Answer:
13.86 MPa
Explanation:
We need to find the change in length ΔL of the steel beam and thus, the strain in the beam, ε.
ΔL = L₀αΔθ
Strain, ε = ΔL/L₀ = αΔθ where α = coefficient of linear expansion of steel = 11 × 10⁻⁶ °C and Δθ = temperature change of steel beam = θ₂ - θ₁ where θ₁ = initial temperature of steel beam = 23 °C and θ₂ = final temperature of steel beam = 42 °C
Substituting the values of the variables into the equation, we have
ε = αΔθ
ε = α(θ₂ - θ₁)
ε = 11 × 10⁻⁶ /°C(42 °C - 23 °C )
ε = 11 × 10⁻⁶ /°C(42 °C)
ε = 462 × 10⁻⁶
ε = 4.62 × 10⁻⁴
Now, the Young's modulus of concrete is Y = 30 GPa = 30 × 10⁹ N/m²
Y = σ/ε where σ = compressional stress from concrete and ε = strain on steel
So, σ = Yε
σ = 30 × 10⁹ N/m² × 4.62 × 10⁻⁴
σ = 138.6 × 10⁵ N/m²
σ = 1.386 × 10² × 10⁵ N/m²
σ = 1.386 × 10⁷ N/m²
σ = 13.86 × 10⁶ N/m²
σ = 13.86 MPa
Why is the newtons law of cooling and explain how to derive it/
Answer:
For small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed. qf = final temperature of object
Explanation:
hope this helps you sorry if it doesn’t help you
For a sixth-order Butterworth high pass filter with cutoff frequency 3 rad/s, compute the following:
a. The locations of the poles.
b. The transfer function H(s).
c. The corresponding LCCDE description.
Solution :
Given :
A six order Butterworth high pass filter.
∴ n = 6, [tex]w_c=1 \ rad/s[/tex]
a). The location at poles :
[tex]$s^6-(w_c)^6=0$[/tex]
[tex]$s^6=(w_c)^6=1^6$[/tex]
∴ [tex]$s^6 = 1$[/tex]
Therefore, it has 6 repeated poles at s = 1.
b). The transfer function H(S) :
Transfer function H(S) [tex]$=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$[/tex]
[tex]$=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$[/tex]
∴ H(S) [tex]$=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$[/tex]
H(S) [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]
c). The corresponding LCCDE description :
[tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]
[tex]$Y(s)(s^6-1) = s^6 \times (s)$[/tex]
[tex]$Y(s)s^6-y(s).1 = s^6 \times (s)$[/tex]
By taking inverse Laplace transformation on BS
[tex]$L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$[/tex]
[tex]$\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$[/tex]
Hence solved.
What must you do when you reach a steady yellow traffic light?
Answer:
When you come up on a steady yellow traffic light you should always yield to cross traffic if you can yield safely. The flashing yellow light is there to inform drivers to be careful and to slow down.
Explanation:
hope it helped!
Problem
In the clevis shown in Fig. find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P= 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi
Answer:
In the clevis shown in Fig. 1-11b, find the minimum bolt diameter and the minimum thickness of each yoke that will support a load P = 14 kips without exceeding a shearing stress of 12 ksi and a bearing stress of 20 ksi.
127-clevis-double-shear-bolt.gif
Solution 127
Hide Click here to show or hide the solution
127-fbd-clevis-double-shear-bolt.gifFor shearing of rivets (double shear)
P=τA
14=12[2(14πd2)]
d=0.8618in → diameter of bolt answer
For bearing of yoke:
P=σbAb
14=20[2(0.8618t)]
t=0.4061in → thickness of yoke answer
An assembly line has 3 fail safe sensors and one emergency shutdown switch.The line should keep moving unless any of the following conditions arise:
(1) If the emergency switch is pressed
(2) If the senor1 and sensor2 are activated at the same time.
(3) If sensor 2 and sensor3 are activated at the same time.
(4) If all the sensors are activated at the same time
Suppose a combinational circuit for above case is to be implemented only with NAND Gates. How many minimum number of 2 input NAND gates are required.
Answer:
1 NAND gate
Explanation:
The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1
The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )
Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )
8. The operation of a TXV is controlled by the
O A. thermostatic spring.
O B. temperature bulb.
O C. external pressure of the evaporator.
O D. modulating valve.
The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.
Answer:
[tex]n=5.36kg[/tex]
Explanation:
From the question we are told that:
Volume [tex]V=2.5m^3[/tex]
Pressure[tex]\rho=500Kpa[/tex]
Temperature [tex]T=28^o[/tex]
Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]
Generally the equation for an Ideal gas is mathematically given by
[tex]PV=nRT[/tex]
Therefore
[tex]n=\frac{500*2.5}{8.314*28}[/tex]
[tex]n=5.36kg[/tex]
Request for proposal (RFP) is a type of document that contains the information and proposals mostly through the bidding process. This document is regarding the valuable assets, services, entity, commodity, etc.
Answer:
Answer to the following is as follows;
Explanation:
A request for proposal is a documentation that invites prospective contractors to submit business opportunities to an agency or corporation interested in procuring a commodities, product, or valuable resource through a bid procedure.
A request for proposal (RFP) is a commercial document that introduces a project, defines it, and invites eligible contractors to compete on its completion.
A stream of ethylene glycol vapor at its normal boiling point and 1atm flowing at a rate of 175 kg/min is to be condensed at constant pressure. The product stream from the condenser is liquid g lycol at the condensation temperature.
a. Calculate the rate at which heat must be transferred from the condenser (kW).
b. If heat were transferred at a lower rate than that calculated in part (A), what would the state of the product stream be? (Dedu ce as much as you can about the phase and the temperature of the stream.)
c. If heat were transferred at a higher rate than that calculated in part (A), what could you deduce about the state of the product stream?
Answer: hello attached below is the question properly written
a) 2670 Kw
b) product will be made up of vapor and liquid
c) Product will be a super cooled liquid
Explanation:
mass Flow rate ( m ) = 175 kg/min
pressure = 1 atm
molecular weight of ethylene glycol ( mw ) = 62.07 g/mol
enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol
Using values from the table 8.1 related to the question
a) Determine the rate at which heat must be transferred from condenser
Using values from the table 8.1 related to the question
ΔH = 2670 Kw
b) If heat is transferred at a lower temperature the product will be made up of vapor and liquid
c) If heat was transferred at a higher temperature the product will be a super cooled liquid
When a voltage (v=353 sin (251t+30) is applied to two elements impedance a current (i =7.07 cos 251 t) is passing. Find the nature and the value of the elements and the circuit power
Answer:
A.C. voltage, V= V0 sin ωt As,t = πω = 12.2πω = 12T, therefore, first half cycle (T/2). Hence, average value of AC voltage, Eav = 2V0π.In a certain pressing operation, the metallic powder fed into the open die has a packing factor of 0.5. The pressing operation reduces the powders to 70% of their starting volume. In the subsequent sintering operation, shrinkage amounts to 10% on a volume basis. Given that these are the only factors that affect the structure of the finished part, determine its final porosity.
Answer:
0.2063
Explanation:
Given data:
packing factor = 0.5
percentage of reduction of powders = 70%
Calculate the final porosity
after sintering Bulk specific volume = 0.9 * 0.7 = 0.63
assuming true specific volume = 1
packing factor = 0.5 , bulk specific volume = 2
packing factor after pressing and sintering
= 1 / ( 2 * 0.63 ) = 0.7937
hence : porosity = 1 - packing factor
= 1 - 0.7937 = 0.2063
A test bar of nonferrous material has a diameter of 0.253 inches. Upon applying a tensile load, the sample exhibited 0.002 plastic strain at 3400 lb and the maximum load during testing was 6200 lb and occurred at an engineering strain of 0.65; and breaking occurred at 4400 lb. The sample diameter at fracture was measured to be 0.15 inches.
Required:
a. The yield strength of the material is :________
b. The UTS of the material is:________
Answer:
a. 67607.9psi
b. 123278.33
Explanation:
to get the yield strength of the material
= load/ cross sectional area
cross sectional area = π * 0.253²/4
= 0.0502927
The yield strenght
= 3400/0.0502927
= 67609.9 psi
b. the uts of the material
= maximum load/cross sectional area
= 6200/0.0502927
= 123278.33
Một máy nghiền bi thùng ngắn đường kính D = 1.6m, dài L = 2m dùng để nghiền VL. Kích thước đầu vào D1 = 20mm, sản phẩm sau nghiền có kết quả phân tích rây sau:
Số mesh
60/80
80/100
100/150
150/200
Khối lượng VL trên rây
0.1
0.3
2.5
0.1
Hỏi
1. Tính kích thước bi nghiền bi.
2. Tính số vòng quay tối ưu.
3. Tính năng suất (biết K1 = 1.01).
4. Tính số lượng bi cần nạp vào thùng. Biết hệ số chứa đầy φ = 0.25, hệ số rỗng = 0.65, ⍴bi = 8000 kg/m3
Answer:
bood ekogcd gcaerh is an American fbnuxc
good housekeeping can increase production in a work place is it true or false
False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.
Hope it helps you❤️
thiết kế ic 555 và code để ic hoạt động
Answer:
here you go.
screenshot 2 should give you some basic idea
In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:
Solution :
Given data :
p = 315612 Pa
[tex]$V_1=7.07 \ m/sec$[/tex]
At exit of B,
p = [tex]$P_{atm}$[/tex]
[tex]$V_B = 26.1 \ m/sec$[/tex]
At exit of A,
[tex]p=P_{atm}[/tex]
[tex]$V_{A} = 26.1 \ m/s$[/tex]
We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.
From figure,
[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]
[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]
[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]
Substitute all the values,
[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]
[tex]$F_x = -18.2733 \ kN$[/tex]
Therefore, the force required to hold the nozzle in its place along horizontal direction.
[tex]$F_x = -18.2733 \ kN$[/tex]
R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of the exit pipe diameter to that of the inlet pipe (Dex/Din) so that the velocity stays constant.
Solution :
For R-134a, we are given :
[tex]$T_i = 25^\circ C$[/tex]
[tex]$P_i=750 \ kPa$[/tex]
[tex]$P_e=165 \ kPa$[/tex]
Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :
[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]
We also know that the gas is throttled and there is no change in the kinetic energy.
So, [tex]$v_e=v_i$[/tex]
Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,
[tex]$h_i=h_e$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.
[tex]$T_e=-15^\circ C$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :
[tex]$h_f = 180.19 \ kJ/kg$[/tex]
[tex]$h_{fg} = 209 \ kJ/kg$[/tex]
Calculating the exit flow quality factor,
[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]
[tex]$=\frac{234.59-180.19}{209}$[/tex]
= 0.26
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :
[tex]$v_f = 0.00746 \ m^3/kg$[/tex]
[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]
Calculating the exit specific volume :
[tex]$v_e=v_f+x_e(v_{fg})$[/tex]
= 0.000746 + 0.26 (0.11932)
= 0.0318 [tex]m^3/kg[/tex]
The mass flow is equal to :
[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]
[tex]$=A_e . \frac{v}{v_e}$[/tex]
So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]
Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]
[tex]$\frac{D_e}{D_i}=6.19$[/tex]
Please label the following statements as either True (T) or False (F).
(a) The true stress is higher than the engineering stress for a sample under tension.
(b) Creep test is carried out with a dynamic stress under elevated temperature.
Answer:
a. True
b. False
Explanation:
a. Since true stress, σ' = σ(1 + ε) where σ = engineering stress and ε = engineering strain.
Also under tension ε > 0, so, (1 + ε) > 1
Since (1 + ε) = σ'/σ > 1, ⇒ σ' > σ
So, the true stress is greater than the engineering stress.
So, the statement is true
b. Creep is a time-dependent deformation under certain applied load.
Creep occurs at high temperatures under different types of stress.
But, Creep test is carried out at constant high temperature and constant stress.
This statement is false.
Các đặc điểm chính của đường dây dài siêu cao áp .
Answer:
Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.
Explanation:
If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mmmm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12.0 MPaMPa .
Answer: hello some data related to your question is missing attached below is the missing data
answer:
T2 = 265°C
Explanation:
First step : calculate sum of vertical forces
∑ y = 0
Fmg - 2(0.5 Fst ) = 0
∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )
= 1884.96 N
Also determine the Compatibility equation in order to determine the change in Temperature
ΔT = 250°C
therefore Temperature at which average normal stress becomes 12.0 MPa
ΔT = T2 - T1
250°C = T2 - 15°C
T2 = 250 + 15 = 265°C
attached below is the detailed solution
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has a power rating of 2250 W. To dry one typical load of clothes the dryer will run for approximately 45 minutes. In Ontario, the cost of electricity is $0.11/kWh. Calculate the costs to run the dryer for your family for one year.
Answer:
The costs to run the dryer for one year are $ 9.03.
Explanation:
Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:
1 watt = 0.001 kilowatt
2250/45 = 50 watts per minute
45 x 365 = 16,425 / 60 = 273.75 hours of consumption
50 x 60 = 300 watt = 0.3 kw / h
0.3 x 273.75 = 82.125
82.125 x 0.11 = 9.03
Therefore, the costs to run the dryer for one year are $ 9.03.
An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after traveling a distance d.
a. What is the magnitude of the electric field?
b. What is the direction of the electric field?
1. in the direction of the electron's motion
2. opposite to the direction of the electron's motion
3. perpendicular to the direction of the electron's motion
Answer:
The answer is below
Explanation:
a) The work done is equal to the loss in kinetic energy (KE)
Change in kinetic energy (ΔKE) = Final kinetic energy - initial kinetic energy
Final KE = [tex]\frac{1}{2}mv_f^2[/tex]
But the final velocity is 0 (at rest). Hence:
Final KE = [tex]\frac{1}{2}mv_f^2=\frac{1}{2}m(0)^2=0[/tex]
ΔKE = 0 - K = -K
W = ΔKE = -K
Also, the work done (W) = charge (q) * distance (d) * electric field intensity (E)
W = qEd
but q = -e, hence:
W = -e * E * d
Using:
W = ΔKE
-e * E * d = -K
E= K / (e * d)
b) The electric field is in the direction of the electrons motion
What is the differences between total revenue and total costs? Make
sure that your answer will cover all aspects related with two
mentioned concepts. With any supported simple example about
petroleum industrial?
Answer:
The basic difference between Total cost and total revenue is that the total cost includes the total expenditure incurred on the production of a commodity whereas total revenue refers to the money received from selling that commodity.
Explanation:
convert 25 inches / min to mm/hour
Answer:
25 mins into hours = 0.416667 hours
25 inches as mm = 635
Explanation:
nơi nào có điện tích thì xung quanh điện tích đó có :
Explanation:
sory sorry sorry sorrysorrysorry
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 297(106)ft2. Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 (106) ft^2, in SI units?
This question is incomplete, the complete question is;
To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.
Required:
What is the area measurement, 293 × 10⁶ ft², in SI units?
293 × 10⁶ ft² = ?km²
Answer:
the area measurement is 27.221 km²
Explanation:
Given the data in the question;
What is the area measurement, 293 × 10⁶ ft², in SI units
we are to the result of the measured area from ft² to km²
we know that;
1 meter = 3.2808 ft
1 km = 1000 m
1 ft = (1 / 3.2808)m
1 m = ( 1/1000 ) km
since our measured are is 293 × 10⁶ ft²
hence
A = 293 × 10⁶ × [ (1 / 3.2808)m ]²
A = 27221252.74 m²
A = 27221252.74 × [ ( 1/1000 ) km ]²
A = 27.221 km²
Therefore, the area measurement is 27.221 km²
In a series circuit, there ______ for current to flow.
a.
are at least two paths
b.
are multiple paths
c.
is only 1 path
d.
are very small paths
Answer:
Option C = Only 1 path for current to flow.
find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters and feet
Answer:
The volume for this is 29.7
Explanation:
Trust me on this I'm an expert
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Just because I seen someone else ask but they didn't have enough information.
If a filesystem has a block size of 4096 bytes, this means that a file comprised of only one byte will still use 4096 bytes of storage. A file made up of 4097 bytes will use 4096*2=8192 bytes of storage. Knowing this, can you fill in the gaps in the calculate_storage function below, which calculates the total number of bytes needed to store a file of a given size?
Answer:
Following are the program to the given question:
def calculate_storage(filesize):#definging a method calculate_storage that takes filesize as a parameter
block_size = 4096#definging block_size that holds value
full_blocks = filesize//block_size#definging full_blocks that divides the value and hold integer part
partial_block_remainder = filesize%block_size#definging partial_block_remainder that holds remainder value
if partial_block_remainder > 0:#definging if that compare the value
return block_size*full_blocks+block_size#return value
return block_size*full_blocks#return value
print(calculate_storage(1)) # calling method by passing value
print(calculate_storage(4096)) # calling method by passing value
print(calculate_storage(4097)) # calling method by passing value
Output:
4096
4096
8192
Explanation:
In this code, a method "calculate_storage" is declared that holds a value "filesize" in its parameters, inside the method "block_size" is declared that holds an integer value, and defines "full_blocks and partial_block_remainder" variable that holds the quotient and remainder value and use it to check its value and return its calculated value. Outside the method, three print method is declared that calls the method and prints its return value.
