Answer:
Time needed to empty the pool is 401.35 seconds.
Explanation:
The exit velocity of the water from the orifice is obtained from the Torricelli's law as
[tex]V_{exit}=\sqrt{2gh}[/tex]
where
'h' is the head under which the flow of water occurs
Thus the theoretical discharge through the orifice equals
[tex]Q_{th}=A_{orifice}\times \sqrt{2gh}[/tex]
Now we know that
[tex]C_{d}=\frac{Q_{act}}{Q_{th}}[/tex]
Thus using this relation we obtain
[tex]Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}[/tex]
Now we know by definition of discharge
[tex]Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}[/tex]
Using the above relations we obtain
[tex]Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\[/tex]
The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0
Solving for 't' we get
[tex]\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.[/tex]
Increasing rake angle will decrease the tendency for Built-Up-Edge to form. a) True b) False
Answer:
a)True
Explanation:
While machining of ductile material and high feed and low cutting speed welding action take place between tool material and chip material this welding action is called built up edge. Built up edge action takes place due to high temperature.
To decrease the built up edge action
1. Increase the rake angle.
2. Increase the cutting speed
3. Decrease the feed rate
4. Use cutting fluid
Define volume flow rate Q of air flowing in a duct of area A with average velocity V
Answer:
The volume flow rate of air is [tex]Q=A\times V[/tex]
Explanation:
A random duct is shown in the below attached figure
The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time
Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters
From the attached figure we can see that
The volume of the prism that the flow occupies in 1 second equals
[tex]Volume=Area\times V=A\times V[/tex]
Hence the volume flow rate is [tex]Q=V\times A[/tex]
The engine of a 2000kg car has a power rating of 75kW. How long would it take (seconds) to accelerate from rest to 100 km/hr at full power on level road. Neglect drag and friction.
Answer: 10.29 sec.
Explanation:
Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.
If the car starts from rest, this means the following:
ΔK = 1/2 m*vf ²
As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:
100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec
Now, we calculate the change in energy:
ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J
If P= ΔK/Δt, Δt= ΔK/P= 771,728 J / 75,000 J/sec= 10.29 sec.Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F.
Answer:
In Rankine 524.07°R
In kelvin 291 K
In Fahrenheit 64.4°F
Explanation:
We have given temperature 18°C
We have to convert this into Rankine R
From Celsius to Rankine we know that [tex]T(R)=(T_{C}+273.15)\frac{9}{5}[/tex]
We have to convert 18°C
So [tex]T(R)=(18+273.15)\frac{9}{5}=524.07^{\circ}R[/tex]
Conversion from Celsius to kelvin
[tex]T(K)=(T_{C}+273)[/tex]
We have to convert 18°C
[tex]T(K)=(18+273)=291K[/tex]
Conversion of Celsius to Fahrenheit
[tex]T(F)=T_{C}\times \frac{9}{5}+32=64.4^{\circ}F[/tex]
A has a mass of 7 kg, object B has a mass of 5 lbm, and object C has a mass of 0.5 slug. (a) Which object has the largest mass? Which object has the smallest mass? (b) Find the weights of objects A, B and C (in both N and lbf) on the surface of Mars.
Answer:
1) Object C has the largest mass.
2)Object has the smallest mass.
3) Weight of A = 26.6 Newtons
4)Weight of B = 8.6184 Newtons
5)Weight of C = 27.7286 Newtons
Explanation:
Since all the given masses have different unit's we shall convert them all into a same base unit for comparison. The base unit is selected to be kilogram.
Hence
1) Mass of object A = 7 kilogram
2) Mass of object B = 5 pounds
We know that 1 pound equals 0.4536 kilograms Hence 5 pounds equals
[tex]0.4536\times 5=2.268kg[/tex]
3) Mass of object C = 0.5 slug
We know that 1 slug equals 14.594 kilograms Hence 0.5 slug equals
[tex]14.594\times 0.5=7.297kg[/tex]
Upon comparing all the 3 masses we conclude that object C has the largest mass and object B has the smallest mass.
Part b)
Weight of an object is given by
[tex]Weight=mass\times g[/tex]
Now on Mars value of g equals [tex]3.8m/s^{2}[/tex]
Thus the corresponding weights are as under:
[tex]W_{A}=3.8\times 7=26.6N\\\\W_{B}=3.8\times 2.268=8.6184N\\\\W_{C}=3.8\times 7.297=27.7286N\\\\[/tex]
The International Space Station operates at an altitude of 350 km. Plans for the final construction show that material of weight 4.22 ✕ 106 N, measured at the Earth's surface, will have been lifted off the surface by various spacecraft during the construction process. What is the weight of the space station when in orbit?
Answer:
The weight of the station becomes [tex]3.756\times 10^{6}N[/tex]
Explanation:
Since the acceleration due to gravity decreases with increase in height we conclude that at a height of 350 kilometers the weight of the material will be lesser.
At the ground we have
[tex]W=mass\times g_{surface}\\\\\therefore mass=\frac{W}{g_{surface}}\\\\mass=\frac{4.22\times 10^{6}N}{9.81}\\\\\therefore mass=430173.292kg[/tex]
Now we know that the variation of acceleration due to gravity with height above surface of earth is given by
[tex]g(h)=g_{surface}(1-\frac{2h}{R})[/tex]
where R = 6371 km is Radius of earth
Applying values we get the value of 'g' at height of 350 kilometers equals
[tex]g(350)=9.81\times (1-\frac{2\times 350}{6371})=8.732ms^{-2}[/tex]
hence the weight in orbit becomes
[tex]W_{orbit}=mass\times g_{orbit}\\\\W_{orbit}=430173.292\times 8.732\\\\ \therefore W_{orbit}=3.756\times 10^{6}N\\[/tex]
An aircraft is flying at 300 mph true airspeed has a 50 mph tailwind. What is its ground speed?
Answer:
304.13 mph
Explanation:
Data provided in the question :
The Speed of the flying aircraft = 300 mph
Tailwind of the true airspeed = 50 mph
Now,
The ground speed will be calculated as:
ground speed = [tex]\sqrt{300^2+50^2}[/tex]
or
The ground speed = [tex]\sqrt{92500}[/tex]
or
The ground speed = 304.13 mph
Hence, the ground speed is 304.13 mph
A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find the minimum thickness of the pipe.
Answer:
8 mm
Explanation:
Given:
Diameter, D = 800 mm
Pressure, P = 2 N/mm²
Permissible tensile stress, σ = 100 N/mm²
Now,
for the pipes, we have the relation as:
[tex]\sigma=\frac{\textup{PD}}{\textup{2t}}[/tex]
where, t is the thickness
on substituting the respective values, we get
[tex]100=\frac{\textup{2\times800}}{\textup{2t}}[/tex]
or
t = 8 mm
Hence, the minimum thickness of pipe is 8 mm
An inventor claims that he wants to build a dam to produce hydroelectric power. He correctly realizes that civilization uses a lot more electricity during the day than at night, so he thinks he has stumbled upon a great untapped energy supply. His plan is to install pumps at the bottom of the dam so that he can pump some of the water that flows out from the generators back up into the reservoir using the excess electricity generated at night. He reasons that if he did that, the water would just flow right back down through the generators the next day producing power for free. What is wrong with his plan?
Answer:
The problem is that the pumps would consume more energy than the generators would produce.
Explanation:
Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.
A pump uses electricity to add energy to the water to send it to a higher potential energy state.
Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.
What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.
The atmospheric pressure reads "35.2". What is the "gage pressure"?
Answer:
No, it absolute pressure.
Explanation:
Gauge pressure is relative to the pressure of the atmosphere. It is the difference between the pressure measured and the pressure of the atmosphere. If it is measuring atmospheric pressure it will always read zero.
The measurement is an absolute pressure, which is the pressure above a total vacuum.
Any change in the system from one equilibrium state to another is called: A) Path B) Process C) Cycle D) None of the above
Answer:
B) Process
Explanation:
In thermodynamics a process is a passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium.
A thermodynamic process path is the series of states through which a system passes from an initial to a final state.
Cycle is a process in which initial and final state are identical.
the force acting on a particle of mass m is given by
F=kvx
in which k is a positive constant. The particle passes
throughthe origin with speed Vo at time t=0. Find x as a function
oft.
Answer with Explanation:
From newton's second law the acceleration produced by a force on a mass 'm' is given by
[tex]Acceleration=\frac{Force}{Mass}[/tex]
Applying the given values in the above equation we get
[tex]Acceleration=\frac{kvx}{m}[/tex]
Also we know that acceelration of a particle can ve mathem,atically written as
[tex]a=\frac{v\cdot dv}{dx}[/tex]
Applying the given values in the above equation we get
[tex]\frac{kvx}{m}=\frac{v\cdot dv}{dx}\\\\\Rightarrow {kx}\cdot dx=m\cdot dv\\\\\int kxdx=\int mdv\\\\\frac{kx^2}{2}=mv-c[/tex]
'c' is the constant of integration
whose value is found that at x =0 v= [tex]v_o[/tex]
Thus
[tex]c={mv_o}[/tex]
Thus the velocity as a function of position is
[tex]v=\frac{1}{m}(\frac{kx^2}{2}+c)[/tex]
Now by definition of velocity we have
[tex]v=\frac{dx}{dt}[/tex]
Using the function of velocity in the above relation we get
[tex]\frac{dx}{kx^{2}+\sqrt{2c}}=\frac{dt}{2m}\\\\\int \frac{dx}{(\sqrt{k})^2x^{2}+(\sqrt{2c})^2}=\int \frac{dt}{2m}\\\\\frac{1}{\sqrt{2kc}}\cdot tan^{-1}(\frac{(\sqrt{k})x}{\sqrt{2c}})=\frac{t}{2m}+\phi \\\\[/tex]
where
[tex]\phi [/tex] is constant of integration
Now it is given that at t = 0 ,x = 0
thus from the above equation of position and time we get [tex]\phi =0[/tex]
Thus the position as a function of time is
[tex]x(t)=\sqrt{\frac{2c}{k}}\cdot tan(\frac{kct}{\sqrt{2}m})[/tex]
where c=[tex]mv_o[/tex]
a piston executes simple harmonic motion with an amplitude of 0.1m. If it passes through the center of it's motion with a speed of 0.5 m/s, what is the period of oscillation?
Answer:
Time period will be 1.26 sec
Explanation:
We have given amplitude A = 0.1 M
Speed [tex]\frac{dx}{dt}=0.5m/sec[/tex]
The displacement equation of simple harmonic motion is given by
[tex]x(t)=Asin\omega t[/tex]
Differentiating both side
[tex]\frac{dx}{dt}=A\omega cos\omega t[/tex]
In question it is given that at t=0, x=0 and [tex]\frac{dx}{dt}=0.5m/sec[/tex]
So [tex]0.5=0.1\omega cos0[/tex]
[tex]\omega =5sec^{-1}[/tex]
Now period of oscillation [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{5}=1.26sec[/tex]
Explain the two advantages and the two disadvantages of fission as an energy source.
Answer with Explanation:
1) The advantages of fission energy are:
a) Higher concentration of energy : Concentration of energy or the energy density is defined as the amount of energy that is produced by burning a unit mass of the fuel. The nuclear energy obtained by fission has the highest energy density among all the other natural sources of energy such as coal,gas,e.t.c.
b) Cheap source of energy : The cost at which the energy is produced by a nuclear reactor after it is operational is the lowest among all the other sources of energy such as coal, solar,e.t.c
2) The disadvantages of fission energy are:
a) Highly dangerous residue: The fuel that is left unspent is highly radioactive and thus is very dangerous. Usually the residual material is taken deep into the earth for it's disposal.
b) It has high initial costs of design and development: The cost to design a nuclear reactor and to built one after it is designed is the most among all other types of energy sources and requires highly skilled personnel for operation.
What is the specific volume of superheated steam at 300 °C and a pressure of 1.2 MPa?
Answer:
The specific volume of the super heated steam is 0.2139 m³/kg.
Explanation:
Super heated steam is the condition where only compressed vapor of water present. This is not containing any liquid form of water. Super heated steam has very high pressure depending upon temperature. If heat supplied increases after saturation vapor condition of the water, the water fuses into steam completely. Properties of super heated steam are taken from the super heated steam table.
Given:
Temperature of superheated steam is 300°C.
Pressure of the super heated steam is 1.2 Mpa.
Calculation:
Step1
Value of specific volume of the super heated steam is taken from superheated steam table at 300°C and 1.2 Mpa as follows:
Specific volume is 0.2139 m³/kg.
Step2
All other properties are also taken from the table as:
Internal energy is 2789.7 kj/kg.
Enthalpy is 3046.3 kj/kg.
Entropy is 7.034 kj/kgK.
The part of table at the temperature 300°C and pressure 1.2 Mpa is shown below:
Thus, the specific volume of the super heated steam is 0.2139 m³/kg.
A random sample of 5 hinges is selected from a steady stream of product from a punch press, and the a. b. proportion nonconforming is 0.10. Sampling is with replacement. What is the probability of zero nonconforming unit in the sample? What is the probability of one nonconforming unit in the sample? hat is the probability of 2 or more nonconforming units in the sample?
Given : Sample size : n= 5
The proportion nonconforming : p= 0.10
Binomial probability formula :-
[tex]P(x)=^nC_x p^{x}(1-p)^{n-x}[/tex]
The probability of zero nonconforming unit in the sample :-
[tex]P(0)=^5C_0 (0.10)^{0}(1-0.1)^{5}\\\\=(1)(0.9)^5\ \ [ \because\ ^nC_0=1]=0.59049[/tex]
∴ The probability of zero nonconforming unit in the sample= 0.59049
The probability of one nonconforming unit in the sample :-
[tex]P(1)=^5C_1 (0.10)^{1}(0.9)^{4}\\\\=(5)(0.1)(0.9)^5\ \ [ \because\ ^nC_1=n]=0.295245[/tex]
∴ The probability of one nonconforming unit in the sample=0.295245
The probability of 2 or more nonconforming units in the sample :-
[tex]P(X\geq2)=1-(P(0)+P(1))=1-(0.59049+0.295245)\\\\=1-0.885735=0.114265[/tex]
∴ The probability of 2 or more nonconforming units in the sample=0.114265
Please find the power (in kW) needed in accelerating a 1000 kg car from 0 to 100 km/hour in 10 seconds on a 5% gradient up-hill road.
Answer:
P= 45.384 kW
Explanation:
given data:
m = 1000 kg
u = 0
v = 100 km/hr = 250/9 m/s
t = 10 sec
g =9.81 m/s2
5% gradient
from figure we can have
[tex]tan\theta = \frac{0.05x}{x}[/tex]
[tex]\theta = tan^{-1}0.05 [/tex]
[tex]\theta = 2.86[/tex] from equation of motion we have
v = u + at
[tex]\frac{250}{9} = 0 + 9*10[/tex]
[tex]a = \frac{25}{9} m/s^2[/tex]
distance covered in 10 sec
from equation of motion
[tex]s = ut + \frac{1}{2}at^2[/tex]
[tex]s = 0*10+ \frac{1}{2}*\frac{25}{9}*10^2[/tex]
s = 138.8 m
from newton's 2nd law of motion along inclined position
[tex]F -mgsin\theta = ma[/tex]
solving for f
[tex]f = mgsin\theta +ma[/tex]
[tex]F = 1000*9.81*SIN2.8624 +1000*\frac{25}{9}[/tex]
F = 3267.67 N
work done is given as W
[tex]W = F* s[/tex]
and power [tex]P = \frac{W}{t}[/tex]
[tex]P = \frac{F*s}{t}[/tex]
[tex]P = \frac{3267.67*\frac{1250}{9}}{10}[/tex]
P = 45384.30 W
P= 45.384 kW
How is the principle of superposition used to determine the stress state for a combined loading?
Answer with Explanation:
For a linearly responsive system principle of superposition states that
"The cumulative response of the given system to forces of different magnitudes is the sum of the individual responses of the system to the individual forces"
In a less formal manner principle of superposition states that the effect of various forces acting together on any body is the sum of all the effects on the body produced when each force acts individually.
Now we know that
[tex]Stress=\frac{Force}{Area}[/tex]
Since stress at any point is in linear relation with the force hence we can conclude that the state of stress at any point due to different forces acting together is the sum of the individual stresses due to individual forces alone.
Mathematically
Let the stress due to a force [tex]F_i[/tex] be [tex]\sigma _i[/tex]
and the stress due to combined forces be [tex]\sigma _f[/tex]
thus according to principle of superposition we have
[tex]\sigma_f=\sum_{i=1}^{n}(\sigma_i)[/tex]
Two forces P and Q are applied as shown at point A of a
hooksupport. Knowing that P = 60 lb and
Q = 25 lb, determine graphically the magnitude and directionof
their resultant using the triangle rule.
The magnitude and direction of the resultant force from two applied forces, P and Q, at a point can be determined graphically using the triangle rule by forming a vector triangle, measuring the resultant vector, and calculating its angle.
Explanation:To determine the magnitude and direction of the resultant force using the triangle rule when two forces, P and Q, are applied at a point, you can follow a graphical method. Picture two vectors representing these forces originating from the same point. Since P = 60 lb and Q = 25 lb, you draw them to scale, with the tail of Q starting at the head of P to form a triangle. By connecting the tail of P to the head of Q, you form the resultant vector.
To find the magnitude of the resultant force, measure the length of the resultant vector using the same scale and apply trigonometry or the Pythagorean theorem if the angle is known. The direction of the resultant force is given by the angle it makes with either P or Q, which can be measured with a protractor in reference to a baseline, such as the horizontal.
What is the difference between Rage pressure and absolute pressure?
Answer:
Rage pressure:
The rage pressure in terms of engineering is the protected shield hard plastic shell and made up of the hard material that is basically used in the protection.
It is the effective material and used in the sewing machine so the pressure developed due to the hard material is known as rage pressure and it has strong elasticity.
Absolute pressure:
The pressure which is relative to the perfect vacuum is known as absolute pressure. The absolute pressure is basically measured against the atmospheric pressure.
The absolute pressure is defined as the total pressure in the fluid at the point is equal to the sum of the atmospheric pressure and gauge.
What's the monomer? Show the structure.
In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.
Draw the Pressure - Temperature Diagram showing the liquid and vapor phases, along with the saturation line and the critical point (no need to include the solid section).
Answer:
Pressure- temperature diagram of the fluid is the phase lines that separate all the phases.
Explanation:
Step1
Pressure temperature diagram is the diagram that represents the all the phases of the fluid by separating a line. There is no phase change region in the pressure temperature diagram out of 15 possible diagrams. There are three lines that separate the phase of the fluid. These three lines are fusion, vaporization and sublimation.
Step2
The intersecting point of these lines is triple point of fluid. Out of 15 possible phase diagram, only pressure temperature diagram has triple point as a point. In other diagrams phase change region is present and triple point is not a point. Critical point is the point in all possible property diagrams.
Pressure temperature diagram is shown below:
Calculate the density of a hydraulic oil in units of kg/m^3 knowing that the density is 1.74 slugs/ft^3. Then, calculate the specific gravity of the oil.
Answer:
Density of oil will be 897.292 kg[tex]m^3[/tex]
And specific gravity of oil will be 0.897
Explanation:
We have given density of oil is 1.74 slugs/[tex]ft^3[/tex]
We have to convert this slugs/[tex]ft^3[/tex] into kg/[tex]m^3[/tex]
We know that 1 slugs = 14.5939 kg
So 1.74 slug = 1.74×14.5939 = 25.3933 kg
And 1 cubic feet = 0.0283 cubic meter
So [tex]1.74slug/ft^3=\frac{1.74\times 14.5939kg}{0.0283m^3}=897.292kg/m^3[/tex]
Now we have to calculate specific gravity it is the ratio of density of oil and density of water
We know that density of water = 1000 kg/[tex]m^3[/tex]
So specific gravity of water [tex]=\frac{897.292}{1000}=0.897[/tex]
What are the processes by which polymer foams are produced.
Answer:
1.Molding
a.Cold molding
b.Hot molding
2.Slab stock
Explanation:
Polymer foam:
When solid and gas phases are mixed then they produce polymer foam.The gas used during forming of polymer foam is know as blowing agent and can be physical or chemical.Physical agent will not react and act as inert while chemical agent take part in reaction.
Blowing agent-Carbon diaoxide and Hydrochlorofluorocarbons.
Example of polymer foams-Polyurethane ,Starch etc.
Polymer foam can be produce by following process
1.Molding
a.Cold molding
b.Hot molding
2.Slab stock
What is 30.25 inHg in psia?
Answer:
30.25 in Hg will be equal to 14.855 psi
Explanation:
We have given 30.25 Hg pressure
We have to convert the pressure of 30.25 Hg into psi
We know that 1 inch of Hg = 0.4911 psi
So to convert 30.25 inch Hg in psi we have to multiply with 0.4911
We have to convert 30.25 in Hg
So [tex]30.25inHg=30.25\times 0.4911=14.855775psi[/tex]
So 30.25 in Hg will be equal to 14.855 psi
What are factor of safety for brittle and ductile materials.
Answer:
For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.
Explanation:
Factor of safety:
When materials are subjected to stress then we have to prevent it from a failure so we multiple stress by a factor and that factor is called factor of safety.
Factor of safety can be given as
[tex]FOS=\dfrac{Maximum\ strength}{Applied\ stress}[/tex]
Factor of safety is not a fixed quantity is varies according to the situation.
For brittle material ,ultimate strength use to determine the factor of safety but on the other hand for ductile material yield strength use to determine the factor of safety.
We know that brittle material did not shows any yield point and gets break without showing a indication but ductile materials shows a yield point and gives indication before fracture.
What is referred to as "Pyroelectric" materials?
Answer and Explanation:
Pyroelectric material
Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.
The potential difference generated is for very less time
The generation of potential difference is due to change in position of atoms after heating or cooling.
Consider a cylindrical nickel wire 1.8 mm in diameter and 2.6 × 104 mm long. Calculate its elongation when a load of 290 N is applied. Assume that the deformation is totally elastic and that the elastic modulus for nickel is 207 GPa (or 207 × 109 N/m2).
Calculate the elongation of a cylindrical nickel wire under a specific load by using the formula for elastic deformation and provided values.
Explanation:To calculate the elongation of the cylindrical nickel wire under a load, we can use the formula for elastic deformation: elongation = (F * L) / (A * E), where F is the load, L is the length, A is the cross-sectional area, and E is the Young's modulus.
Substitute the given values: diameter = 1.8 mm (radius = 0.9 mm), load = 290 N, length = 2.6 × 10^4 mm, and Young's modulus for nickel = 207 × 10^9 N/m^2. Solve for the elongation to find the answer.
The elongation of the cylindrical nickel wire under the given load is found to be **insert answer here**.
The elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.
Step 1
To calculate the elongation of the nickel wire under the applied load, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] of a material is directly proportional to the applied force ([tex]\( F \)[/tex]) and the material's elastic modulus ([tex]\( E \)[/tex]), and inversely proportional to its cross-sectional area ([tex]\( A \)[/tex]) and original length ([tex]\( L_0 \)[/tex]). Mathematically, it's expressed as:
[tex]\[ \Delta L = \frac{F \cdot L_0}{A \cdot E} \][/tex]
Where:
- F is the applied force (290 N)
- [tex]\( L_0 \)[/tex] is the original length of the wire (2.6 × [tex]10^4[/tex] mm = 26,000 mm)
- A is the cross-sectional area of the wire
- E is the elastic modulus of nickel (207 × [tex]10^9[/tex] [tex]N/m^2[/tex])
Step 2
First, let's calculate the cross-sectional area (A) of the wire using its diameter (d ):
[tex]\[ A = \frac{\pi d^2}{4} \][/tex]
Given that the diameter [tex](\( d \))[/tex] is 1.8 mm, we have:
[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \][/tex]
Now, let's calculate the elongation ([tex]\( \Delta L \)[/tex]) using Hooke's Law:
[tex]\[ \Delta L = \frac{290 \times 26,000}{A \times 207 \times 10^9} \][/tex]
Step 3
Finally, we can substitute the values and solve for [tex]\( \Delta L \).[/tex] Let's do the calculations.
First, let's calculate the cross-sectional area A:
[tex]\[ A = \frac{\pi \times (1.8 \times 10^{-3})^2}{4} \]\[ A = \frac{\pi \times 3.24 \times 10^{-6}}{4} \]\[ A = 2.55 \times 10^{-6} \, \text{m}^2 \][/tex]
Step 4
Now, let's calculate the elongation [tex](\( \Delta L \))[/tex] using Hooke's Law:
[tex]\[ \Delta L = \frac{290 \times 26,000}{2.55 \times 10^{-6} \times 207 \times 10^9} \]\[ \Delta L = \frac{7,540,000}{528.15} \]\[ \Delta L = 14.295 \, \text{m} \][/tex]
So, the elongation of the nickel wire when a load of 290 N is applied is approximately 14.295 meters.
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
Answer:
[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]
Explanation:
Reynolds number:
Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .
Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.
[tex]Re=\dfrac{F_i}{F_v}[/tex]
For plate can be given as
[tex]Re=\dfrac{\rho\ v\ l}{\mu }[/tex]
Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.
Flow on plate is a external flow .The values of Reynolds number for different flow given as
[tex]Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.[/tex]
[tex]Reynolds\ number\is \ <\ 5 \times 10 ^5\ then\ flow\ will\ be\ laminar.[/tex]
Area under the strain-stress curve up to fracture:______
Answer:
Area under the strain-stress curve up to fracture gives the toughness of the material.
Explanation:
When a material is loaded by external forces stresses are developed in the material which produce strains in the material.
The amount of strain that a given stress produces depends upon the Modulus of Elasticity of the material.
Toughness of a material is defined as the energy absorbed by the material when it is loaded until fracture. Hence a more tough material absorbs more energy until fracture and thus is excellent choice in machine parts that are loaded by large loads such as springs of trains, suspension of cars.
The toughness of a material is quantitatively obtained by finding the area under it's stress-strain curve until fracture.