Answer:
2.9237 ft/s
Explanation:
Given the data in the question;
A₁ = 9-inch × 9-inch = 81 in² = 81 / 144 = 0.5625 ft²
V₁ = 6.1 ft/s
A₂ = 13 in × 13 in = 169 in² = 1.17361 ft²
v₂ = ?
using the the equation if continuity
( Rate of volumetric flow is constant )
A₁V₁ = A₂V₂
we substitute
0.5625 ft² × 6.1 ft/s = 1.17361 ft² × V₂
3.43125 ft³/s = 1.17361 ft² × V₂
V₂ = 3.43125 ft³/s / 1.17361 ft²
V₂ = 2.9237 ft/s
Therefore, the speed of the air flowing into the room is 2.9237 ft/s
Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion products outside. A woman in a household lights a fire in her cookstove at 4 pm. The fire emits benzo(a)pyrene, a carcinogenic compound, at a rate of 0.01 ng/min. The house has a size of 40 m3 and a ventilation rate of 1 m3/h. The ambient indoor air concentration of benzo(a)pyrene was 0.2 ng/m3 when she started the fire, but no significant concentration of benzo(a)pyrene is found in the outdoor air. What is the indoor benzo(a)pyrene concentration in the household at 6 pm?
Answer:
C = 0.22857 ng / m³
Explanation:
Let's solve this problem for part the total time in the kitchen is
t = 2h (60 min / 1h) = 120 min
The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant
amount generated
C = co + time_generation rate / (area_house + area_flow)
C = 0.2 + 0.01 120 / (40+ 2)
C = 0.22857 ng / m³
A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The normal ground water level was at the ground surface. Observation wells were located at distances of 10 and 25 ft. from the pumping well. At a discharge of 761 ft3 per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft. Compute the hydraulic conductivity in ft. /sec.
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per million Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.
Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
c. If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Answer:
a) 570 kWh of electricity will be saved
b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) $1.296 can be earned by selling the SO₂ saved by a single CFL
Explanation:
Given the data in the question;
a) How many kilowatt-hours of electricity would be saved?
first, we determine the total power consumption by the incandescent lamp
[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh
next, we also find the total power consumption by the fluorescent lamp
[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh
So the value of power saved will be;
[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex] - [tex]P_{fluorescent}[/tex]
[tex]P_{saved}[/tex] = 750 - 180
[tex]P_{saved}[/tex] = 570 kWh
Therefore, 570 kWh of electricity will be saved.
now lets find the heat of electricity saved in Bituminous
heat saved = energy saved per CLF / efficiency of plant
given that; the utility has 36% efficiency
we substitute
heat saved = 570 kWh/CLF / 36%
we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)
so
heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (
heat saved = 5.4 × 10⁶ Btu/CLF
i.e eat of electricity saved per CLF is 5.4 × 10⁶
b) How many 2,000-lb tons of SO₂ would not be emitted
2000 lb/tons = 5.4 × 10⁶ Btu/CLF
0.6 lb SO₂ / million Btu = x
so
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]
x = 3.24 × 10⁶ / 2 × 10⁹
x = 0.00162 ton/CLF
Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Amount = ( SO₂ saved per CLF ) × ( rate per CFL )
we substitute
Amount = 0.00162 ton/CLF × $800
= $1.296
Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.
(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring cup and flows into the downsprue. The cross section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 20 cm long, determine the proper diameter at its base so as to main- tain the same volume flow rate.
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top [tex]d_{top}[/tex] = 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
[tex]V_{base}[/tex] = √2gh
we substitute
[tex]V_{base}[/tex] = √(2 × 9.81 m/s² × 0.2 m )
[tex]V_{base}[/tex] = 1.98091 m/s
[tex]V_{base}[/tex] = 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π[tex]d_{bottom}^2[/tex]/4) × [tex]V_{base}[/tex]
[tex]d_{bottom}[/tex] = √(4Q/π[tex]V_{base}[/tex])
we substitute our values into the equation;
[tex]d_{bottom}[/tex] = √(4(400 cm³/s) / (π×198.091 cm/s))
[tex]d_{bottom}[/tex] = 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm
A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half its volume, determine the resulting pressure.
Answer:
The resulting pressure is 300 kilopascals.
Explanation:
Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:
[tex]P_{1}\cdot V_{1} = P_{2} \cdot V_{2}[/tex] (1)
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressure, measured in kilopascals.
[tex]V_{1}[/tex], [tex]V_{2}[/tex] - Initial and final volume, measured in litres.
If we know that [tex]\frac{V_{1}}{V_{2}} = 2[/tex] and [tex]P_{1} = 150\,kPa[/tex], then the resulting pressure is:
[tex]P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}[/tex]
[tex]P_{2} = 300\,kPa[/tex]
The resulting pressure is 300 kilopascals.
The output of a first order transducer is to be connected to a signal conditioner which also has first order dynamic characteristics. The transducer has known time constant of x milliseconds and static sensitivity of xx V/kPa while the signal conditioner has a time constant of y milliseconds and a static sensitivity of yy V/V. Determine the steady state response of this measurement system to an input signal of the form: P(t)=(A+B*sin(w*t) ) kPa.
Answer:
hello your question is poorly written attached below is the correct question
answer : y(t) = 1.5 + 0.025 sin( 600t - 0.259 ) v
Explanation:
Given data:
Time constant = 5msec
static sensitivity = 0.05 v/°c
f = 100 hz
ε = 0.8
T(t) = ( 30 + 2.5 sin600t )°C
attached below is a detailed solution to the question above
Initially, 1 lbm of water is at rest at 14.7 psia and 70 F. The water then undergoes a process where the final state is 30 psia and 700 F with a velocity of 100 ft/s and an elevation of 100 ft above the starting location. Determine the increases in internal energy, potential energy, and kinetic energy of the water in Btu/lbm. Compare the increases in potential energy and kinetic energy individually to the change in the internal energy.
Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]\text{The mass (m) = 1 lbm}[/tex]
Suppose: g = 32.2 ft/s²
At the inlet conditions:
[tex]\text{mass (m) = 1 \ lbm water} \\ \\ P_1 = 14.7 \ psia \\ \\ T_1 = 70 F \\ \\ z_1 = 0 \ ft[/tex]
At the outlet conditions:
[tex]P_2 = 30 \ psia \\ \\ T_2 = 700\ F \\ \\ v_2 =100 \ ft/s\\ \\ z_2 = 100 \ ft[/tex]
[tex]\text{Using the information obtained from saturated water table at P1 = 14.7 \ psia \ and \ T1 = 70 F }[/tex]
[tex]u_1 =u_f = 38.09 \ Btu/lbm[/tex]
[tex]\text{Applying informations from superheated water vapor table:}[/tex]
[tex]P_2 = 30 \ psia \ and \ T_2 = 700 \ F \\ \\ u_ = 1256.9 \ kJ/kg[/tex]
The change in the internal energy is:
[tex]\Delta U = U_2 -U_1 \\ \\ \Delta U = 1256.9 -38.09 \\ \\ \Delta U = 1218.81 \ Btu/lbm[/tex]
For potential energy (P.E):
Initial P.E = mgz
P.E = 1 × 32.2 × 0 = 0 ft²/s²
Final P.E = mgz
P.E = 1 × 32.2 × 100 = 3220 ft²/s²
The change in the potential energy = PE₂ - PE₁
ΔPE = (3220 - 0) ft²/s²
ΔPE = 3220 ft²/s²
ΔPE = (3220 × 3.9941 × 10⁻⁵) Btu/lbm
ΔPE =0.12861 Btu/lbm
Initial Kinetic energy (K.E)
[tex]KE_1 = \dfrac{1}{2}mV_1[/tex]
[tex]KE_1 = \dfrac{1}{2}(1)(0) = 0 \ lbm \ ft^2/s^2[/tex]
FInal K.E
[tex]KE_2= \dfrac{1}{2}mV_2[/tex]
[tex]KE_2= \dfrac{1}{2}(1)(100)^2_2 = 50000 \ lbm \ ft^2/s^2[/tex]
Change in K.E [tex]\Delta K.E[/tex] = [tex]KE_2-KE_1[/tex]
[tex]\Delta K.E = 50000 -0 = 50000 \ lbm.ft^2/s^2[/tex]
[tex]\mathbf{\Delta K.E = 0.199 \ Btu/lbm}[/tex]
Choose two other elements from the periodic table that you predict should react to form something like table salt
Please pleassssss helppp
I give branlistttttt
Resistors are used to reduce current flow, adjust signal levels to divide voltages, bias active elements and terminate transmission line.true or false
Answer:
True
Explanation:
Those are the exact uses of a resistor
Shorter lines are faster than longer lines is an example of an algorithm.
Answer:
Programmers count the number of lines of code in an algorithm.
Programmers count the number of lines of code in an algorithm. Thus, shorter lines are faster than longer lines is an example of an algorithm.
What do you mean by algorithm?An algorithm is a finite sequence of exact instructions that is used in mathematics and computer science to solve a class of particular problems or carry out a computation.
For performing calculations and processing data, algorithms are employed as specifications. Conditionals can be used by more sophisticated algorithms to divert code execution along several paths and draw reliable inferences, ultimately leading to automation.
Alan Turing was the first to use terminology like "memory," "search," and "stimulus" to describe human traits as metaphorical descriptions of machines.
A heuristic, on the other hand, is a method for addressing problems that may not be fully articulated or may not provide accurate or ideal solutions, particularly in problem domains where there isn't a clearly defined proper or ideal conclusion.
Learn more about algorithm, here
https://brainly.com/question/22984934
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the uniform beam has a mass of 50 kg determine the reaction at the support
Answer:
1.4KN 4KN - m C 3KN 0.6m A 0 B 0.6m R 0.6m 0.6m 30° R Resolving Forces Horizontally, we get X=0 Rx
Find the minimum stages at total reflux for separating ethanol-acetone mixture to achieve 95 mol% purity of acetone in the distillate and 95 mol% purity of ethanol in the bottoms product. Use equilibrium data from problem 4D7. x y Equilibrium data for acetone (A) and ethanol at 1 atm (Perry et al., 1963, pp. 13-4) are
0.100.150.200.250.300.350.40 0.50 0.60 0.70 0.80 0.90 0.262 0.348 0.417 0.478 0.524 0.566 0.605 0.674 0.739 0.80 20.865 0.929
Answer:
minimum number of stages ≈ 9 ( from Image number 1 )
Explanation:
Attached below is a detailed solution to the question
Total reflux = R Tending to infinity
we can determine the operation equation by stripping and rectifying section in total reflux condition
minimum number of stages ≈ 9 ( from Image number 1 )
In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.8) is 1.2 and there may be a 2 mm crack present in a bone specimen, what would be the maximum tensile stress that can be applied before fracture occurs?
Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon alloy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.
Answer:
the carburizing time necessary to achieve a carbon concentration is 31.657 hours
Explanation:
Given the data in the question;
To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:
(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)
where Cs is Concentration of carbon at surface = 0.90
Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )
C0 is Initial concentration of carbon = 0.10
ERF() = Error function at the given value
D = Diffusion of Carbon into steel
t = Time necessary to achieve given carbon concentration ,
so
(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)
= 0.6 / 0.8
= 0.75
now, ERF(z) = 0.75; using ERF table, we can say;
Z ~ 0.81; which means ( x / 2√Dt) = 0.81
Now, Using the table of diffusion data
D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K
now we calculate the carbonizing time by using the following equation;
z = (x/2√Dt)
t is carbonizing time
so we we substitute in our values
0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)
0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t
0.81 × 1.4628 × 10⁻⁵ × √t = 0.004
1.184868 × 10⁻⁵ × √t = 0.004
√t = 0.004 / 1.184868 × 10⁻⁵
√t = 337.5903
t = ( 337.5903)²
t = 113967.21 seconds
we convert to hours
t = 113967.21 / 3600
t = 31.657 hours
Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours
A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.
Answer:
the entrained fluid flowrate is 150 liters/s
Explanation:
Given the data in the question;
we determine the flow rate of water though the jet by using the following expression;
Q₂ = A₂ × V₂
where Q₂ is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )
so we substitute
Q₂ = 0.01 m² × 30 m/s
Q₂ = 0.3 m³/s
Next we determine the flow rate of water through the pump by using the following expression
Q₃ = A₃ × V₃
where Q₃ is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )
so we substitute
Q₃ = 0.075 m² × 6 m/s
Q₃ = 0.45 m³/s
so to calculate the flow pumping rate of water into the water jet pump, we use the expression;
Q₁ + Q₂ = Q₃
we substitute
Q₁ + 0.3 m³/s = 0.45 m³/s
Q₁ = 0.45 m³/s - 0.3 m³/s
Q₁ = 0.15 m³/s
we know that 1 m³/s = 1000 Liter/second
so
Q₁ = 0.15 × 1000 Liter/seconds
Q₁ = 150 liters/s
Therefore, the entrained fluid flowrate is 150 liters/s
Write a python program to get the following output. 1-----99 2-----98 3-----97 . . . . . . 98-----2 99-----1
Answer:
i dont know th answer can u help ?
Explanation:
We intend to measure the open-loop gain (A open ) of an actual operational amplifier. The magnitude of A_open is in the range of 10^6 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the A_open?
Answer:
circuit diagram is attached below
Explanation:
Magnitude of A_ open = 10^6 v/v
Vp = Vin / 1001
where: Vin = 1 mV.
hence Vp = 1 mV / 1001 ≅ 1 μ V
while Vout = Aopen( Vp )
∴ Aopen = Vout / Vp
since Vp = 1 μV then we can can measure Aopen with Vout ranging up to
10 V
attached below is the measuring circuit for measuring open loop gain of amplifier
A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a final diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20% is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW
Answer:
procedure attached below
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
Explanation:
Given data:
Minimum tensile strength = 865 MPa
Ductility = 10%EL
Desired Final diameter = 6.0 mm
20% cold worked 7.94 mm diameter 1040 steel stock
Describe the procedure you would follow to obtain this material.
assuming 1040 steel experiences cracking at 40%CW
attached below is a detailed procedure of obtaining the material
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
The diameter of a cylindrical water tank is Do and its height is H. The tank is filled with water, which is open to the atmosphere. An orifice of diameter D with a smooth entrance (i.e., negligible losses) is open at the bottom. Develop a relation for the time required for the tank (a) to empty halfway (5-point) and (b) to empty completely (5-point).
Answer:
a. The time required for the tank to empty halfway is presented as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
b. The time it takes for the tank to empty the remaining half is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time 't', is presented as follows;
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
Explanation:
a. The diameter of the tank = D₀
The height of the tank = H
The diameter of the orifice at the bottom = D
The equation for the flow through an orifice is given as follows;
v = √(2·g·h)
Therefore, we have;
[tex]\dfrac{P_1}{\gamma} + z_1 + \dfrac{v_1}{2 \cdot g} = \dfrac{P_2}{\gamma} + z_2 + \dfrac{v_2}{2 \cdot g}[/tex]
[tex]\left( \dfrac{P_1}{\gamma} -\dfrac{P_2}{\gamma} \right) + (z_1 - z_2) + \dfrac{v_1}{2 \cdot g} = \dfrac{v_2}{2 \cdot g}[/tex]
Where;
P₁ = P₂ = The atmospheric pressure
z₁ - z₂ = dh (The height of eater in the tank)
A₁·v₁ = A₂·v₂
v₂ = (A₁/A₂)·v₁
A₁ = π·D₀²/4
A₂ = π·D²/4
A₁/A₂ = D₀²/(D²) = v₂/v₁
v₂ = (D₀²/(D²))·v₁ = √(2·g·h)
The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;
dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh
We have;
[tex]dt = \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } dh[/tex]
The time for the tank to drop halfway is given as follows;
[tex]\int\limits^{t_1}_0 {} \, dt = \int\limits^h_{\frac{h}{2} } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_1 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{\frac{H}{2} }^{H} =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{\frac{H}{2} }^{H} = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{2 \cdot D_0^2 }{D^2\cdot \sqrt{2\cdot g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right)[/tex]
[tex]t_1 = { \dfrac{\sqrt{2} \cdot D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{H} - \sqrt{\dfrac{H}{2} } \right) = { \dfrac{D_0^2 }{D^2\cdot \sqrt{ g} } \cdot \left(\sqrt{2 \cdot H} - \sqrt{{H} } \right) =\dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]The time required for the tank to empty halfway, t₁, is given as follows;
[tex]t_1 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right)[/tex]
(b) The time it takes for the tank to empty completely, t₂, is given as follows;
[tex]\int\limits^{t_2}_0 {} \, dt = \int\limits^{\frac{h}{2} }_{0 } { \dfrac{D_0^2}{D} \cdot\dfrac{1}{\sqrt{2\cdot g \cdot h} } } \, dh[/tex]
[tex]t_2 =\left[{ \dfrac{D_0^2}{D\cdot \sqrt{2\cdot g} } \cdot\dfrac{h^{-\frac{1}{2} +1}}{-\frac{1}{2} +1 } \right]_{0}^{\frac{H}{2} } =\left[ { \dfrac{D_0^2 \cdot 2\cdot \sqrt{h} }{D\cdot \sqrt{2\cdot g} } \right]_{0 }^{\frac{H}{2} } = { \dfrac{2 \cdot D_0^2 }{D\cdot \sqrt{2\cdot g} } \cdot \left( \sqrt{\dfrac{H}{2} } -0\right)[/tex]
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The time it takes for the tank to empty the remaining half, t₂, is presented as follows;
[tex]t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }[/tex]
The total time, t, to empty the tank is given as follows;
[tex]t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}[/tex]
[tex]t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }[/tex]
Determine the maximum shear stress in the beam and the minimum yield strength that should be considered to obtain a minimum factor of safety of 2 based on the distortion-energy theory. The maximum shear stress in the beam is 1875 Numeric ResponseEdit Unavailable. 1875 correct.psi. The minimum yield strength based on the distortion-energy theory is
Answer: hello some parts of your question is missing attached below is the missing part
max shear stress = 1875 psi
minimum yield strength = 14911.76 psi
Explanation:
Since The maximum shear stress in the Beam is already given as 1875 psi , I will calculate for the minimum yield strength
Determine minimum yield strength
attached below is the detailed solution
minimum yield strength = 14911.76 psi
If an improvement creates no significant change in a product’s performance, then it is a(n) design improvement.
Answer:
Following are the responses to these question:
Explanation:
They might believe that it was an enhanced layout because the quality is not updated. For instance, its new XS Max iPhone does have a better display than the iPhone X, however, the performance wasn't enhanced. It also has the same processor or graphic cards however a bigger pixel every centimeter ratio. When its output AND is not altered, the layout doesn't change, basically the very same item.
Answer:
If an Improvement creates no significant change in a product’s performance, then it is a(n) Superficial design improvement.
Explanation:
DIFFERENT BREED!
What's the ampacity of a No. 10 type TW copper Wire in a raceway containing six wires and located in an area where the ambient Temperature is 50 C
A. 15.1 A
B. 8.6 A
C. 13.9 A
D. 10.8 A
A detailed image of a brain scan with height, width, and depth is an example of a(n) 3D _________ model.
Answer:
is a mathematical representation of something three-dimensional.
Explanation:The typical base of a the model is a 3D mesh; the structural build consists of polygons.
A three-story structure is to be constructed over an 8000-m2 site. The initial subsurface exploration indicates the presence of sinkholes and voids due to dissolution of the limestone formation. The predominant soil type is a silty fine sand grading to a fine sand with seams of sandy clay. The design indicates that shallow foundations can be used for this project provided the soils were made more homogeneous as far as load support and no voids were present within the depth up to 7.6 m below the ground surface. Assume groundwater is not a concern. Dynamic compaction is proposed to improve the ground. The local contractor doing dynamic compaction has a 15-ton tamper with the diameter of 2.0 m and the height of 1.4 m. You are requested to conduct the preliminary design for this dynamic compaction project including drop height, spacing, number of drops, number of passes, estimated crater depth, and settlement
Answer:
a) 24.07 m
b) 4 m
c) 14 number of drops
d) p = number of passes
e) Dcd = 2.27
0.69 m
Explanation:
Given data:
Depth ( D )= 7.6 m below ground surface
dynamic compaction ( w ) = 15-ton , diameter of tamper = 2.0 m , thickness = 1.4 m
Determine :
A) drop height ( H )
D = n √wH
therefore H = 361 / 15 = 24.07 m
where : D = 7.6 m , n = 0.4 , w = 15
B) Drop spacing
drop spacing = average of ( 1.5 to 2.5 ) * diameter of tamper
= 2 * 2.0m = 4 m
C) number of drops
since the applied energy for fine grained soils and day fills range from 250 - 350 kj/m^2 the number of drops can be calculated using the relation below
AE = [tex]\frac{NWHP}{SPACING ^2}[/tex]
w = 15, H = 24.07 , Np = ? , AE = 300 kj/m^2
∴ Np = 4800 / 361.05 = 13.3
the number of drops at one pass = 14
D) number of passes
p = number of passes
E) estimated crater depth and settlement
crater depth ( Dcd ) = 0.028 [tex]N_{d} ^{0.55} \sqrt{wtIt}[/tex]
Nd = 14 , wt = 15, It = 24.07
therefore : Dcd = 2.27
estimate settlement is within 3 to 5% therefore the improved settlement
= 2.27 * 0.04 * 7.6 = 0.69 m
write to change past tense
Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural
Answer:
B. schematic design
Explanation:
This correct for Plato/edmentum
Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.
What is a schematic design?A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.
The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.
The schematic design consists of a rough sketch with markings and measurements.
Therefore, the correct option is B. schematic design.
To learn more about schematic design, refer to the link:
https://brainly.com/question/14959467
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Consider the formation of p-nitrophenol from p-nitrophenyl trimethyl acetate. The process is known as enzymatic hydrolysis and it occurs in the presence of the enzyme elastase. Along with the formation of p-nitrophenol, trimethyl acetic acid is also formed which is an undesired byproduct. p-nitrophenol is an important intermediate in the manufacture of several pharmaceuticals. Your role as a Chemical Engineer is to maximize the production of p-nitrophenol. The reactions can be denoted as:
E+S → P+ES R1
ES+PE+A R2
where e denotes the enzyme elastase, denotes the substrate p-nitrophenyl trimethyl acetate, es denotes enzyme-substrate intermediate and A denotes the trimethyl acetic acid. The rate of the reactions 1 and 2 are given by:
kg Cs KM + C r2 = kxCp
where Cs and Cp denote the concentrations of the substrate and the product, k, and ky are the rate constants given by 0.015 s' and 0.0026 s. Ky is the Michaelis - Menten constant and is given by 5.53 mol/m!. All the reactants and products are in the liquid phase. The initial concentrations of S and E are 0.5 mol/m3 and 0.001 mol/m..Consider the above reaction to occur in a batch reactor for 15 minutes.
a. Plot the concentration profiles of S, P and A as a function of time in a single figure.
b. Plot the selectivity of P with respect to 5 as a function of time b.
Solution :
cs=zeros(9001);
ca=zeros(9001);
cp=zeros(9001);
psi=zeros(9001);
t=[0:0.1:900];
cs(1)=0.5;
ce(1)=0.001;
cp(1)=0;
ca(1)=0;
psi(1)=0;
for i=1:1:9000
cs(i+1)=cs(i)-0.1*((0.015*cs(i))/(5.53+cs(i)));
cp(i+1)=cp(i)+0.1*((0.015*cs(i))/(5.53+cs(i))-0.0026*cp(i));
ca(i+1)=ca(i)+0.1*0.0026*cp(i);
psi(i+1)=((cp(i+1)-cp(i)))/((cs(i)-cs(i+1)));
end
plot(t,cs,t,cp,t,ca);
plot(t,psi);
Please help fast
What would happen if the air outside the hot air balloon were as hot as the air inside.
Dry air does NOT contain
Explanation:
Dry air doesn't contain water vapor .
A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Answer:
N0
Explanation:
It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Given data :
Diameter ( d ) = 10 mm
length ( l ) = 150 mm
elasticity ( ∈ ) = 100 GPa
longitudinal strain ( б ) 200 MPa
Poisson ratio ( μ ) ( assumed ) =0.3
Assumption : deformation totally elastic
attached below is the detailed solution to why it is not reasonable .
The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen