Answer:
true
Explanation:
true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 30 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Answer:
power input = -66.2798 kW
Explanation:
Steady state pressure ( Cp ) = 1.05 bar,
Temperature ( T1 ) = 300 K
Volumetric flow rate = 30 m^3/min
exit pressure = 12 bar
exit temperature ( T2 ) = 400 K
Heat transfer rate ( Q ) = 5 kW
Calculate the power input in kW
p1v1 = m RT
m = p1v1 / RT1
= (1.05 * 10^2 * ( 30/60 )) / ( 0.287 * 300 )
= 0.60975 m^3/sec
also
h1 + Q = h2 + w
∴ w = m ( h1 - h2 ) + Q
= mCp ( t1 - t2 ) + Q ----- ( 1 )
where : ( Q ) = 5 kW , Cp = 1.05 bar, t1 = 300 K, t2 = 400 k ( input values into equation 1 )
w = -66.2798 kW
A long cylindrical black surface fuel rod of diameter 25 mm is shielded by a surface concentric to the rod. The shield has diameter of 50 mm, and its outer surface is exposed to surrounding air at 300 K with a convection heat transfer coefficient of 15 W/m2.K. Inner and outer surfaces of the shield have an emissivity of 0.05, and the gap between the fuel rod and the shield is a vacuum. If the shield maintains a uniform temperature of 335 K, determine the surface temperature of the fuel rod
Answer:
surface temp of fuel rod = 678.85 K
Explanation:
Given data :
D1 = 25 mm
D2 = 50 mm
T2 = 335 k
T∞ = 300 k
hconv = 0.15 w/m^2.k
ε2 = 0.05
ε1 = 1
Determine energy at Q23
Q23 = Qconv + Qrad
attached below is the detailed solution
Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )
To which part of the system does the actuator connect to in an automobile cruise control system?
A. Piston
B. Wheel
C. Throttle
D. Gearbox
Answer:
C. throttle
Explanation:
see the picture hope this helps someone
A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V rms to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary
Answer:
a. 41
b. i. 15 mA ii. 625 mA
c. 192 Ω
Explanation:
Here is the complete question
A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?
Solution
(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?
For a transformer N₂/N₁ = V₂/V₁
where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V
So, N₂/N₁ = V₂/V₁
N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)
(b) What are the rms currents in the primary and secondary coils of the transformer?
i. The rms current in the secondary
We need to find the current in the secondary from
P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V
P = IV
I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA
ii. The rms current in the primary
Since N₂/N₁ = V₂/V₁ = I₁/I₂
where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA
So, V₂/V₁ = I₁/I₂
V₂I₂/V₁ = I₁
I₁ = V₂I₂/V₁
= P/V₁
= 75.0 W/120 V
= 0.625 A
= 625 mA
(c) What is the effective resistance that the 120-V source is subjected to?
Using V = IR where V = voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil
R = V/I
= 120 V/0.625 A
= 192 V/A
= 192 Ω
g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured
This question is not complete, the complete question is;
A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is [tex]88 Mpam^\frac{1}{2}[/tex] , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured
Answer:
length of crack is 5.585 cm
we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures
Explanation:
Given the data in the question;
vessel thickness = 6 cm
fracture toughness k = [tex]88 Mpam^\frac{1}{2}[/tex]
yield strength = 1250 MPa
hoop stress equal = 300 MPa
we know that, the relation between fracture toughness and crack length is expressed as;
k = (1.1)(2/π)(r√(πa))
where k is the fracture toughness, r is hoop stress and a is length of crack
so we rearrange to find length of crack
a = 1/π[( k / 1.1(r)(2/π)]²
a = 1/π[( kπ / 1.1(r)(2)]²
so we substitute
a = 1/π [( 88π / 1.1(300)(2/π)]²
a = 1/π[ 0.1754596 ]
a = 0.05585 m
a = 0.05585 × 100 cm
a = 5.585 cm
so, length of crack is 5.585 cm
we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures
Consider a 3-km2 urban catchment. The main channel has a slope of 0.9% and a Manning n of 0.10. The catchment is 50% impervious. The distance along the main channel from the catchment boundary to the outlet is 1100 m. The urban catchment has an average curve number of 60. Determine the peak flow (m3 /s) of the NRCS unit hydrograph for a 20-min rainfall excess.
Answer:
The right answer is "5.105×10⁸ m³/sec".
Explanation:
The given values are:
Catchment area,
A = 3 km²
Length to watershed,
L = 1100 m
Average watershed slope,
S = 0.9% i.e., 0.009
Curve number,
CN = 60
Rainfall duration,
D = 20 min
Let,
Time form beginning of the rainfall will be "[tex]t_p[/tex]".Lag time will be "[tex]t_1[/tex]".Now,
⇒ [tex]t_1=\frac{L^{0.8}\times (\frac{1000}{CN} -9)^{0.7}}{19000S^{0.5}}[/tex]
On substituting the values, we get
⇒ [tex]t_1=\frac{1100^{0.8}\times (\frac{1000}{60} -9)^{0.7}}{19000\times 0.009^{0.5}}[/tex]
⇒ [tex]=0.625 \ hours[/tex]
then,
⇒ [tex]T_p=\frac{D}{2}+t_1[/tex]
⇒ [tex]=\frac{0.33}{2}+0.625[/tex]
⇒ [tex]=\frac{0.33+1.25}{2}[/tex]
⇒ [tex]=\frac{1.58}{2}[/tex]
⇒ [tex]=0.79 \ hr[/tex]
hence,
The peak flow will be:
⇒ [tex]Q_p=\frac{484A}{t_p}[/tex]
⇒ [tex]=\frac{484\times 3}{0.79}[/tex]
⇒ [tex]=\frac{1452}{0.79}[/tex]
⇒ [tex]=1837.97 \ km^3/hr[/tex]
or,
⇒ [tex]=5.105\times 10^8 \ m^3/sec[/tex]
3
Select the correct answer
Which statement is true about a corporation?
A
B.
The shareholders hold no liability for the corporation's debts.
The shareholders hold limited liability for the corporation's debts.
The shareholders hold complete liability for the corporation's debts.
C.
D.
There are no shareholders in a corporation.
OE.
A single individual owns the corporation,
Which industry does a shoe manufacturer belong?
Diseña un mecanismo multiplicador con un engranaje motriz cuya relación de transmisión sea de 0.5 y que transmita el movimiento entre ejes distantes. Inserta una captura de pantalla indicando la relación entre los diámetros y la velocidad de giro del engranaje motriz.
A hypothetical metal alloy has a grain diameter of 1.7 102 mm. After a heat treatment at 450C for 250 min, the grain diameter has increased to 4.5 102 mm. Compute the time required for a specimen of this same material (i.e., d 0 1.7 102 mm) to achieve a grain diameter of 8.7 102 mm while being heated at 450C. Assume the n grain diameter exponent has a value of 2.1.
Answer:
the required time for the specimen is 1109.4 min
Explanation:
Given that;
diameter of metal alloy d₀ = 1.7 × 10² mm
Temperature of heat treatment T = 450°C = 450 + 273 = 723 K
Time period of heat treatment t = 250 min
Increased grain diameter 4.5 × 10² mm
grain diameter exponent n = 2.1
First we calculate the time independent constant K
dⁿ - d₀ⁿ = Kt
K = (dⁿ - d₀ⁿ) / t
we substitute
K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250
K = (373032.163378 - 48299.511117) / 250
K = 1298.9306 mm²/min
Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )
dⁿ - d₀ⁿ = Kt
t = (dⁿ - d₀ⁿ) / K
t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306
t = ( 1489328.26061158 - 48299.511117) / 1298.9306
t = 1441028.74949458 / 1298.9306
t = 1109.4 min
Therefore, the required time for the specimen is 1109.4 min
) If the blood viscosity is 2.7x10-3 Pa.s, length of the blood vessel is 1 m, radius of the blood vessel is 1 mm, calculate the tubular resistance of the blood vessel (in GPa.S/m3 ). If the blood pressure at the inlet of the above vessel is 43 mm Hg and if the blood pressure at the outlet of the above vessel is 38 mm Hg. Calculate the flow rate (in ml/min).
Answer:
a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.
b) the flow rate is 5.8 ml/min
Explanation:
Given the data in the question;
Length of blood vessel L = 1 m
radius r = 1 mm = 0.001 m
blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s
Now, we know that Resistance = 8μL / πr⁴
so we substitute
Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]
Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]
Resistance = 6.8755 ≈ 6.88 Gpa.s/m³
Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.
b)
blood pressure at the inlet of the vessel = 43 mm Hg
blood pressure at the outlet of the vessel = 38 mm Hg
flow rate = ?
we know that;
flow rate Q = ΔP / R
where ΔP is change in pressure and R is resistance.
ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg = 665 pa
R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹
so we substitute
Q = 665 pa / 114.5916 pa.m.ml⁻¹
Q = 5.8 ml/min
Therefore, the flow rate is 5.8 ml/min
A 3-ft-diameter duct is used to carry ventilating air ( , ) into a vehicular tunnel at a rate of 11000 ft3/min. Tests show that the pressure drop is 1.2 in. of water per 1500 ft of duct. What is (a) the value of the friction factor for this duct and (b) the approximate size of the equivalent roughness of the surface of the duct
Answer:
a) Friction factor for this duct = 0.0239
b) ε = 0.006 ft
Explanation:
Given data :
Flow rate = 11000 ft^3 /min
Pressure drop = 1.2 in per 1500 ft of duct
a) Determine the value of the friction factor for this duct
Friction factor for this duct = 0.0239
b) Determine the approximate size of the equivalent roughness of the surface of the duct
ε = 0.006 ft
attached below is the detailed solution to the given problem
14. The flow water in a 10-in Schedule 40 pipe is to be metered. The temperature of the water is
100oF, and the static pressure upstream of the meter is 20 psig, The density of the water is 62
Ibm/ft3
, Assume the flow meter is a square-edged orifice with a diameter ratio of 0.5 . When the
flow rate is 1,200 gpm and the flow coefficient is 0.5. the pressure drop across the orifice
(ibf/in2
) is Pipe ID = 10.002 in.
Answer:
I don't know plead hrdffffdddffff
Fracture Mechanics: A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa m ( ) is exposed to a stress of 1000 MPa (145,000 psi). Assume that the parameter Y has a value of 1.0. a) Calculate the critical stress for brittle fracture of this specimen if it is known that the largest surface crack is 0.75 mm (0.03 in.) long. b) Will this specimen experience fracture
Answer:
a) 927 MPa
b)The specimen will experience fracture
Explanation:
a) Calculate critical stress for brittle fracture
σ = fracture toughness / (y √ π * surface crack)
= 45 / ( 1 [tex]\sqrt{\pi * ( 0.75*10^-3)}[/tex] )
= 927 MPa
b) since critical stress( 927 MPa) < 1000 MPa
hence : The fracture will occur
Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assumeuniform mixing of the fluids occurs within the 4 m3 tank.
This question is incomplete, the complete question as well as the missing diagram is uploaded below;
Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.
Take [tex]p_o[/tex] = 880 kg/m³ and [tex]p_{glycerol[/tex] = 1260 kg/m³
Answer:
the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
Explanation:
Given that;
Inlet velocity of Glycerin, [tex]V_A[/tex] = 6 m/s
Inlet velocity of oil, [tex]V_B[/tex] = 3 m/s
Density velocity of glycerin, [tex]p_{glycerol[/tex] = 1260 kg/m³
Density velocity of glycerin, Take [tex]p_o[/tex] = 880 kg/m³
Volume of tank V = 4 m
from the diagram;
Diameter of glycerin pipe, [tex]d_A[/tex] = 100 mm = 0.1 m
Diameter of oil pipe, [tex]d_B[/tex] = 80 mm = 0.08 m
Diameter of outlet pipe [tex]d_C[/tex] = 120 mm = 0.12 m
Now, Appling the discharge flow equation;
[tex]Q_A + Q_B = Q_C[/tex]
[tex]A_Av_A + A_Bv_B = A_Cv_C[/tex]
π/4 × ([tex]d_A[/tex])²[tex]v_A[/tex] + π/4 × ([tex]d_B[/tex] )²[tex]v_B[/tex] = π/4 × ([tex]d_C[/tex])²[tex]v_C[/tex]
we substitute
π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²[tex]v_C[/tex]
0.04712 + 0.0150796 = 0.0113097[tex]v_C[/tex]
0.0621996 = 0.0113097[tex]v_C[/tex]
[tex]v_C[/tex] = 0.0621996 / 0.0113097
[tex]v_C[/tex] = 5.5 m/s
Now we apply the mass flow rate condition
[tex]m_A + m_B = m_C[/tex]
[tex]p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C[/tex]
so we substitute
1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5
1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335
59.3712 + 13.27 = 0.06220335p
72.6412 = 0.06220335p
p = 72.6412 / 0.06220335
p = 1167.8 kg/m³
Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11mm. The pressure at inlet is 110 mm of Hg and pressure at outlet is 95 mm of Hg. Determine the forces required to keep the artery in place (consider steady state, ignore the mass of blood in the vessel and the mass of blood vessel; blood density is 1050 kg/m3)
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
the forces required to keep the artery in place is 1.65 N
Explanation:
Given the data in the question;
Inlet velocity V₁ = 50 cm/s = 0.5 m/s
diameter d₁ = 15 mm = 0.015 m
radius r₁ = 0.0075 m
diameter d₂ = 11 mm = 0.011 m
radius r₂ = 0.0055 m
A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²
A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²
pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal
pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal
Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s
given that; blood density is 1050 kg/m³
mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s
Now, using continuity equation
A₁V₁ = A₂V₂
V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁
we substitute
V₂ = (0.015 / 0.011 )² × 0.5
V₂ = 0.92975 m/s
from the diagram, force balance in x-direction;
0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )
so we substitute in our values
0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )
0 - 0.6014925 + Rₓ = 0.043106929 - 0
Rₓ = 0.043106929 + 0.6014925
Rₓ = 0.6446 N
Also, we do the same force balance in y-direction;
P₁A₁ - P₂A₂ × sin(60°) + R[tex]_y[/tex] = m'( V₂sin(60°) - 0.5 )
we substitute
⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R[tex]_y[/tex] = 0.092728( 0.92975sin(60°) - 0.5 )
⇒ 1.5484 + R[tex]_y[/tex] = 0.092728( 0.305187 )
⇒ 1.5484 + R[tex]_y[/tex] = 0.028299
R[tex]_y[/tex] = 0.028299 - 1.5484
R[tex]_y[/tex] = -1.52 N
Hence reaction force required will be;
R = √( Rₓ² + R[tex]_y[/tex]² )
we substitute
R = √( (0.6446)² + (-1.52)² )
R = √( 0.41550916 + 2.3104 )
R = √( 2.72590916 )
R = 1.65 N
Therefore, the forces required to keep the artery in place is 1.65 N
At steady state, a thermodynamic cycle operating between hot and cold reservoirs at 1000 K and 500 K, respectively, receives energy by heat transfer from the hot reservoir at a rate of 1500 kW, discharges energy by heat transfer to the cold reservoir, and develops 1000 kW of power. Determine the rate of entropy production and then comment on whether this cycle is possible or impossible, and why.
Answer:
Wmax = 750 kw < power developed ( 1000kw ) for a reversible the cycle is Impossible
Explanation:
Hot reservoir Temperature = 1000 K
Cold reservoir Temperature = 500 K
Heat transfer ( energy received by Hot reservoir ) ( Q ) = 1500 kW
Heat transfer ( energy received by Cold reservoir via Hot reservoir ) = 1000 Kw
Calculate the rate of entropy production
The higher the entropy production the less efficient the system
Δs = Cp In ( T2 / T1 )
power developed = 1000 kW
considering that the cycle is reversible and the constant volume or constant pressure of the substance in the thermodynamic cycle is not given we will use the efficiency to determine if the cycle is possible or not
Л = efficiency
∴Л = 1 - T2 / T1 = 1 - ( 500 / 1000 ) = 0.5
note as well that; Л = work output / work input = Wmax / Q
= 0.5 = Wmax / Q
∴ rate of entropy production = Q ( 0.5 ) = 1500 * 0.5 = 750 kw
Given that Wmax = 750 kw < power developed ( 1000kw ) for a reversible the cycle is Impossible
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Garth is a recruitment executive in a firm and knows the eight stages of recruitment. What activity or incident should Garth carry out or
expect to occur at each stage of the process?
place an advertisement in a job portal
vacancyWhat activity should Garth
Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.
What is recruitment?The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.
Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:
Identifying the Need for the Position: Garth should review the company's staffing needs and determine if a position needs to be filled. Once a decision has been made, he should create a job description and identify the position's requirements.Garth should create a recruitment plan that includes a timeline for the recruitment process, a list of recruitment sources, and an advertising strategy.Garth should actively seek qualified candidates through various recruitment channels such as job boards, social media, referrals, and recruiting events.Screening Candidates: Garth should go over resumes, cover letters, and other application materials to see if candidates meet the job requirements. Garth should conduct interviews with the most qualified candidates to assess their skills, experience, and fit for the position. Garth should review all of the information gathered during the recruitment process and choose the best candidate for the position. Garth should ensure that the new hire has all of the necessary information and resources to succeed in their new role.Evaluating the Recruitment Process: Garth should go over the recruitment process to see where he can improve.Thus, these are the stages of recruitment.
For more details regarding recruitment, visit:
https://brainly.com/question/30086296
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Describe ways Texas and its citizens contributed to the Allied war effort during World War II. (Site 1)
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mistboundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate
Answer:
hello your question has some missing values below are the missing values
Mirror Radius (mm) Bending Failure Stress (MPa)
.603 225
.203 368
.162 442
answer : 191 mPa
Explanation:
Determine the stress present at the time of fracture for the original plate
Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )
at 0.603 bending stress = 225
at 0.203 bending stress = 368
at 0.162 bending stress = 442
applying equation 1 determine the value of n for several combinations
( 225 / 368 ) = ( 0.203 / 0.603 )^n
hence : n = 0.452
also
( 368/442 ) = ( 0.162 / 0.203 ) ^n
hence : n = 0.821
also
( 225 / 442 ) = ( 0.162 / 0.603 ) ^n
hence : n = 0.514
Next determine the average value of n
n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596
Calculate estimated stress present at the time of fracture for the original plate
= bending stress at x = 0.796 / bending stress at x = 0.603
= x / 225 = ( 0.603 / 0.796 ) ^ 0.596
therefore X ( stress present at the time of fracture of original plate )
= 225 * 0.84747
= 191 mPa
A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be deformed using a tensile load of 18,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.5 x 10-2 mm. Would the 1040 steel be a possible candidate for this application
Answer:
1040 steel will be a possible candidate for this application since : Yield strength > stress
Explanation:
The 1040 steel would be a possible candidate for this application because the stress experienced by the load is Lesser than its Yield strength
Given that 1040 steel has the following parameter values
Modulus of elasticity ( GPa ) = 205
Yield strength ( Mpa ) = 450
Poisson's ratio = 0.27
limitation of = 1.5 x 10^-2 mm.
stress = Tensile load / area of steel
= 18,000 N / 4.418 * 10^-5 m^2
= 407 .424 Mpa
A wastewater treatment plant discharges 1 m3/s of effluent with an ultimate BOD of 40 mg/L into a stream flowing at 10 m3/s. Just upstream of the discharge point, the stream has an ultimate BOD of 3 mg/L. The deoxygenation constant (kd) is estimated to be 0.22 1/d. (a) Assuming complete and instantaneous mixing, find the ultimate BOD of the mixture of the waste and the river just downstream of the outfall. (b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find at a point 10,000 m downstream
Answer:
(a) 6.36 mg/L
(b) 5.60 mg/L
Explanation:
(a)
Using the formula below to find the required ultimate BOD of the mixture.
[tex]L_o = \dfrac{Q_wL_w+ Q_rL_r}{Q_W+Q_r}[/tex]
where;
Q_w = volumetric flow rate wastewater
Q_r = volumetric flow rate of the river just upstream of the discharge point
L_w = ultimate BOD of wastewater
Replacing the given values:
[tex]L_o = \dfrac{(1 \ m^3/L ) (40 \ mg/L) + (10 \ m^3/L) (3 \mg/L)}{(1m^3/L) +(10 \ m^3/L)} \\ \\ L_o = 6.36 \ mg/L[/tex]
(b)
The Ultimate BOD is estimated as follows:
Recall that:
[tex]time(t) = \dfrac{distance }{speed}[/tex]
replacing;
distance with 10000 m and speed with [tex]\dfrac{11 \ m^3/s}{55 \ m^2}[/tex]
[tex]time =\dfrac{10000 \ m}{\Bigg(\dfrac{11 \ m^3/s}{55 \ m^2}\Bigg)}\Bigg(\dfrac{1 \ hr}{3600 \ s}\Bigg) \Bigg(\dfrac{1 \ day}{24 hr}\Bigg)[/tex]
time (t) = 0.578 days
Finally; [tex]L_t = L_oe^{-kt}[/tex]
here;
k = rate of coefficient reaction
[tex]L_ t= (6.36) \times e^{-(0.22/day)(0.5758 \ days)}\\ \\ \mathbf{L_t =5.60 \ mg/L}[/tex]
Thus, the ultimate BOD = 5.60 mg/L
When framing a wall, temporary bracing is
used to support, plumb, and straighten the wall.
used to support, level, and straighten the wall.
used to square the wall before it is erected.
removed before the next level is constructed.
Calculate the rms value.
Answer:
(√6)/3 ≈ 0.8165
Explanation:
The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...
[tex]\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}[/tex]
g Three unequal point masses are attached to a vertical shaft with light rods. Two masses, m and 3m, are at a distance a from the shaft. The third mass, 2m, is at a distance 2a from the shaft. The three masses and the shaft rotate as a single unit about the vertical axis through the shaft. What is the moment of inertia of this system about the vertical axis
Answer:
12ma²
Explanation:
The moment of inertia I = ∑mr² where m = mass and r = distance from shaft
Since we have three masses,
So, I = m₁r₁² + m₂r₂² + m₃r₃² where m₁ = first mass = m, m₂ = second mass = 3m and m₃ = third mass = 2m. Also, r₁ = distance of first mass from shaft = a, r₂ = distance of second mass from shaft = a and r₃ = distance of third mass from shaft = 2a.
I = m₁r₁² + m₂r₂² + m₃r₃²
I = ma² + 3ma² + 2m(2a)²
I = ma² + 3ma² + 2m(4a²)
I = ma² + 3ma² + 8ma²
I = 12ma²
A low-resistance path in a circuit, commonly called a _____ can cause a circuit breaker to trip
A. short circuit
B. closed circuit
C. parallel connection
D. series connection
Answer:
b
Explanation:
Solar azimuth is the horizontal angle of the sun as measured from a predetermined direction. For the northern hemisphere, the 0°
direction is due
Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
6) A deep underground cavern Contains 980 cuft
of methane gas (CH4) at a pressure of 230
psia and temperature of 150°F. How many
(omllbmol of methane does this gas
deposit contain?
Answer:
15625 moles of methane is present in this gas deposit
Explanation:
As we know,
PV = nRT
P = Pressure = 230 psia = 1585.79 kPA
V = Volume = 980 cuft = 27750.5 Liters
n = number of moles
R = ideal gas constant = 8.315
T = Temperature = 150°F = 338.706 Kelvin
Substituting the given values, we get -
1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin
n = (1585.79*27750.5)/(8.315 * 338.706) = 15625
A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is 40 kh at 520 rev/min. The application factor is 1.4. The radial load is 2600 lbf. The reliability goal is 0.90.
Required:
Determine the C10 value in kN for this application and design factor.
Answer:
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
Explanation:
From the information given:
Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]
Speed (N) = 520 rev/min
Reliability goal [tex](R_D)[/tex] = 0.9
Radial load [tex](F_D)[/tex] = 2600 lbf
To find C10 value by using the formula:
[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]
where;
[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]
[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]
The Weibull parameters include:
[tex]x_o = 0.02[/tex]
[tex](\theta - x_o) = 4.439[/tex]
[tex]b= 1.483[/tex]
∴
Using the above formula:
[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]
[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]
[tex]C_{10} = 30962.449 \ lbf[/tex]
Recall that:
1 kN = 225 lbf
∴
[tex]C_{10} = \dfrac{30962.449}{225}[/tex]
[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]
A 20 mm diameter rod made of ductile material with a yield strength of 350 MN/m2 is subjected to a torque of 100 N.m, and a bending moment of 150 N.m. An axial tensile force is then gradually applied. What is the value of the axial force when yielding of the rod occurs using: a. The maximum-shear-stress theory b. The maximum-distortional-energy theory.
Answer:
a) 42.422 KN
b) 44.356 KN
Explanation:
Given data :
Diameter = 20 mm
yield strength = 350 MN/m^2
Torque ( T ) = 100 N.m
Bending moment = 150 N.m
Determine the value of the applied axial tensile force when yielding of rod occurs
first we will calculate the shear stress and normal stress
shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / [tex]\pi /32[/tex] * ( 20)^4
= 63.662 MPa
Normal stress( Гb + Гa ) = MY/ I + P/A
= [( 150 * 10^3) * 10 ] / [tex]\pi /32[/tex] * ( 20)^4 + 4P / [tex]\pi * 20^2[/tex]
= 190.9859 + 4P / [tex]\pi * 20^2[/tex] MPa
a) Using MSS theory
value of axial force = 42.422 KN
solution attached below
b) Using MDE theory
value of axial force = 44.356 KN
solution attached below
What disadvantages can a resort come across
Answer:
the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you
Explanation: