Answer:
there are generally two types of toilet bowl types- round and elongated.
Identify the first step in preparing a spectrophotometer for use.
A. Make sure all samples and the blank are ready for measurement.
B. Prepare a calibration curve.
C. Measure the absorbance of the blank.
D. Turn on the light source and the spectrophotometer.
Answer:
D. Turn on the light source and the spectrophotometer.
Explanation:
A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.
In a CNC machining operation, the has to be moved from point (5, 4) to point(7, 2)along a circular path with center at (7,2). Before starting operation, the tool is at (5, 4).The correct G and M code for this motion is
Answer: hello your question is incomplete below is the complete question
answer:
N010 GO2 X7.0 Y2.0 15.0 J2.0 ( option 1 )
Explanation:
Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path
GO2 = circular interpolation in a clockwise path
G91 = incremental dimension
hence the correct option is :
N010 GO2 X7.0 Y2.0 15.0 J2.0
Bainite has finer grains because the transformation takes place at a lower temperature where the nucleation rate is high relative to the growth rate. True or False
Answer:
False
Explanation:
Bainite is a type of steel. It is formed by the decomposition of austenite. This happens at a temperature above MS but given that MS temperature is below the one where fine pearlite is formed.
In other words, when iron goes through a metastable crystallization phase Bainite is formed. It takes the rapid cooling, or quenching, of austenite for that to happen.
Metastability is the phenomenon where matter remains in an apparent state of equilibrium or stability even though it can achieve a more stable state. It can also be referred to as the condition of remaining in that pseudo unstable state for a very protracted period of time.
Bainite steel is used for the construction of components that require frequent usage and where plasticity, or tendency for deformity, as well as wear, cannot be tolerated.
Cheers
Lab 5A Problem Input two DWORD values from the keyboard. Determine which number is larger or if they are even. Your program should look like the following: First number larger Enter a number 12 Enter a number 10 12 is the larger number Press any key to close this window... Second number larger Enter a number 10 Enter a number 12 12 is the larger number Press any key to close this window... Numbers Equal Enter a number 12 Enter a number 12 Numbers are equal Press any key to close this window...
Answer:
Explanation:
#include<iostream>
using namespace std;
int main()
{
int n1,n2;
cout<<"Enter a number:"<<endl; //Entering first number
cin>>n1;
cout<<"Enter a number:"<<endl; //Entering second number
cin>>n2;
if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not
{
if(n1>n2)
{
cout<<n1<<" is the larger number"<<endl;
}
else if(n1==n2)
{
cout<<"Numbers are equal"<<endl;
}
else
{
cout<<n2<<" is the larger number"<<endl;
}
}
else
{
cout<<"The number are not even"<<endl;
}
}
Using 1.5 V batteries, a switch, and three lamps, devise a circuit to apply 4.5 V across eitherone lamp, two lamps in series, or three lamps in series with a single-control switch. Draw theschematic.
Answer: the attached picture is the answer.
Explanation:
Assuming:
the switch position connect to 1, hence 4.5V exist at across lamp1
the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2
the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.
Reinforced concrete is a raw material that has always been available, but it was unappreciated by architects until the nineteenth century.
a. True
b. False
Answer: False
Explanation:
Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.
Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.
Your shifts productivity is Slow because one person is not pulling his share. The rest of the team is Getting upset.
Answer:
you are right but then you ddnt ask a question
Explain why veracity, value, and visualization can also be said to apply to relational databases as well as Big Data.
Answer:
Veracity, Value and Visualization are not only the characteristics of Big Data but are also the characteristics of relational databases. Veracity of data is issue with smallest data stores this is the reason that it is important in relation...
g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
All of the following are instruments involved in changing a tire EXCEPT:
Answer:
can you please give us the option for this question
How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree
Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Answer:
a. 164 °F b. 91.11 °C c. 1439.54 kJ
Explanation:
a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.
So, Δ°F = 212 °F - 48 °F = 164 °F
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
To find the degree change in Celsius, we convert the initial and final temperature to Celsius.
°C = 5(°F - 32)/9
So, 48 °F in Celsius is
°C₁ = 5(48 - 32)/9
°C₁ = 5(16)/9
°C₁ = 80/9
°C₁ = 8.89 °C
Also, 212 °F in Celsius is
°C₂ = 5(212 - 32)/9
°C₂ = 5(180)/9
°C₂ = 5(20)
°C₂ = 100 °C
So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.
So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?
Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise = 15.8 kJ/°C × 91.11 °C = 1439.54 kJ
James the Pilot James is a pilot. He is wearing a flight suit. He flies to Paris. He loves flying. 1. James is a a) teacher b) doctor c) pilot. whatisthe 2. He is wearing a a) shirt b) t-shirt c) flight suit. 3. Where does he fly to? a) Italy b) Luxembourg c) Paris http https://whatistheurl.com Please visit our site for worksheets and charts
Answer:
1.c
2.c
3.c
Explanation:
James is a pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.
What is the point of a flight suit?When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.
The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.
The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.
Thus, option C, C, and C are correct.
For more information about point of a flight suit, click here:
https://brainly.com/question/12302183
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The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Answer:
The explanation according to the given query is summarized in the explanation segment below.
Explanation:
If somehow the fin has become too lengthy, this same fin tip temperature approaches the temperature gradient and maybe we'll ignore heat transmission out from end tips.Additionally, effective heat transmission as well from the tip could be ignored unless the end tip surface is relatively tiny throughout comparison to its overall surface.A 5.74 kg rock is thrown upwards with a force of 317 N at a location where the local gravitational acceleration is 9.81 m/s^2. What is the net acceleration of the rock?
Answer:
[tex]a=45.31m/s^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=5.74[/tex]
Force [tex]F=317N[/tex]
Gravitational Acceleration [tex]g=9.81m/s^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F-mg=ma[/tex]
[tex]317-5.74*9.81=5.74 a[/tex]
[tex]a=\frac{260.7}{5.74}[/tex]
[tex]a=45.31m/s^2[/tex]
WILL MARK BRAINLIST I need help on this asap thanks
Determine the dimensions for T if T = M V^2 A / L^3 where M is a mass, V is a velocity, A is an area, and L is a length.
L / T
M
M L / T^2
M / (L T^2)
No dimensions
Explanation:
ask your dad please and use your brain
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temperature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.
Answer:
i did not known answer but anobody help you
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]
in this expression the densities are constant
v = A √h
A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h = [tex]\frac{1}{454.55^2} \ v^2[/tex]
Δh = [tex]\frac{dh}{dv}[/tex] Δv
[tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
[tex]\frac{\delta v}{v} = 0.05[/tex]
[tex]\frac{\Delta h}{h}[/tex] = 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
[tex]\frac{\Delta h}{h}[/tex] = 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
Suppose a causal CT LTI system has bilateral Laplace transform H(s) 2s - 2 $2 + (10/3)s + 1 (8)
(a) Write the linear constant coefficient differential equation (LCCDE) relating a general input x(t) to its corresponding output y(t) of the system corresponding to this transfer function in equation (8).
(b) Suppose the input x(t) = e-tu(t). Find the output y(t). In part (c), the output signal can be expressed as y(t) = - e-(1/3)t u(t) + e-tu(t) e-3tu(t), - 019 Where a, b, and care positive integers. What are they? a = b = C=
Solution :
Given :
[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]
[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]
Apply Inverse Laplace Transforms,
[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]
The above equation represents the differential equation of transfer function.
Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]
We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]
[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]
[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]
[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]
[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]
So,
[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]
[tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]
Applying Inverse Laplace Transform,
[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]
[tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]
where, a = 2
b = 1
c= 2
A flow inside a centrifuge can be approximated by a combination of a central cylinder and a radial line source flow, giving the following potential function:
Ø= a2/r -cosØ + aßlnr = r
Where a is the radius of the central base of the centrifuge and ß is a constant.
a) Provide expressions for the velocities Vr and vo .
b) Find the expression for the stream function.
Answer:
a) Vr = - a^2/r cosθ + aß / r
Vθ = 1/r [ -a^2/r * sinθ ]
b) attached below
Explanation:
potential function
Ø= a^2 /r cosØ + aßlnr ----- ( 1 )
a = radius , ß = constant
a) Expressions for Vr and Vθ
Vr = dØ / dr ----- ( 2 )
hence expression : Vr = - a^2/r cosθ + aß / r
Vθ = 1/r dØ / dθ ------ ( 3 )
back to equation 1
dØ / dr = - a^2/r sinθ + 0 --- ( 4 )
Resolving equations 3 and 4
Vθ = 1/r [ -a^2/r * sinθ ]
b) expression for stream function
attached below
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
For a steel alloy it has been determined that a carburizing heat treatment of 3-h duration will raise the carbon concentration to 0.38 wt% at a point 2.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.1 mm position for an identical steel and at the same carburizing temperature.
Answer:
The right answer is "16.5 hrs".
Explanation:
Given values are:
[tex]x_1=2.6 \ mm[/tex]
[tex]t_1=3 \ hrs[/tex]
[tex]x_2=6.1 \ mm[/tex]
As we know,
⇒ [tex]\frac{x^2}{Dt}=constant[/tex]
or,
⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]
⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]
By putting the values, we get
[tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]
[tex]=5.5\times 3[/tex]
[tex]=16.5 \ hrs[/tex]
A venturimeter of 400 mm × 200 mm is provided in a vertical pipeline carrying oil of specific gravity 0.82, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm. The differential U-tube mercury manometer shows a gauge deflection of 300 mm. Calculate: (i) The discharge of oil, and (ii) The pressure difference between the entrance section and the throat section.Take the coefficient of meter as 0.98 and specific gravity of mercury as 13.6
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ =
A₂ = Area at the throat
A₂ =
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Hope This Helps :D
Match the test to the property it measures.
a. Rockwell
b. Inston
c. Charpy
d. Fatigue
e. Brinell
f. Izod
1. impact strength
2. stress vs strain
3. hardness
4. Endurance Limit
Answer:
a. Rockwell 3. hardness
b. Instron 2. stress vs strain
c. Charpy 1. impact strength
d. Fatigue 4. Endurance Limit
e. Brinell 3. hardness
f. Izod 1. impact strength
Explanation:
Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.
Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.
Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.
Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.
A horizontal water jet impinges against a vertical flat plate at 30 ft/s and splashes off the sides in the verti- cal plane. If a horizontal force of 500 lbf is required to hold the plate against the water stream, determine the volume flow rate of the water.
Answer:
8.6 ft³/s
Explanation:
The force due to the water jet F = mv where m = mass flow rate = ρQ where ρ = density of water = 62.4 lbm/ft³ and Q = volume flow rate. v = velocity of water jet = 30 ft/s
So, F = mv
F = ρQv
making Q subject of the formula, we have
Q = F/ρv
Since F = force due to water jet = force needed to hold the plate against the water stream = 500 lbf = 500 × 1 lbf = 500 × 32.2 lbmft/s² = 16100 lbmft/s²
Since
Q = F/ρv
Substituting the values of the variables into the equation for Q, we have
Q = F/ρv
Q = 16100 lbmft/s²/(62.4 lbm/ft³ × 30 ft/s)
Q = 16100 lbmft/s²/1872 lbm/ft²s
Q = 8.6 ft³/s
So, the volume flow rate is 8.6 ft³/s.
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
In low speed subsonic wind tunnels, the value of test section velocity can be controlled by adjusting the pressure difference between the inlet and test-section for a fixed ratio of inlet-to-test section cross-sectional area.
a. True
b. false
Answer:
Hence the given statement is false.
Explanation:
For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.
The speed within the test section is decided by the planning of the tunnel.
Thus by adjusting the pressure difference won't change the worth of test section velocity.
Answer:
The given statement is false .
An intersection with a four phase signal has a displayed red time of 35 seconds, a start-up lost time of 2 seconds, a yellow time of 4 seconds, and an all red time of 1 second per phase. The total lost time is typically calculated as ____ seconds per cycle.
Answer:
53 sec / cycle
Explanation:
Displayed red time = 35 seconds
Start up lost time = 2 seconds
Yellow time = 4 seconds
Red time = 1 second
Total lost time L = 2n + r
L = lost time
n = number of phase
R = red time
35+2+4+4*1
= 45
L = 2x4+45
= 53 sec/cycle
The total lost time is typically calculated as 53 seconds per cycle