Answer:
30.12345° can be written as : 30°7'20.42''
Or,
30 degrees 7 minutes and 20.42 seconds.
Explanation:
1 degree consists of 60 arc minutes.
1 arc minutes consists of 60 arc seconds.
Thus, 30.12345° can be written as:
30.12345°= 30° + 0.12345°
1° = 60'
So,
0.12345° = 0.12345*60' = 7.407'
Thus, 7.407' can be written as:
7.407' = 7' + 0.407'
1' = 60''
So,
0.407' = 0.407*60'' = 20.42''
Thus,
30.12345° can be written as : 30°7'20.42''
Convert the temperature of 451 degree Fahrenheit to the units requested: a. Rankine b. Kelvin c. Celsius
Answer:
(a) 910.67°R (b) 505.9277 (c) 232.777
Explanation:
FAHRENHIET TO RANKINE: T(R°)=T(F°)+459.67
we have to change 451°F
T(R°)=451+459.67
=910.67°R
FAHRENHET TO KELVIN: T(K)=(T(F°)+459.67) ×[tex]\frac{5}{9}[/tex]
we have to convert 451°F
T(K)=(451+459.67)×[tex]\frac{5}{9}[/tex]
=505.9277
FAHRENHET TO CELSIUS: T(C°)=[tex]\frac{F-32}{9}[/tex]×5
we have to convert 451°F
T(C°)=[tex]\frac{451-32}{9}[/tex]×5
=232.777
In a gas turbine, air (kinematic viscosity of 1x104-5 m 2/s) flows over a 2 cm long turbine blade at 100 m/s. How long should the blade be in my lab's wind tunnel (air, kinematic viscosity of 1.5x10A-5 mA2/s, velocity of 10 m/s), to match the Reynolds number of the gas turbine? a)-2cm b)-30cm c)-0.3cm
Answer:
30 cm
Explanation:
For Reynold's number similarity between model and prototype we should have
[tex]R_{e} _{model} =R_{_{e prototype}} \\\\\frac{V_{model} L_{model} }{kinematic viscosity in model} =\frac{V_{proto}L_{proto} }{kinematic viscosity in prototype}[/tex]
Given L(prototype)= 2cm
V(prototype) = 100m/s
V(model) = 10m/s
Thus applying values in the above equation we get
[tex]\frac{100m/s^{} X2cm^{} }{1X10^{-5}m^{2}/s } =\frac{L_{M}X10m/s }{1.5X10^{-5}m^{2}/s }[/tex]
Solving for Lmodel we get Lm = 30cm
In general, this of the following methods yields the most conservative fatigue strength proof (a) Saderberg method (b)-Goodman method (c)-Gerber line (d)-The ASME elliptic curve.
Answer:
a). Soderberg method
Explanation:
A straight line joining the endurance limit, [tex]S_{e}[/tex] on the ordinate and to the yield strength,[tex]S_{yt}[/tex] on the abscissa is know as Soderberg line.
The Soderberg line is the most conservative failure criteria and in this there is no need to consider yielding point in this case.
The equation for Soderberg is given by
[tex]\frac{\sigma _{m}}{S_{yt}}+\frac{\sigma _{a}}{S_{e}}=1[/tex]
where [tex]\sigma _{m}[/tex] is mean stress
[tex]\sigma _{a}[/tex] is amplitude stress
How much power is needed to operate a Carnot heat pump if the pump receives heat 10°C and delivers 50 kW of heat at 40°C? at A) 5.30 kw B) 151 kw C) 37.5 kW D) 4.79 kw
Answer:
Power needed to pump=4.79 KW.
Explanation:
Given that:[tex]T_{1}=283K,T_{2}=313K,Q_{H}=50KW[/tex]
We know that coefficient of performance of heat pump
COP=[tex]\dfrac{T_{H}}{T_{H}-T_{L}}[/tex]
So COP=[tex]\dfrac{313}{313-283}[/tex]
COP=10.43
COP=[tex]\frac{Q_{H}}{W_{in}}[/tex]
10.43 =[tex]\frac{50}{W_{in}}[/tex]
[tex]W_{in}[/tex]=4.79 KW
So power needed to pump=4.79 KW.
A 350 gal air storage tank is initially at 100 psig. For how long can the tank supply 30 cfm of air to a machine that requires at least 80 psig to operate?
Answer:
93.8 sec
Explanation:
it is given that tab has 350 gallon
we know that 1 gallon = 0.134 cubic foot
350 gallon = 350×0.134=46.9 cubic foot
the delivery pressure is 100 psi which is greater than 80 psi to operate machine
it is given that supply volume is 30 cubic foot per minute
= [tex]\frac{30}{60}=0.5[/tex] [tex]ft^{3}/sec[/tex]
[tex]time\ required\ =\frac{tab\ air }{supply\ volume}[/tex]
[tex]time\ required\= [tex]\frac{46.9}{0.5}[/tex]
=93.8 sec
1kg of air (R 287 J/kgK) fills a weighted piston-cylinder device at 50kPa and 100°C. The device is cooled until the temperature is 0°C. Determine the work done during this cooling.
Answer:
the work done during this cooling is −28.7 kJ
Explanation:
Given data
mass (m) = 1 kg
r = 287 J/kg-K
pressure ( p) = 50 kPa
temperature (T) = 100°C = ( 100 +273 ) = 373 K
to find out
the work done during this cooling
Solution
we know the first law of thermodynamics
pv = mRT ....................1
here put value of p, m R and T and get volume v(a) when it initial stage in equation 1
50 v(a) = 1 × 0.287 × 373
v(a) = 107.051 / 50
v(a) = 2.1410 m³ .......................2
now we find out volume when temperature is 0°C
so put put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1
50 v(b) = 1 × 0.287 × 273
v(a) = 78.351 / 50
v(a) = 1.5670 m³ .......................3
by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.
work done = [tex]\int\limits^a_b {p} \, dv[/tex]
integrate it and we get
work done = p ( v(b) - v(a) ) ................4
put the value p and v(a) and v(b) in equation 4 and we get
work done = p ( v(b) - v(a) )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 (−0.574)
work done = −28.7 kJ
here we can see work done is negative so its mean work done opposite in direction of inside air
'
The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal
Answer:
Isobaric process
Explanation:
The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.
At constant pressure, the work done is given by :
[tex]W=p\times \Delta V[/tex]
Where
W is the work done by the system
p is the constant pressure
[tex]\Delta V[/tex] is the change in volume
So, the correct option is (c) " isobaric process ".
Name three major heat transfer mechanisms giving one example of each from day-to-day life. Also explain the physical mechanism behind these modes of heat transfer.
Answer:
The major heat transfer mechanisms are:
Conduction: When a body at higher temperature comes in direct contact with a body at lower temperature flow of heat takes place from higher temperature to lower temperature due to the Kinetic energy of particles in motion and this motion continues till equilibrium is reached. Heat transfer by this method is called conduction. For example: When a hot metal comes in direct contact with a cold metal, heat is transferred by conduction.Convection: This method of heat transfer applies to fluid motion of particles. Here, the heat transfer is due to thermal energy of the fluid particles. Due to the differences in their density the liquid over hot surface expands and rises up and heat flows from high to low temperatures. For example: When earth's surface is heated by the sun, the warm air comes up and cool air comes in Radiation: Thermal radiations are generated by electromagnetic waves. These are the result of motion of random molecules in the matter which carry the energy from the emitting body and provides motion to the charge particles of EM waves. At high temperature, shorter wavelength is produced and vice-versa. For example: visible light, UV, IR, etcA structural steel shaft with an outer diameter of 1.9 inches and an applied torque of 82.7 ft*lbs. Find: The maximum torsional shear stress in the shaft. Select one: a)- 736.88 ksi b)- 61.41 psi c)- 1473.76 ksi d)- 736.88 psi e)- 368.44 psi
Answer:
Answer is part d -736.88 psi
Explanation:
We know that for a bar subjected to pure torsion the shear stresses that are generated can be calculated using the following equation
[tex]\frac{T}{I_{P} } =\frac{t}{r}[/tex]....................(i)
Where
T is applied Torque
[tex]I_{P}[/tex] is the polar moment of inertia of the shaft
t is the shear stress at a distance 'r' from the center
r is the radial distance
Now in our case it is given in the question T =82.7 ft*lbs
converting T into inch*lbs we have T = 82.7 x 12 inch*lbs =992.4 inch*lbs
We also know that for a circular shaft polar moment of inertia is given by
[tex]I_{P}=\frac{\pi D^{4} }{32}[/tex]
[tex]I_{P}= \frac{\pi\ 1.9^{4} }{32} =1.2794 inch^{4}[/tex]
Since we are asked the maximum value of shearing stresses they occur at the surface thus r = D/2
Applying all these values in equation i we get
[tex]\frac{992.4 inch*lbs}{1.2794 inch^{4} } \frac{1.9 inches}{2}[/tex] = t
Thus t = 736.88 psi
All bodies at a temperature above absolute zero emit thermal radiation. a)-True b)-False
Answer:
a). TRUE
Explanation:
Absolute zero temperature is the lowest possible temperature that can be achieved where no heat energy remains in the body. Absolute zero temperature is 0 k in the Kelvin scale and -273.16 degree Celsius in Centigrade scale.
All bodies with temperature greater than absolute zero emits energy in the form of electro magnetic radiation. Two laws namely Stefan Boltzmann law and Wein's law gives the basis of the fact that bodies with temperature greater than absolute zero temperature emits electromagnetic radiation.
Stefan Boltzmann law : It states the relationship between temperature of the body and radiations that it can emit.
E = σ. [tex]T^{4}[/tex]
where E = radiation emissions
σ = Stefans Boltzmann constant
t is temperature
Wein's Law : It states the temperature of the object and the wavelength at which the body emits maximum radiations.
[tex]\lambda _{max} = \frac{b}{T}[/tex]
where λ is wavelength
b is a constant
T is temperature
A finished, tapered workpiece has an included angle of 70 degrees. If the taper is nade vith the compound, how many degrees should tha setting be on the swivel base? A. 70 B. 140 C. 20 D. 35
Answer:
Tapering is basically the process of thinning or reducing a work piece according to the set standards. and the final product after tapering is known as tapered workpiece.
Solution:
Included angle = 70 degrees
setting on the swivel base is given by:
2α = 70°
α = 35°
Therefore, the setting on the swivel base should be 35°
An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the gravitational potential energy of the object, in kJ, relative to the surface of the earth.
Answer:
Gravitational Potential =58.914 KJ
Explanation:
We know that
[tex]Gravitational Potential Energy = mass\times g\times Height[/tex]
Given mass = 251 kg
Height= 24 m
g is acceleration due to gravity = [tex]9.78m/s^{2}[/tex]
Applying values in the equation we get
[tex]Gravitational Potential Energy=251X9.78X24 Joules[/tex]
[tex]Gravitational Potential Energy=58914.72 Joules[/tex]
[tex]Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ[/tex]
What are the mechanisms of energy transfer in an open system?
Answer:
mechanism of energy transfer in system is depend on Heat and Work:
Explanation:
Heat :Heat is described as the type of energy transmitted by a temperature difference between two structures (or a system and its environment).
Work:it is is an interaction of energy between a system and its environment. In the form of work it can cross the boundaries of a closed system. if energy crossing boundary of the system is not heat then it must be work..
Provide main reasons for the short shot during the injection molding.
Answer:
some cause of short shot is
1) due to the restriction in the flow
2) air pockets
3) high viscosity.
Explanation:
short shot is a word defined for major defect, it is actually occur when molten material does not fully occupy the cavities in a mold. Due to which mold remained incomplete after cooling. short shot may be because of restriction in the flow of molten material through the cavities and other main cause is present of large percentage of entrapped air.
Tool life testing on a lathe under dry cutting conditions gauge 'n' and 'C' of Taylor tool life equation as 0.12 and 130 m/min. respectively. When a coolant was used, 'C' increased by 10%. The increased tool life with the use of coolant at a cutting speed of 90 m/min is
Answer:
So % increment in tool life is equal to 4640 %.
Explanation:
Initially n=0.12 ,V=130 m/min
Finally C increased by 10% , V=90 m/min
Let's take the tool life initial condition is [tex]T_1[/tex] and when C is increased it become [tex]T_2[/tex].
As we know that tool life equation for tool
[tex]VT^n=C[/tex]
At initial condition [tex]130\times (T_1)^{0.12}=C[/tex]------(1)
At final condition [tex]90\times (T_2)^{0.12}=1.1C[/tex]-----(2)
From above equation
[tex]\dfrac{130\times (T_1)^{0.12}}{90\times (T_2)^{0.12}}=\dfrac{1}{1.1}[/tex]
[tex]T_2=47.4T_1[/tex]
So increment in tool life =[tex]\dfrac{T_2-T_1}{T_1}[/tex]
=[tex]\dfrac{47.4T_1-T_1}{T_1}[/tex]
So % increment in tool life is equal to 4640 %.
What is a rotary actuator and give an example of how it is used?
Answer Explanation:
ROTARY ACTUATOR: A rotary actuator is an actuator that produces a rotary motion. An actuator requires a control signal and a source of energy.the linear motion in one direction gives rise to rotation.
EXAMPLE OF ROTARY ACTUATOR: the most used rotary actuators are rack and pinion, vane and helical
HOW IT IS USED: an actuator requires a control signal and its energy sources are current, fluid pressure when it receives a control signal it responds by converting signal energy into mechanical motion
What is the difference between pump and turbine? Write the first law of thermodynamics for both (pump & turbine)?
Answer:
Pumps converts mechanical energy into hydraulic energy while turbines convert hydraulic energy into mechanical energy.
Explanation:
The machines which converts and transfers mechanical energy in the form of torque on the shaft into hydraulic energy in the form of water under pressure are called pumps whereas those machines which converts water pressure or hydraulic energy into mechanical energy that is further converted into electrical energy are called turbines.
The pump impeller rotates in the opposite direction to the turbine runner.
A turbine delivers work as output whereas a pump consumes work.
First law of thermodynamics for a pump :
W = ( H₁-H₂) +Q , where H₁ > H₂
First law of thermodynamics for a turbines :
W = ( H₂-H₁) +Q , where H₁ < H₂
Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high density b. good lubricity, high viscosity, low density c. chemically stable, compatible with system materials, good heat insulative capability d, readily available, high density, large bulk modulus e. fire resistant, inexpensive, non-toxic.
Answer:
e.Fire resistance,Inexpensive,Non-toxic.
Explanation:
Desirable hydraulic property of fluid as follows
1. Good chemical and environment stability
2. Low density
3. Ideal viscosity
4. Fire resistance
5. Better heat dissipation
6. Low flammability
7. Good lubrication capability
8. Low volatility
9. Foam resistance
10. Non-toxic
11. Inexpensive
12. Demulsibility
13. Incompressibility
So our option e is right.
Describe the process that you would use to hot forge an automotive connecting rod, indicating why each of the steps is used.
Answer:
Hot forging is a process which is carried at a temperature that is higher than the recrystalization temperature.
Explanation:
A connecting rod is used in a reciprocating engine which links the piston to the crankshaft. Connecting rods are made of steel which are hot forged.
The various steps that are used to hot forged a connecting rod are :
1. Rods are made to cut in the required size from the billet by billet shearing machine or saw band.
2. Heating of the billets in the furnace upto its recrystalization temperature.
3. Placing the billets in both upper and lower dies and doing the forging operation.
4. Rolling forging : it is important for the quality of the forged component.
5. Finishing and trimming : finishing is done to improve the surface quality and provide a smooth finish.
6. Inspection : Visual inspection is done for any defects.
Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow rate is 5.5 kg/s, and the power developed is 1120 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine
(a) The temperature of the air at the turbine exit, in K.
(b) The isentropic turbine efficiency.
Answer:
a) [tex]T_{2}=837.2K[/tex]
b) [tex]e=91.3[/tex] %
Explanation:
A) First, let's write the energy balance:
[tex]W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})[/tex] (The enthalpy of an ideal gas is just function of the temperature, not the pressure).
The Cp of air is: 1.004 [tex]\frac{kJ}{kgK}[/tex] And its specific R constant is 0.287 [tex]\frac{kJ}{kgK}[/tex].
The only unknown from the energy balance is [tex]T_{2}[/tex], so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.
[tex]T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K[/tex]
B) The isentropic efficiency (e) is defined as:
[tex]e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}[/tex]
Where [tex]{h_{2s}[/tex] is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because [tex]h_{2}-h_{1}[/tex] can be obtained from the energy balance [tex]\frac{W}{m}=h_{2}-h_{1}[/tex]
[tex]h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}[/tex]
An entropy change for an ideal gas with constant Cp is given by:
[tex]s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})[/tex]
You can review its deduction on van Wylen 6 Edition, section 8.10.
For the isentropic process the equation is:
[tex]0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})[/tex]
Applying logarithm properties:
[tex]ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}[/tex]
Then,
[tex]T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K[/tex]
So, now it is possible to calculate [tex]h_{2s}-h_{1}[/tex]:
[tex]h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}[/tex]
Finally, the efficiency can be calculated:
[tex]e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3[/tex] %
Different types of steels contain different elements that alter the characteristics of the steel. For each of the following elements, explain what the element does when alloyed with steel.
Answer:
The presence of element Carbon.
Explanation:
The diagrams Steel- Carbon usually show the percent the carbon vs the phases of the steel.
In the middle you increase the carbon percent the steels are not commercial because they are no malleables ( Hardennes).
By the other hand according the application of the steel you need to look the diagram Fe-Cr.
It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca criterion; b) Distortion energy or von Mises criterion; (c) Mohr-Coulomb criterion; (d) Any of the above
Answer:
(b)Distortion energy theory.
Explanation:
The best suitable theory for ductile material:
(1)Maximum shear stress theory (Guest and Tresca theory)
It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.
(2)Maximum distortion energy theory(Von Mises henkey's theory)
It states that maximum shear train energy per unit volume at any point is equal to strain energy per unit volume under the state of uni axial stress condition.
But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.
The universe is sometimes described as an isolated system. Why?
Answer and Explanation :The universe means it includes everything, even the things which we can not see is an isolated system because universe has no surroundings. an isolated system does not exchange energy or matter with its surroundings.Sometime universe is treated as isolated system because it obtains lots of energy from the sun but the exchange of matter or energy with outside is almost zero.
the total energy of an isolated system is always constant means total energy of universe is also constant there is no exchange of matter or energy in an isolated systemName 3 types of hydraulic cylinder mountings.
Answer:
Flanges MountingTrunnions MountingClevises MountingAnswer:
1.Flange mounting:
2.Foot mounting:
3.Mounting on end joint:
extra.4. Trunnion mounting
Crystal lattice can be characterized with a) angle, geometry and coordination number b) the color, size, lattice type; c) hardness, geometry d) atomic size, plasticity
Answer: a) angle, geometry and coordination number
Explanation: Crystal lattice is described as arrangement of groups of atoms inside three dimensional structure of crystal. It has a particular geometry in which atoms are placed in a symmetry. The also have angle in which placement of atoms are done. Co-ordination number also determines the crystal lattice by counting the atoms with which it is bonded. Thus, option (a) is the correct option.
Explain the following terms; i.Water content in air ii. Relative humidity iii. Enthalpy
Answer:
Explanation:
WATER CONTENT IN AIR-the water content of the air varies from place to place and from time to time because water content in air is dependent on temperature if temperature is change then water content also change water exist in air as a solid liquid and gas
RELATIVE HUMIDITY-Relative humidity is the ratio of partial pressure of water vapor to the equilibrium vapor pressure of water at a given temperature relative humidity depends on temperature and pressure of the system
Enthalpy-when a substance changes at constant pressure enthalpy tells how much heat and work was added or removed from the substance
enthalpy is equal to the sum of system internal energy and product of its pressure and volume.it is denoted H
_______On what basis composites are classified a)- shape of dispersed phase b)-matrix materials c)-chemistry of dispersed phase d)-a & b
Answer: d) a & b
Explanation: Composite materials are made up of two or more different types of phases which include dispersed phase and matrix phase as most important phases.
Matrix phase is a types of continuous phase which is responsible for holding of the dispersed phase.It shows good property of ductility.Dispersed phase is also known as the secondary phase which is harder in nature than matrix phase.A reciprocating engine of 750mm stroke runs at 240 rpm. If the length of the connecting rod is 1500mm find the piston speed and acceleration when the crank is 45 past the top dead center position.
Answer:
speed = 16.44 m/s
Acceleration = 71.36 m/s²
Explanation:
Given data
Speed ( N) = 240 rpm
angle = 45°
stoke length(L) = 750 mm
length of rod ( l ) = 1500 mm
To find out
the piston speed and acceleration
Solution
we find speed by this formula
speed = r ω (sin(θ) + (sin2(θ)/ 2n)) ...................1
here we have find r and ω
ω = 2[tex]\pi[/tex] N / 60
so ω = 2[tex]\pi[/tex] × 240 / 60
ω = 25.132 rad/s
n = l/r = 1500/750 = 2
we know L = 2r
so r = L/2 = 750/2 = 375 mm
put these value in equation 1
speed = 375 × 25.132 (sin(45) + (sin2(45)/ 2×2))
speed = 16444.811823 mm/s = 16.44 m/s
Acceleration = r ω² (cos(θ) + (cos2(θ)/ n)) ...................2
put the value r, ω and n in equation 2
Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))
Acceleration = 375 × (25.132)² (cos(45) + (cos2(45)/2))
Acceleration = 71361.363659 = 71.36 m/sec²
A centrifugal pump provides a flow rate of 0.03 m/s when operating at 1750 rpm against 60 m head. Determine the pump's flow rate and developed head if the pump speed is increased to 3500 rpm.
Answer:240m
[tex]Q=0.06m^3/s[/tex]
Explanation:
Given rpm increases from 1750 rpm to 3500 rpm
initial head 60 m and flow rate=[tex]0.03 m^{3}/s[/tex]
Since unit speed remains same
therefore
[tex]N_u=\frac{N}{\sqrt{H}}[/tex]
[tex]\frac{1750}{\sqrt{60}}[/tex]=[tex]\frac{3500}{\sqrt{H}}[/tex]
H=240m
Also unit Flow remains same
[tex]\frac{Q}{\sqrt{H}}[/tex]=[tex]\frac{Q}{\sqrt{H}}[/tex]
[tex]\frac{0.03}{\sqrt{60}}[/tex]=[tex]\frac{Q}{\sqrt{240}}[/tex]
[tex]Q=0.06m^3/s[/tex]
What are the three elementary parts of a vibrating system?
Answer:
the three part are mass, spring, damping
Explanation:
vibrating system consist of three elementary system namely
1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.
2) Spring - the part that has elasticity and help to hold mass
3) Damping - this part considered to have zero mass and zero elasticity.