Answer:
1500Ω
Explanation:
Given data
voltage = 15 V
total Resistance = 4000Ω
potential drop V = 9.375 V
To find out
R2
Solution
we know R1 +R2 = 4000Ω
So we use here Ohm's law to find out current I
current = voltage / total resistance
I = 15 / 4000 = 3.75 × [tex]10^{-3}[/tex] A
Now we apply Kirchhoffs Voltage Law for find out R2
R2 = ( 15 - V ) / current
R2 = ( 15 - 9.375 ) / 3.75 × [tex]10^{-3}[/tex]
R2 = 1500Ω
Work done by a system during a process can be considered as a property of the system. a)True b) False
Answer:
b) False
Explanation:
Work done by a system is not a property because it doesn't define the system's state. Work is mechanical energy exchanged across the system's boundaries.
Which of the following quenching materials is LEAST severe in its quenching action (slowest cooling)? a. Air b. Brine c. Oil d. Water
Answer:
(a) air
Explanation:
in the following given quenching materials air is the least serve quenching action quenching is a process of heating the material and then rapidly cooling it. Quenching freezes the structure of the material including stresses. Oil has the most sever in its quenching action the commonly used used oil for quenching process is peanut and canola oil
Define the work Envelope for a Robot
Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive machining tool.
Answer:
Explained
Explanation:
Cutting tools:
1. Cutting tools can either be single point or multi point.
2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.
3.Cutting tools have definite shapes and geometry.
Abrasive machining tools
1. Abrasive tools are always multi point tools.
2. Abrasive tools composed of abrasives bounded in medium of resin or metal.
3. They do not have definite geometry of shape
Cutting tools engage materials with a sharp edge for aggressive material removal, while abrasive machining tools employ hard particles to wear away material for high precision and smoothness in finishing operations.
Explanation:The differences between cutting tools and abrasive machining tools are fundamental in the processes they are used for and their operational principles. Cutting tools are typically used in operations like turning, milling, and drilling where the tool itself engages the material to be cut with a sharp edge, removing materials in the form of chips. These tools are often made of high-speed steel, carbide, ceramics, or other hard materials and are precisely shaped according to the specific cutting operation.
On the other hand, abrasive machining tools, which include grinding wheels, sandpaper, and abrasive belts, remove material through the action of hard, abrasive particles that are either bonded to the tool's surface or are used as loose grains. Abrasive machining is used for finishing surfaces to a high degree of smoothness, precision, and complex shapes that cutting tools cannot achieve. These tools are made of materials like aluminum oxide, silicon carbide, diamond, or cubic boron nitride.
To summarize, cutting tools use a sharp edge to remove material in a defined shape, while abrasive tools use hard particles to wear away material for finishing and shaping. Moreover, cutting tools are applied in more aggressive material removal processes and quicker operations like shaping or roughing out material, whereas abrasive machining is associated with finishing operations that require high precision and smoothness.
Define the isentropic efficiency for each of the following 3. a. i. Adiabatic turbine ii. Adiabatic compressor iii. Adiabatic nozzle
Answer:
a)[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
Explanation:
a)
Adiabatic turbine
Adiabatic turbine means turbine can not reject or take heat from surrounding.
Isentropic efficiency of turbine can be define as the ratio of actual work out put to the Ideal or isentropic work out put.Ideal means when turbine will give maximum work and there is no any friction losses we can say when process is isentropic.
Isentropic efficiency of turbine=(Actual work output)/(Ideal work output)
[tex]\eta_{st}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
b)
Adiabatic compressor
Isentropic efficiency of compressor can be define as the ratio of ideal or isentropic work in put to the actual work in put.
Isentropic efficiency of turbine=(Ideal work input)/(actual work input)
[tex]\eta_{sc}=\dfrac{\Delta h_{ideal}}{\Delta _{actual}}[/tex]
c)Adiabatic nozzle
We know that nozzle is device which used to accelerate the fluid.
Basically it covert pressure energy to kinetic energy.
Isentropic efficiency of nozzle can be define as the ratio of actual enthalpy drop put to the Ideal or isentropic enthalpy drop.
[tex]\eta_{sn}=\dfrac{\Delta h_{actual}}{\Delta _{ideal}}[/tex]
For all substances, Cp>C. Why is that?
How does the thermal efficiency of an ideal cycle, in general, compare to that of a Carnot cycle operating between the same temperature limits?
Answer:
Rankine cycle less efficient as compare to Carnot cycle operating betwwen same temperature limit.
Explanation:
We know that Carnot's cycle is an ideal cycle for all heat engine which operating between same temperature.It is a reversible cycle which have all process reversible that is why it have maximum efficiency.
On the other hand Rankine cycle is a practical working cycle so it is impossible to make all process reversible .In practical there will be always loss due to this any process can not make 100 % reversible.That is why Rankine cycle have low efficiency as compare to Carnot cycle operating between same temperature limits.
An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal efficiency of 70%. Comment on the validity of this claim.
Answer Explanation:
the efficiency of the the engine is given by=1-[tex]\frac{T_2}{T_1}[/tex]
where T₂= lower temperature
T₁= Higher temperature
we have given efficiency =70%
lower temperature T₂=27°C=273+27=300K
higher temperature T₁=627°C=273+627=900K
efficiency=1-[tex]\frac{T_2}{T_1}[/tex]
=1-[tex]\frac{300}{900}[/tex]
=1-0.3333
=0.6666
=66%
66% is less than 70% so so inventor claim is wrong
An example of Ferrous alloy is Brass a)-True b)-False
Answer: False
Explanation: No, brass is not a ferrous alloy.
Ferrous alloys are those alloy which contain iron like cast iron, steel, strain-less steel, high carbon steel. Brass on the other hand does not contain any composition. of iron hence it can not be considered as a ferrous alloy. Brass comes under the category of non- ferrous made with a composition of copper and zinc, however their proportion is not strict and we can add other elements like aluminium or lead to alter its durability or corrosiveness.
A piston-cylinder device contains 2.8 kg of water initially at 400 °C and 1.2 MPa. The water is allowed to cool at constant pressure until 28% ofits mass condenses into liquid. a) Evaluate the final temperature. b) Calculate the initial and final volumes (m3) c) Calculate the enthalpy at the initial and final states (kJ)
Answer:
a).Final temperature, [tex]T_{2}[/tex] = 180°C
b).Initial Volume, [tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
Final Volume, [tex]V_{2}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). Initial enthalpy,[tex]H_{1}[/tex] =9129.68 kJ
Final enthalpy, [tex]H_{1}[/tex] =6234.76 kJ
Explanation:
Given :
Total mass, m= 2.8 kg
Initial temperature, [tex]t_{i}[/tex] = 400°C
Initial pressure, [tex]p_{i}[/tex] = 1.2 MPa
Therefore from steam table at 400°C, we can find--
[tex]h_{1}[/tex] = 3260.6 kJ/kg
[tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
Now it is mentioned that 28% of the mass is condensed into liquid.
So, mass of liquid, [tex]m_{l}[/tex] = 0.28 of m
= 0.28 m
mass of vapour, [tex]m_{v}[/tex] = 0.72 m
∴ Dryness fraction, x = [tex]\frac{m_{v}}{m_{l}+m_{v}}[/tex]
= [tex]\frac{0.72 m}{0.28 m+0.72 m}[/tex]
= 0.72
a). The final temperature can be evaluated from the steam table at 1.2 MPa,
[tex]h_{2}[/tex] = 2226.7 kJ/kg
[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
Final temperature, [tex]T_{2}[/tex] = 180°C
b). We know [tex]v_{1}[/tex] = 0.25479 [tex]m^{3}[/tex] / kg
∴ Initial Volume, [tex]V_{1}[/tex] = [tex]v_{1}[/tex] x m
[tex]V_{1}[/tex] = 0.25479 x 2.8
[tex]V_{1}[/tex] = 0.713412 [tex]m^{3}[/tex]
We know,[tex]v_{2}[/tex] = 0.1179 [tex]m^{3}[/tex] / kg
∴ Final Volume, [tex]V_{2}[/tex] = [tex]v_{2}[/tex] x m
[tex]V_{2}[/tex] = 0.1179 x 2.8
[tex]V_{1}[/tex] = 0.33012 [tex]m^{3}[/tex]
c). We know,
[tex]h_{1}[/tex] = 3260.6 kJ/kg
∴ Initial enthalpy,[tex]H_{1}[/tex] = [tex]h_{1}[/tex] x m
= 3260.6 x 2.8
= 9129.68 kJ
[tex]h_{2}[/tex] = 2226.7 kJ/kg
∴ Final enthalpy, [tex]H_{1}[/tex] = [tex]h_{2}[/tex] x m
= 2226.7 x 2.8
= 6234.76 kJ
Assume that light of wavelength 6000A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Answer:
θ=0.0288 radian
Explanation:
resolution limit is the minimum angular separation of two sources that can be viewed distinctly by telescope
[tex]\theta =\frac{1.22\times \lambda}{D}[/tex]
[tex]\lambda=6000\times 10^{-8} cm=6 \times 10^{-5} cm[/tex]
[tex]d=100 inch=100\times 2.54=254cm[/tex]
[tex]\theta = \frac{1.22 \times 6 \times 10^{-5}}{254}[/tex]
θ=0.0288 radian
_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares
e)-consumables
Answer:
a)-Bin stock items free issue
Explanation:
Bin stock items free issue items are similar to the free issue items, but their access is limited.
Bin stock items free issue items are similar to the free issue items, but their access is limited.
An hydraulic cylinder has a piston diameter of 150mm and strokes at 300mm in 10 seconds. Calculate- A) The swept volume of the actuator in liters B) The pump displacement in Liters/minute
Answer:
(a)Volume in liters=5.3 liters.
(b)Volume in liters/minute=31.8 liters/minute.
Explanation:
Given:
Diameter of cylinder ,D=150 mm
Stroke,L=300 mm
Time ,t=10 sec
we know that swept volume of cylinder
[tex]V_{s}=\dfrac{\pi }{4}\times D^2\times L[/tex]
So [tex]V_{s}=\dfrac{\pi }{4}\times 0.15^2\times 0.3 m^{3}[/tex]
[tex]V_{s}=0.0053 m^3[/tex]
(a) Volume in liters =5.3 liters ( 1[tex]m^3[/tex]=1000 liters)
(b) When we divide swept volume by time(in minute) we will get liters/minute.
We know that 1 minute=60 sec
⇒10 sec=[tex]\frac{10}{60}[/tex] minute
So volume displace in liters/minute=31.8 liters/minute.
A mass of 2 kg is suspended from a vertical spring of stiffness 15 kN/m and subject to viscous damping of 5 Ns/m. What is the amplitude of the forced oscillations produced when a periodic force of amplitude 25 N and angular frequency 100 rad/s acts on the mass? What is the maximum force transmitted to the support of the spring?
Answer:
Amplitude of A is 4.975 mm and total force is 94.3 N
Explanation:
given data in question
mass (m) = 2 kg
stiffness (k) = 15 kN/m
viscous damping (c) = 5Ns/m
amplitude (F) = 25 N
angular frequency (ω) = 100 rad/s
to find out
amplitude of the forced and maximum force transmitted
Solution
static force for transmitted is mg i.e 2 × 9.81 = 19.6 N .............. 1
we know the amplitude formula i.e.
Amplitude of A = amplitude / [tex]\sqrt{c^{2}\omega^{2} + (k - m \omega^{2})^{2}[/tex]
now put the value c k m and ω and we find amplitude
Amplitude of A = 25 / [tex]\sqrt{5^{2} * 100^{2} + (15000 - 2 * 100^{2})^{2}[/tex]
Amplitude of A = 4.975 mm
now in next part we know the maximum force value when amplitude is equal displacement i.e.
maximum force = amplitude of A [tex]\sqrt{k^{2}+c^{2}\omega^{2}}[/tex]
now put all these value c , ω k and amplitude and we get
maximum force = 4.975 [tex]\sqrt{15000^{2}+5^{2} * 100^{2}}[/tex]
maximum force = 74.7 N .......................2
total force is combine equation 1 and 2 we get
total force 19.6 + 74.7 = 94.3 N
A particle moves along a straight line such that its acceleration is a=(4t^2-2) m/s, where t is in seconds. When t = 0, the particle is located 2 m to the left of the origin, and when t = 2, it is 20 m to the left of the origin. Determine the position of the particle when t=4s.
Answer with Explanations:
We are given:
a(t)=4*t^2-2............................(1)
where t= time in seconds, and a(t) = acceleration as a function of time.
and
x(0)=-2 .................................(2)
x(2) = -20 ............................(3)
where x(t) = distance travelled as a function of time.
Need to find x(4).
Solution:
From (1), we express x(t) by integrating, twice.
velocity = v(t) = integral of (1) with respect to t
v(t) = 4t^3/3 - 2t + k1 ................(4)
where k1 is a constant, to be determined.
Integrate (4) to find the displacement x(t) = integral of (4).
x(t) = integral of v(t) with respect to t
= (t^4)/3 - t^2 + (k1)t + k2 .............(5) where k2 is another constant to be determined.
from (2) and (3)
we set up a system of two equations, with k1 and k2 as unknowns.
x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 ......................(6)
substitute (6) in (3)
x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20
16/3 -4 + 2k1 -2 = -20
2k1 = -20-16/3 +4 +2 = -58/3
=>
k1 = -29/3 ....................................(7)
Thus substituting (6) and (7) in (5), we get
x(t) = (t^4)/3 - t^2 - 29t/3 + 2 ..............(8)
which, by putting t=4 in (8)
x(4) = (4^4)/3 - (4^2 - 29*4/3 +2
= 86/3, or
= 28 2/3, or
= 28.67 (to two places of decimal)
The answer is "28.87 m" and the further calculation can be defined as follows:
Given:
[tex]\to a=(4t^2-2)\ \frac{m}{s}\\\\ [/tex]
When
[tex] \to t=0 \ \ \ \ \ \ \ v= 2\ m\\\\ \to t=2 \ \ \ \ \ \ \ v= 20\ m\\\\ \to t=4\ \ \ \ \ \ \ v=?[/tex]
To find:
value=?
Solution:
[tex]V(t)=\int a(t)\ dt=\int (4t^2-2)\ dt=\frac{4}{3}t^3-2t+C_1\\\\ x(t)=\int V(t)\ dt =\int (\frac{4}{3}t^3-2t+C_1)\ dt=\frac{1}{3}t^4-t^2+C_1t+C_2\\\\ x(0)=-2=\frac{1}{3} 0^4-0^2+C_1(0)+C_2\to C_2=-2\ m\\\\ x(2)=-20=\frac{1}{3} 2^4-2^2+C_1(2)-2\to C_1=-\frac{29}{3}\ \frac{m}{s}\\\\ x(4) = \frac{1}{3}4^4-4^2 -\frac{29}{3}4-2 = 28.87 \ m \\\\[/tex]
The particle will be at 28.87 m at the right of the origin.
Learn more about velocity:
brainly.com/question/24376522
If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250 K, what is the amount of heat added to the system? a)-5J b)-125J c)- 600 J d)-5000 J e)-8000 J
Solution:
Given:
Change in entropy of the system, ΔS = 12J/K
Temperature of the system, [tex]T_{o}[/tex] = 250K
Now, we know that the change in entropy of a system is given by the formula:
ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex]
Amount of heat added, ΔQ = [tex]\Delta S\times T_{o}[/tex]
ΔQ = 3000J
What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d). 74%
Answer:
(b) 56%
Explanation:
the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature
here we have given T₁ (Higher temperature)= 600+273=873
lower temperature T₂=110+273=383
Efficiency of power cycle is given by =1-[tex]\frac{T2}{T1}[/tex]
=1-[tex]\frac{383}{873}[/tex]
=1-0.43871
=.56
=56%
Convert 10.25 degrees into radians; and π, π/2 and π/3 radians into degrees.
Answer:
0.1788 ,180°,90°,60°
Explanation:
CONVERSION FROM DEGREE TO RADIANS: For converting degree to radian we have to multiply with [tex]\frac{\pi}{180}[/tex]
using this concept 10.25°=10.25×[tex]\frac{\pi}{180}[/tex]=0.1788
CONVERSION FROM RADIAN TO DEGREE: For converting radian to degree we have to multiply with[tex]\frac{180}{\pi}[/tex]
using this concept π=π×[tex]\frac{180}{\pi}[/tex]
=180°
[tex]\frac{\pi}{2}[/tex]= [tex]\frac{\pi}{2}[/tex×[tex]\frac{180}{\pi}[/tex]
=90°
[tex]\frac{\pi}{3}[/tex]= [tex]\frac{\pi}{3}[/tex]×[tex]\frac{180}{\pi}[/tex]
=60°
Answer:
10.25° = 0.1790 radians
π radians = 180°
π/2 radians = 90°
π/3 radians = 60°
Explanation:
The conversion of degree into radians is shown below:
1° = π/180 radians
So,
10.25° = (π/180)*10.25 radians
Also, π = 22/7
So,
[tex]10.25^0=\frac{22\times10.25}{7\times180}radians[/tex]
Solving it we get,
10.25° = 0.1790 radians
The conversion of radians into degree is shown below:
1 radian = 180/π°
(a)
π radians = (180/π)*π°
Thus,
π radians = 180°
(b)
π/2 radians = (180/π)*(π/2)°
[tex]\frac {\pi }{2} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{2}^0[/tex]
π/2 radians = 90°
(c)
π/3 radians = (180/π)*(π/3)°
[tex]\frac {\pi }{3} radians=\frac{180}{\not {\pi }} \times \frac{\not {\pi }}{3}^0[/tex]
π/3 radians = 60°
What is the name of the instrument that measures or senses the vibration and is commonly referred to as a pickup or sensor.
Answer Explanation :
ACCELEROMETERS: Accelerometers are used for sensing the vibration. basically accelerometers are used for measuring high as well as low frequency. when a transducer is used in addition with another device to measure vibration is called pickups. Seismic instruments are commonly used vibration pickups
PROXMITY PROBE: Proximity probe is used for the measurements of vibration the vibration sensitivity is highest around 8 to 16 hertz
The larger the Bi number, the more accurate the lumped system analysis. a)-True b)- False
Answer:
b). False
Explanation:
Lumped body analysis :
Lumped body analysis states that some bodies during heat transfer process remains uniform at all times. The temperature of these bodies is a function of temperature only. Therefor the heat transfer analysis based on such idea is called lumped body analysis.
Biot number is a dimensionless number which governs the heat transfer rate for a lumped body. Biot number is defined as the ratio of the convection transfer at the surface of the body to the conduction inside the body. the temperature difference will be uniform only when the Biot number is nearly equal to zero.
The lumped body analysis assumes that there exists a uniform temperature distribution within the body. This means that the conduction heat resistance should be zero. Thus the lumped body analysis is exact when biot number is zero.
In general it is assume that for a lumped body analysis, Biot number [tex]\leq[/tex] 0.1
Therefore, the smaller the Biot number, the more exact is the lumped system analysis.
Typical metals show elastic-plastic behaviour in tension and shear but not in compression. a)True b)- False
Answer: True
Explanation:Typical metals have a property of ductility and malleability that is metals can be drawn into wires or any other shape by beating or stretching the metal by putting the tensile strength or shear strength that pulls them apart . But while compression the metals are squeezed together which affects the hardness of a metal and they are not able to bear the compression force well and thus cannot show elastic-plastic behavior while compression .Therefore the statement given is true typical metals show elastic-plastic behavior in tension and shear but not in compression.
Why the velocity potential Φ(x,y,z,t) exists only for irrotational flow
Answer:
[tex]\omega_y,\omega_x,\omega_Z[/tex] all are zero.
Explanation:
We know that if flow is possible then it will satisfy the below equation
[tex]\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0[/tex]
Where u is the velocity of flow in the x-direction ,v is the velocity of flow in the y-direction and w is the velocity of flow in z-direction.
And velocity potential function [tex]\phi[/tex] given as follows
[tex]u=-\frac{\partial \phi }{\partial x},v=-\frac{\partial \phi }{\partial y},w=-\frac{\partial \phi }{\partial z}[/tex]
Rotationality of fluid is given by [tex]\omega[/tex]
[tex]\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega_Z[/tex]
[tex]\frac{\partial v}{\partial z}-\frac{\partial w}{\partial y}=\omega_x[/tex]
[tex]\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=\omega_y[/tex]
So now putting value in the above equations ,we will find
[tex]\omega =\frac{\partial \phi }{\partial x},u=\frac{\partial \phi }{\partial x},[/tex]
[tex]\omega_y=\dfrac{\partial^2 \phi }{\partial z\partial x}-\dfrac{\partial^2 \phi }{\partial z\partial x}[/tex]
So [tex]\omega_y[/tex]=0
Like this all [tex]\omega_y,\omega_x,\omega_Z[/tex] all are zero.
That is why velocity potential flow is irroational flow.
What are the two types of pumps in compressors?
Answer: The two different types of pumps in compressors are:
1) Centrifugal Pumps
2) Reciprocating Pumps
Explanation:
Compressor is defined as a mechanical device which increases the gas pressure by reducing its volume.The main action of a pump is to transport liquids and pressurize and it coverts rotational energy. As, centrifugal pumps are the most common pumps used for transfer of fluids and it works on simple mechanism.
Reciprocating Pumps is used for low volumes of flow at a high pressure. It is the positive displacement pump. As, it works on the principle of pushing of liquid which executes a reciprocating motion in a closed cylinder.
Answer:
Centrifugal Pumps and Positive Displacement Pumps
Explanation:
The two of pumps listed above is based on the mode of operation. The centrifugal pumps works by increasing the velocity of the liquid through the machine while the displacement pumps works by alternating, filling a cavity and then displace some amount of liquid.
An inventor claims to have devised a cyclical power engine that operates with a fuel whose temperature is 750 °C and radiates waste heat to a sink at 0 °C. He also claims that this engine produces 3.3 kW while rejecting heat at a rate of 4.4 kW. Is this claim valid?
Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=[tex]\frac{Work produced}{Heat supplied}[/tex]
and
Heat Supplied [tex]{\left (Q_s\right)}=Work Produced(W)+Heat rejected\left ( Q_r \right )[/tex]
[tex]{Q_s}=3.3+4.4=7.7KW[/tex]
η=[tex]\frac{3.3}{7.7}[/tex]
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-[tex]\frac{T_ {sink}}{T_ {source}}[/tex]
η=1-[tex]\frac{273}{1023}[/tex]
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.
Quantity of erystallization centres in crystallization may increase a) insert imitation of crystallization b) use mechanical mixing: c) change cooling rate; d) all of the above are correct.
Answer: d) All of the above
Explanation: Crystallization is the process in which a solid crystalline structured material is obtained from the liquid substance. the quantity of crystallization center in crystal may increase due to several reasons like changing the cooling rate or mechanical mixing of the substances or imitation of crystals etc.
These processes end up adding mineral atoms to get attached to the center of crystal and hence increasing the size. Thus, the correct option is option (d).
A system that is not influence anyway by the surroundings is called a)- control mass system b)- Isothermal system c)-- isolated system d)- open system
Answer:
Isolated system
Explanation:
By definition of a closed system it means that a system that does not interact with it's surroundings in any manner
The other options are explained as under:
Isothermal system : It is a system that does not allow it's temperature to change
Control Mass system : It is a system whose mass remains conserved which means the mass entering the system equals the mass leaving the system
Open system: It is a system that allows transfer of mass and energy across it's boundary without any opposition i.e freely.
For a 4-bar linkage with ri =7-in, r2 =3-in, r3= 9-in, and r =8-in, determi the minimum and maximum transmission angles.
Answer:
[tex]\mu_{min}[/tex]=[tex]26.38^{\circ}[/tex]
[tex]\mu_{max}[/tex]=[tex]71.79^{\circ}[/tex]
Explanation:
[tex]r_{1}[/tex]=7 in, [tex]r_{2}[/tex]=3 in, [tex]r_{3}[/tex]=9in
,[tex]r_{4}[/tex]=8 in
Transmission angle (μ ):
It is the acute angle between coupler and the output (follower) link.
Here we consider link [tex]r_{1}[/tex] as fixed link ,[tex]r_{2}[/tex] as input link ,link [tex]r_{3}[/tex] as coupler and link [tex]r_{4}[/tex] as output link.
As we know that
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
When link [tex]r_{2}[/tex] will be horizontal in left side direction then transmission angle will be minimum and when link [tex]r_{2}[/tex] will be horizontal in right side direction then transmission angle will be maximum.
Now by putting the values we will find
[tex]\cos\mu_{max}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}-\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{max}=0.3125[/tex]
[tex]\mu_{max}=71.79^\circ[/tex]
[tex]\cos\mu_{min}=\frac{r_{4}^2+r_{3}^2-r_{1}^2-r_{2}^2}{2r_{3}r_{4}}+\frac{r_{1}r_{2}}{{r_{3}r_{4}}}[/tex]
[tex]\cos\mu_{min}=0.8958[/tex]
[tex]\mu_{min}=26.38^\circ[/tex]
Hence, The minimum and maximum angle of transmission angle is 26.38° and 71.79° respectively.
Citations must be contested within_____working days of the notice of proposed penalty. a)-15 b)-10 c)-30 d)-7
Answer:
30
Explanation:
Legally that's when you have to respond
Explain why different types of equipment are required for proper conditioning of air
Answer:
Different types of equipment are required for proper conditioning of air because every air conditional space faces some geometrical and environmental issues or problems. There are some different types of equipment used for conditioning of air that are air system, water system and air-water system. In many cases the air conditioning of the system varies with size of the equipment.
A rigid tank initially contains 2 kg of steam at 500 kPa and 350°C. The steam is then cooled until it is at 100°C. Determine the final pressure and the heat transferred during this process.
Answer:
299.36 kPa
Explanation:
given mass of steam =2 kg
initial pressure that is [tex]P_1=500kPa[/tex]
initial temperature that is [tex]T_1=350^{\circ} C=350+273=623 K[/tex]
final temperature that is [tex]T_2=100^{\circ} C=100+273=373 K[/tex]
it is a rigid tank so volume is constant
for constant volume process [tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
[tex]P_2=\frac{T_2}{T_1}\times P_1=\frac{373}{623}\times 500=299.36 kPa[/tex]
so the final pressure will be 299.36 kPa