may be
A mixture of octane, C8H18, and air flowing into a combustor has 60% excess air and 1 kmol/s of octane. What is the mole flow rate (kmol/s) of CO2 in the product stream?
Answer:
8 kmol/s
Explanation:
From the given information:
The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:
[tex]C_{8}H_{18} +12.5(O_2 + \dfrac{79}{21} N_2) \to 9H_2O +8CO_2 + 12.5(\dfrac{79}{21}N_2)[/tex]
[tex]C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)[/tex]
In the combustor, it is said that 60% of excess air and 1 mole of Octane is present.
Thus;
the air supplied = 1.6 × 12.5 = 20
The equation can now be re-written as:
[tex]C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2[/tex] because for 1 mole of Octane, 8 moles of CO2 can be found in the combustion product.
Thus, for 1 kmol/s of Octane also produce 8 kmol/s of CO2.
∴
The mole flow rate in Kmol/s of CO2 in the product stream = 8 kmol/s
An incremental encoder is rotating at 15 rpm. On the wheel there are 40 holes. How many degrees of rotation would 1 pulse be?
Answer:
1 pulse rotate = 9 degree
Explanation:
given data
incremental encoder rotating = 15 rpm
wheel holes = 40
solution
we get here first 1 revolution time
as 15 revolution take = 60 second
so 1 revolution take = [tex]\frac{60}{15}[/tex]
1 revolution take = 4 seconds
and
40 pulse are there for 1 revolution
40 pulse for 360 degree
so 1 pulse rotate is = [tex]\frac{360}{40}[/tex]
1 pulse rotate = 9 degree
For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is:_______
a. The same as that of the larger-diameter cylinder
b. Larger than that of the larger-diameter cylinder
c. Smaller than that of the larger-diameter cylinder
Answer:
B. Larger than that of the larger-diameter cylinder
Explanation:
By looking at the equation for the Nusselt number for a cylinder in cross flow, Nu = hd/k, and assuming both cylinders are made of the same material, you can see that the heat transfer coefficient h will have to be much larger for the smaller cylinder. You can verify this by using random numbers for each variable (shown below).
Verification:
ds = 1
db = 10
k = 12
Nu = 10
h small = ?
h big = ?
Nu = hd/k
10 = h small * ds / 12
120 = 1 * h small
h small = 120
Nu = hd/k
10 = h big*db / 12
120 = 10 * h big
h big = 12
This shows that the heat transfer coefficient for the smaller diameter, ds, must be bigger than the heat transfer coefficient for the larger diameter, db.
For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is " Smaller than that of the larger-diameter cylinder." (Option C)
What does the information confirm about this?The given information states that for two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same.
The Nusselt number is a dimensionless number that relates the convective heat transfer rate to the conductive heat transfer rate.
Since the Nusselt number is the same for both cylinders, and the heat transfer coefficient is directly related to the Nusselt number, the smaller-diameter cylinder will have a smaller average heat transfer coefficient compared to the larger-diameter cylinder.
Thus option C is the right answer.
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An equal-tangent sag vertical curve is designed with a PVC as station 109+00 and elevation 950ft, the PVI has a station of 110+77 and elevation of 947.34ft, and the low point at station 110+50. Determine the design speed of the cure.
Answer:
K = 96 and the design speed of the curve = 50mph
Explanation:
109+00 = 10900
Elevation = 950ft
110+77 = 11077
Elevation = 947.34ft
Station of low point = 110+50 = 11050
To get grade of curve
Gi = 947.34-950/11077-10900
= -2.66/177
= -0.015x100
= -1.5%
Locate of low point (XL)
= 11050-10900
= 150
To get the value of K
XL = |GL| x K
When we substitute values
150 = 1.5 x K
150 = 1.5K
K = 150/1.5
K = 100
The suitable and most nearest value is K = 96
Then we use the standard chart to get the design speed for K = 96
On this chart, the design speed for the curve = 50mph
Therefore K = 96 and speed = 50mph
A spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]
In the case of critical damping we have the relation
[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]
Mass that would produce critical damping is 0.75 kg.
0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.
What is zero velocity?A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.
For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.
Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.
Thus, it is 0.75 kg.
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We put capacitors on our voltage supplies in order to filter out high frequency noise. Which is better. a 10uF capacitor or a 0.1uF capacitor? Why?
Answer:
10uF
Explanation:
A higher value of capacitance is the best option when we are trying to filter power supply outputs in other to reduce hum.
The greater the capacitance or the voltage of a circuit is, the more energy it can the particular circuit can store. When capacitors are being connected in series, the total value of the capacitance reduces but contrarily, the voltage of the same system increases anyway. Connecting circuits in parallel helps to keep the voltage rating the same but on the other hand, it increases the total capacitance.
A 10 μF capacitor is better.
This is because, to filter out high frequency noise, our capacitor is connected in parallel with the voltage supply. This parallel connection causes the capacitance of the circuit to increase but the voltage stays constant.
Since there is an increase in capacitance, this causes the circuit to filter out high frequency noise.
So, a high value capacitance connected in parallel with the voltage source is a better filter for high frequency noise.
So, the 10 μF capacitor is better.
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The big ben clock tower in london has clocks on all four sides. If each clock has a minute hand that is 11.5 feed in length, how far does the tip of each hand travel in 52 minutes?
Answer:
Updated question
The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?
The distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Explanation:
Let us assume the shape of the clock is circular.
the minute hand is equal to the radius = 11.5 ft
Diameter = radius x 2
Diameter = 11.5 x 2 = 23 ft
The distance traveled by the tip of the minute hand can be calculated thus;
the fraction of the circumference traveled by the minute hand would be;
52/60 = 0.8667
Circumference of the clock would be;
C = pi x d
where C is the circumference
pi is a constant
d is the diameter
C = 3.14 x 23
C = 72.22 ft
Therefore the fraction of the circumference covered by the minute hand would be;
72.22 ft x 0.8667 = 62.59 ft
Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft
is a process that is used to systematically solve problems.
design
engineering
brainstorming
O teamwork
Answer:
design
Explanation:
Design is a process used to solve problems systematically.
Human beings have specific needs and desires, which require a design process to interpret those needs and make them real from a product or service.
Design uses specific methods and techniques integrating ideals, creativity, technology and innovation to satisfy users' needs and solve problems.
what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.
Answer:
robotic technology
Explanation:
Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.
Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.
One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.
Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.
Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.