Answer:
the flow rate of the oil is 2.5 m³/s
Explanation:
Given data
relative density (S) = 0.8
diameter (d1) = 60 mm = 0.06 m
diameter (d2) = 35 mm = 0.035 m
height (h) = 22 mm = 0.022 m
discharge coefficient (Cd) = 0.98
To find out
the flow rate of the oil
solution
we know the formula for rate of flow i.e.
flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] / [tex]\sqrt{a1^{2} a2^{2} }[/tex] ...............1
here first we find area a1 and a2 i.e.
a1 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.06² = 0.002827 m²
a2 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.035² = 0.000962 m²
and now we find n = (density of mercury / density of oil) - 1 × h
n = ((13.56 / 0.8) - 1) × 0.022 = 0.3509
put all these value in equation 1
flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] / [tex]\sqrt{a1^{2} a2^{2} }[/tex]
flow rate = 0.98× 0.002827× 0.000962 [tex]\sqrt{2*9.81*0.3509}[/tex] / [tex]\sqrt{0.002827^{2} 0.000962^{2} }[/tex]
flow rate = 2.571386 m³/s
If the compound swivel base is set on 60 degrees at the lathe centerline index, how many degrees will the reading be at the cross slide index? A. 45 B. 150 C. 30 D. 90
Answer:
C.30°
Explanation:
Given that compound swivel base is set on 60° at the lathe center line index.
We need to find reading on cross slide index
We know that relationship between center line index and cross slide index in angle 2∝=β
Where ∝ Angle of swivel and β is the reading on cross line index.
So by using above relationship between center line index and cross slide index
2∝=60°⇒∝=30°
So our option is C.
Intrinsic semiconduction is a property of a pure material. (True , False )
True.
An intrinsic semiconductor is a pure semiconductor. At room temperature it behaves as an insulator because it only has a few free and hollow electrons due to thermal energy.
In an intrinsic semiconductor there are also electron fluxes and gaps, although the total current resulting is zero. This is because the action of thermal energy produces free electrons and gaps in pairs, so there are as many free electrons as there are gaps with which the total current is zero.
A roller support acts like a contact boundary condition as it can produce a reaction force as a push response to a body but will not produce a pull force to hold a body from moving away. a)True b)- False
Answer:
a) True
Explanation:
Roller can provide reaction for push support but it can not provide reaction for pull support.
From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.
On the other hand hinge support can provide reaction in both the direction.
So we can say that roller can not proved reaction for pull support.
Give two methods on how powder is produced in powder metallurgy.
Answer:
Explanation:
Atomization using gas streamMolten metal is forced through a small orifice and is shatter by a jet of compressed air,inert gas .
In Atomization, the particles shape is analysed by the rate of solidification and varies from spherical to highly irregular shape.
Reductionoxide of metals are transformed to pure metal powder when undefended to under melting point gases results in a product of spongy material.
It is used for Iron,copper,tungsten,Nickel etc.
Answer:
(1)Atomizing process
(2)Gaseous reduction
Explanation:
The first step in powder metallurgy is the production of powder,because the property of the final product depends on the powder.
The methods for the production of powder are as follow
(1)Atomizing process
(2)Gaseous reduction
(1)Atomizing process:
In the Atomizing process the molten metal is passing through an orifice into a stream of inert gas.Due to this rapid cooling of metal occurs and then it will in very fine particle .
(2)Gaseous reduction:
In this process powder is producing by grinding of metallic oxide to a fine state,after that reducing it by carbon mono oxide.
A gas contained within a piston-cylinder undergoes the follow change in states: Process 1: Constant volume from p1 = 1 bar V1 = 2.6 m3 to state 2 with p2 = 2.7 bar Process 2: Compression to V3 = 1.5 m3, which the pressure-volume relationship is pV = constant. Process 3: Constant pressure to state 4, where V4 = 0.5 m3. Sketch the processes on p-V graph and evaluate the work for each process in kJ.
Answer:
Process 1:W=0
Process 2:W= -386.13 KJ
Process 3:W= -468 KJ
Explanation:
Process 1:[tex]P_1=1 bar,V_1=2.6m^3[/tex]
Process 2:[tex]P_2=2.7bar,V_2=2.6m^3[/tex]
Process 3:[tex]V_3=1.5 m^3[/tex]
[tex]V_4=0.5 m^3[/tex]
Process 1:
Work (W)=0 ,because it is constant volume process.
Process 2:
It is constant temperature process so PV=C
[tex]P_2V_2=P_3V_3[/tex]
[tex]P_3=\dfrac{P_2V_2}{V_3}[/tex]
[tex]P_3=\dfrac{2.7\times 2.6}{1.5}[/tex]
[tex]P_3=4.68 [/tex]bar
So work in constant temperature process
W=[tex]P_2V_2\ ln\dfrac{V_3}{V_2}[/tex]
W=[tex]270\times 2.6\ ln\dfrac{1.5}{2.6}[/tex] (1 bar=100KPa)
W= -386.13 KJ
Negative sign means it is compression process.
Process 3:
It is a constant pressure.
So work W=[tex]P_3(V_4-V_3)[/tex]
W=468(0.5-1.5) KJ
W= -468 KJ
Negative sign means it is compression process.
A properly installed window quilt can provide an additional insulation of R-4 to windows. For a window with a U-factor of 0.33, determine the percentage of heat lost for a quilt-covered win- dow compared to a bare window.
Answer:
Heat loss 67%.
Explanation:
We know that heat transfer
Q=AUΔT
Where U is the overall heat transfer coefficient ,A is the area and ΔT is the temperature difference.
Now heat transfer in terms of U-factor
[tex] Q_1=AU_1ΔT[/tex]
[tex] Q_2=AU_2ΔT[/tex]
Given that temperature difference is same in both condition so
[tex]\dfrac{Q_1}{U_1}=\dfrac{Q_2}{U_2}[/tex]
[tex]\dfrac{Q_1}{Q_2}=\dfrac{U_1}{U_2}[/tex]
[tex] heat\ loss=\dfrac{Q_2-Q_1}{Q_1}[/tex]
[tex] heat\ loss=\dfrac{U_2-U_1}{U_1}[/tex]
Given that[tex]U_2=0.33U_1[/tex]
[tex] heat\ loss=\dfrac{0.33U_1-U_1}{U_1}[/tex]
Heat loss 67%.
What is the theoretical density in g/cm3 for Lead [Pb]?
Answer:
11.34 g/cm3
Explanation:
At room temperature, where it is in a solid state, it is 11.34 [tex]\frac{g}{cm^{3}}[/tex]. While at melting temperature, at 327.5 ° C, it is 10.66 [tex]\frac{g}{cm^{3}}[/tex]
Heat conduction is a function of material property, temperature difference and fluid the geometry of the object. a) True b) False
Answer:
a)True
Explanation:
According to Fourier's law,It states that Heat transfer is directly proportionate to the temperature gradient.For unidirectional conduction
[tex]q''\alpha\dfrac{dT}{dx}[/tex]
Where q'' is the heat flux.
When This directly proportionate is remove then one constant will on that place and that constant is called conductivity of material.
[tex]q''=K\dfrac{dT}{dx}[/tex]
So total heat transfer Q
[tex]Q=KA\dfrac{dT}{dx}[/tex]
When conductivity (K) of material is not constant
K=[tex]K_0[/tex](a+bT) where a,b are constant.
So from the above expression we can say that heat transfer in material will depend on temperature difference,material property and on on geometry of object.
A lake contains water at a maximum depth of 237.3 meters. This water has a density of 989 kg/m3 and the acceleration due to gravity at this location is 9.806 m/s2. The air above the water has a barometric pressure of 29.83 in of Hg where the altitude correct for pressure is -1.87 in of Hg to get absolute pressure. What is the absolute pressure at the bottom of the lake?
Answer:690.21 mm of Hg
Explanation:
Given data
depth of lake[tex]\left (h\right )[/tex]=237.3m
density of lake water[tex]\left ( \rho\right )[/tex]=[tex]989kg/m^3[/tex]
acceleration due to gravity[tex]\left ( g\right )[/tex]=[tex]9.806m/s^2[/tex]
air above water has a barometric pressure of 29.83in of Hg=711.2 mm of Hg
Altitude correct for pressure=1.87in of Hg=47.5 mm of Hg
Absolute pressure above water =711.2-47.5mm of Hg=663.7 mm of Hg
Absolute pressure at bottom of Lake=Pressure at surface level +[tex]\left ( \rho \times g\times h\right )[/tex]
Absolute pressure at bottom of Lake=663.7+[tex]\frac{989\times 9.81\273.3}{10^{5}}[/tex]
Absolute pressure at bottom of Lake=690.21 mm of Hg
From your cooling load (8890.007 Btu/hr = 2.605kW, determine mass flow rate of refrigerants. Use the following "rule of thumb" estimate of a standard room of size 180 ft^2: 1 ton of refrigerant = 12,000Btu = 3.517 kW Rule of thumb: 1 ton cooling load = 300-400 ft^2
Answer:
0.740833917 ton/hr
Explanation:
Given:
Cooling load, 8890.007 Btu/hr = 2.605 kW
Room size = 180 [tex]ft^{2}[/tex]
According to the thumb rule
1 ton of refrigerant = 12000Btu
Hence for 8890.007 Btu/hr,
the mass flow rate of the refrigerant is =8890.007 / 12000
= 0.740833917 ton per hr
Hence, mass flow rate is 0.740833917 ton/hr
A circular plate with diameter of 20 cm is placed over a fixed bottom plate with a 1 mm gap between two plates filled with Kerosene at 40 degree C, shown in the following figure. Find the torque needed to rotate the top plate at 5 rad/s. The velocity distribution in the gap is linear and the shear stress on the edge of the rotating plate can be neglected.
Answer:
T = 1.17 x [tex]10^{-3}[/tex] N-m
Explanation:
Given :
Gap between the two plates , dy = 1 mm
dy = 1 x [tex]10^{-3}[/tex] m
Angular velocity of the top plate , ω = 5 rad/s
Diameter of the plate, D = 20 cm
Radius of the plate, R = 10 cm
= 0.1 m
Temperature of the kerosene = 40°C
Viscosity of kerosene at 40°C = 0.0015 Pa-s
Now let us take a small elemental ring of thickness dr at a radius r.
Therefore, area of this elemental ring of dr = 2πrdr
Now linear velocity at radius r = ω x r
5r m/s
Now applying Newtons law of viscosity we get,
Shear stress, τ = μ.[tex]\frac{du}{dy}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{du}{dy}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu .\frac{5r-0}{1\times 10^{-3}}[/tex]
[tex]\Rightarrow \frac{F_{s}}{A}=\mu \times 10^{3}\times 5r[/tex]
[tex]F_{s}=\mu \times 10^{3}\times 5r\times 2\pi rdr[/tex]
[tex]F_{s}=5 \times 10^{3}\times \mu \times r\times 2\pi rdr[/tex]
[tex]F_{s}=\frac{18849}{400}\times r^{2}dr[/tex]
Now we know torque due to small strip,
dT = [tex]F_{s}[/tex] x r
dT = [tex]\frac{18849}{400}\times r^{3}dr[/tex]
Therefore total torque for r=0 to r=R can be calculated. So by integrating,
[tex]\int dT=\int_{0}^{R}\frac{18849}{400}\times r^{3}dr[/tex]
[tex]T = \frac{18849}{400}\times \frac{R^{4}}{4}[/tex]
[tex]T = 47.1225\times \frac{0.1^{4}}{4}[/tex]
T = 1.17 x [tex]10^{-3}[/tex] N-m
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature of 1623°C. The initial volume is 0.05 m^3 and the gas expands through a volume ratio of 20 according to the law PV^1.25 = constant. Calculate: a) Work transfer b) Heat transfer
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
[tex]P_{1}[/tex] = 5 Mpa
[tex]T_{1}[/tex] = 1623°C
= 1896 K
[tex]V_{1}[/tex] = 0.05 [tex]m^{3}[/tex]
Also given [tex]\frac{V_{2}}{V_{1}} = 20[/tex]
Therefore, [tex]V_{2}[/tex] = 1 [tex]m^{3}[/tex]
R = 0.27 kJ / kg-K
[tex]C_{V}[/tex] = 0.8 kJ / kg-K
Also given : [tex]P_{1}V_{1}^{1.25}=C[/tex]
Therefore, [tex]P_{1}V_{1}^{1.25}[/tex] = [tex]P_{2}V_{2}^{1.25}[/tex]
[tex]5\times 0.05^{1.25}=P_{2}\times 1^{1.25}[/tex]
[tex]P_{2}[/tex] = 0.1182 MPa
a). Work transfer, δW = [tex]\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
[tex]\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}[/tex]
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= [tex]\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
=[tex]\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W[/tex]
=[tex]\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200[/tex]
= 197.7 kJ
An electric heater draws 12 A from a 120 V power source. How much powe heater dissipate? (a) 10 W (b) 1200 W (c) 1440 W (d) 17,280 W
Answer:
c) 1440 W
Explanation:
P = V*i = (12A)*(120V) = 1440W
According to the scenario, the amount of power that is dissipated by the heater is 1440 W. Thus, the correct option for this question is C.
What is Power?Power may be defined as the amount of energy that is significantly transferred or converted per unit of time. The unit of power is the watt (W). It may be typically calculated with the help of the given formula:
P = W/Δt.According to the context of this question, the amount of power that is dissipated within the heater is calculated by the following formula:
P = V × i = 12 × 120 = 1440 W.Therefore, according to the scenario, the amount of power that is dissipated by the heater is 1440 W. Thus, the correct option for this question is C.
To learn more about Power, refer to the link:
https://brainly.com/question/1634438
#SPJ5
The pressure at any point in a static fluid depends only on the Select one a)- depth, surface pressure, and specific weight. b)- specific weight. c)- surface pressure and depth. d)- depth and container shape
Answer:
c). surface pressure and depth
Explanation:
We know that fluid pressure is measured in two different ways namely --
1. Pressure measured above complete vacuum or absolute zero is called Absolute Pressure.
2.Pressure measured above atmospheric pressure is called Gauge Pressure.
In the figure below, we can find the pressure at the point A in the static fluid inside the tank which is at a depth of h from the water surface.
Let the atmospheric pressure which is acting on the water surface be [tex]P_{atm}[/tex].
Let ρ be the density of water and g be the acceleration due to gravity.
Therefore we know that pressure at a point in a fluid is
P = ρ[tex]\times[/tex]g[tex]\times[/tex]h
Therefore total pressure acting on the point A in a fluid is
[tex]P_{tot}[/tex] = [tex]P_{atm}[/tex] + P
[tex]P_{tot}[/tex] = [tex]P_{atm}[/tex] + ρ[tex]\times[/tex]g[tex]\times[/tex]h
Thus, pressure at a point A in a static fluid depends on the surface pressure and the depth of the point from the free surface.
In the SI system of units, the absolute temperature is measured to be 303 K Its value in Fahrenheit is a-) 76 F b)- 86F c)-79 F d)- 82 F
Answer:
The correct answer is option B i.e. 86 f
Explanation:
The freezing point of water is 32°f and boiling point is 212 °f. Thus difference in boiling and freezing point is exactly come out to be 180°. therefore fahrenheit scale is 1/180 interval of both scale ( freezing and boiling point)
given data:
absolute temperature is 303 K
We know by general formula of temperature in fahrenheit
°F = [tex]\frac{9}{5}[/tex] × (K-273) +32
°F = [tex]\frac{9}{5}[/tex] × (303-273) +32
°F = 86 F
Why is it important (in the context of systems engineering management) to become familiar with analytical methods? Provides some specific examples.
Understanding analytical methods is essential in systems engineering for problem-solving and system comprehension, with real-world applications in areas like environmental monitoring.
Explanation:The importance of analytical methods in systems engineering management is crucial for problem-solving and comprehending the physics of the situation. Analytical methods help in selecting the right system and developing solutions efficiently. For instance, in the field of environmental monitoring programs, the use of analytical methods aids in assessing system health and trends accurately.
Water, in a 150 in^3 rigid tank, initially has a temperature of 70°F and an enthalpy of 723.5 Btu/lbm. Heat is added until the water becomes a saturated vapor. Calculate: a) The initial quality of the water b) The total mass of the water c) The temperature of the water at the final state.
Answer:
a)initial quality of water 0.724.
b)mass of water =2.45 kg
c)T=70°F
Explanation:
h=732.5 Btu/lbm
h=763.33 KJ/kg (1 Btu=1.05 kj)
T=70°F⇒T=21.1°C
[tex]V=150 in^3=0.002458m^3[/tex]
(a)From steam table
Properties of saturated steam at 21.1°C
[tex]h_f= 88.56\frac{KJ}{Kg} ,h_g= 2539.49\frac{KJ}{Kg}[/tex]
To find dryness fraction
[tex]h=h_f+x(h_g-h_f)\frac{KJ}{Kg}[/tex]
[tex]763.33=88.56+x(2539.49-88.56)\frac{KJ}{Kg}[/tex]
x=0.27
So initial quality of water 0.724.
(b)
[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]
where v is specific volume
From steam table at 21.1°C
[tex]v_f= 0.001\frac{m^3}{Kg} ,v_g= 54.13\frac{m^3}{Kg}[/tex]
V=[tex]m_f\times v_f[/tex]
0.002458=[tex]m_f\times 0.001[/tex]
So mass of water =2.45 kg
(c)
Actually water will take latent heat ,it means that heating of will take place at constant temperature and constant pressure.So we can say that final temperature of water will remain same (T=70°F).
What is a cascade refrigeration system?
Answer Explanation:
CASCADE REFRIGERATION SYSTEMS :
cascade refrigeration systems are commonly used in the liquefaction of natural gas and some other gases
cascade refrigeration system is also known as freezing system. It uses two types of refrigerants having different boiling points.this system is employed to obtain temperature of -40 to -80It allows stable ultra low temperature operationexamples of cascade refrigeration system: LNG (Liquefied natural gas ) plants mostly used cascade refrigeration system
ADVANTAGES OF CASCADE REFRIGERATION SYSTEMS:
energy is saved because the system allows the use of refrigerant gasesthe running cost is inexpensive repair is easyA cascade refrigeration system utilizes multiple linked refrigeration cycles to achieve ultra-low temperatures not possible with a single cycle. It features a cascade control mechanism for precise temperature management, using heat exchangers to interconnect and sequentially cool each stage.
Explanation:Understanding Cascade Refrigeration SystemsA cascade refrigeration system is a complex refrigeration strategy that employs multiple refrigeration cycles linked together to achieve cooling. Typically, these systems are utilized in environments requiring very low temperatures, which cannot be efficiently or cost-effectively achieved through a single refrigeration cycle. The essence of a cascade system lies in its deployment of two or more refrigeration units in series, each referred to as a "stage," with each successive stage operating at a lower temperature range than its predecessor.
The interconnection between these stages involves a heat exchanger, where a refrigerant from one cycle cools the condenser of the next lower temperature cycle. This process is crucial for achieving the desired low temperatures. One common application is in industrial refrigeration, where maintaining precise low temperatures is critical for process efficiency and product quality.
A key feature of cascade refrigeration systems is the use of cascade control, which enhances system responsiveness to temperature changes and maintains tight control over the process. This control mechanism consists of a master-slave relationship, where the output of one control loop (master) serves as the setpoint for the next (slave), creating a highly integrated control environment. Such setup enables rapid adjustment to changing operating conditions, ensuring the system swiftly responds to disturbances while maintaining the desired temperature levels.
What are units for heating capacity? Mark all that apply: a)- Tons b)- kJ/kg c)- kW d)- Btu
Answer:
(b) kJ/kg
Explanation:
The ratio of amount of energy required to change the temperature of the substance by certain magnitude and this magnitude of temperature change is known as heat capacity of the substance.
The expression for Heat capacity is:
C=E/ΔT
Where,
C is the Heat capacity
E is the energy absorbed/released
ΔT is the change in temperature
The SI unit of heat capacity is J/K.
(a) Tons represents the unit of mass (1000 kg)
(c) kW represents the unit of power (1000 W)
(d) Btu represents the unit of heat (1055 J)
The units from the options that can be a unit of heat capacity is (b) kJ/kg.
An aircraft increases its speed by 2% in straight and level flight. If the total lift remains constant determine the revised CL as a percentage of its original value to three significant figure
Answer:
96.1%
Explanation:
We know that lift force
[tex]F_L=\dfrac{1}{2}C_L\rho AV^2[/tex]
------------(1)
Where [tex]C_L[/tex] is the lift force coefficient .
ρ is the density of fluid.
A is the area.
V is the velocity.
Now when speed is increased by 2 % and all other parameter remains constant except [tex]C_L[/tex] .
Let;s take new value of lift force coefficient is [tex]C_L'[/tex] .
[tex]F_L=\dfrac{1}{2}C_L'\rho A(1.02V)^2[/tex]
-----------(2)
Now from equation 1 and 2
[tex]C_L\times V^2=C_L'\times1.0404 V^2[/tex]
⇒[tex]C_L'=0.961C_L[/tex]
So we can say that revised value of lift force coefficient is 96.1% of original value.
List the main activities of exploration??
Answer: Exploration includes plethora of activities and depend upon the kind of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.
Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.
Which of the following describes the action of a capacitor? (a) creates a de resistance (b) converts ac into de (c) opposes changes in the flow of current (d) stores electrical energy
Answer:
From the multiple choices provided for the action of capacitor, option
(d) stores electrical energy
is correct
Explanation:
A capacitor is basically a two terminal device that stores electrical energy in the electric field in its vicinity. It is apassive element.
The property of a capacitor to store electrical energy or the effect of a capacitor is known as its capacitance. The capacitance of a capacitor is given by:
Q = CV or C = [tex]\frac{Q}{V}[/tex]
It is originally known as condensor and it can be said that a capacitor adds capacitance to the circuit.
The resistivity of mercury drops suddenly to zero at a critical temperature, maki mercury a superconductor below that temperature. ( True , False )
Resistance zero meaning superconductor, so True.
Product service life is determined by a. estimates b. market forces c. liability d. property tests e. failure analysis f. all of the above
Answer: d) property tests
Explanation: Product service life can be referred as the life that define the service that can be provided by the product manufactured.The service life contains the testing and calculation of the product's quality, reliability, maintenance factor etc. These factors are known as the property of the product and so is calculated by the property test. Therefore option (d) is the correct option because other option does not define the factors for defining the product service life.
The response of an inductor to current is most directly similar to the response of the capcitor (a) current (b) voltage (c) resistance (d) inductor
Answer:
(b) vpltage
Explanation:
we know the expression for voltage across the inductor V=L[tex]\frac{di}{dt}[/tex] which clearly shows voltage across the inductor is directly proportional to rate of change of current similarly current across the capacitor I=C[tex]\frac{dv}{dt}[/tex] from the expression we can see that current across the capacitor is directly proportional to rate of change of voltage. so from above discussion it is clear that response of an iductor to current is similar to response of capacitor to voltage
Thin film deposition is a process where: a)-elemental, alloy, or compound thin films are deposited onto a bulk substrate! b)-Phosphatization is also applied c)-Zinc plating is also applied d)-None of the above
Answer:
(A) elemental, alloy, or compound thin films are deposited on to a bulk substrate
Explanation:
In film deposition there is process of depositing of material in form of thin films whose size varies between the nano meters to micrometers onto a surface. The material can be a single element a alloy or a compound.
This technology is very useful in semiconductor industries, in solar panels in CD drives etc
so from above discussion it is clear that option (a) will be the correct answer
Microchips found inside most electronic devices today are examples of what material A. Polymers B. Alloys C. Composites D. None of the above. E. Metals
Answer: A
Explanation:
Microchips are made out of silicone witch is a polymer.
Answer:
its a
Explanation:
The specific heat of aluminum is approximately 900 J/kg°C. If the temperature of a 5 kg specimen needs to be raised from 30°C to 1000°C, the amount of heat required will be equal to Select one: a)- 4365 kJ b)- 4365 Btu c)- 4365W d)- None of the above
Answer:
a) 4365 kJ
Explanation:
In any thermodynamic system, any heat change is accompanied by the change in temperature. The relation between heat released/gained in a system and the temperature is:
Q=mcΔT
where,
Q is the amount of heat absorbed or released
m is the mass
ΔT is the change in temperature
c is called the specific heat.
Specific heat is defined as heat gained by 1 unit mass of any sample to raise the temperature of the sample by 1 °Celsius.
Thus, from the question:
Mass of aluminum =5 kg
Final temperature = 1000°C
Initial temperature = 30°C
ΔT = (1000 -30)°C = 970°C
Specific heat of aluminum = 900 J/kg°C
Thus, Amount of heat required:
Q = 5 kg×900 J/kg°C×970°C = 4365000 J
The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus, Heat gained by aluminum =4365000 ×10⁻³ J = 4365 kJ
Is it possible that two types of dislocation coexist. a)-True b)- False
Answer:
yes it is possible
Explanation:
dislocation are if two type edge and screw dislocations
edge dislocation is a defect where an extra half plane is inside the lattice.
and screw dislocation is one in which can be assumed as the first half of the crystal slips over another.
These dislocation can coexist together where the line direction and burger vectors are neither parallel nor perpendicular then at that condition both dislocation screw and edge will coexist
If the specific gravity of a fluid is 0.721, its mass density is a. 44.99 lb/ft3 -b. 44.99 slugs/ft. c. 1.397 lb/ft d. 1.397 slugs/t e. none of these 11. A fluid a. is a gas. b. is a liquid. _c. is incompressible d. has a definite volume regardless of the size of the container e. a. and b. above.
Answer:
mass density of fluid is = 1000*0.721 = 721 kg/m3
so option e is correct
11. option e is correct ( a and b).
Explanation:
given data
specific gravity of fluid = 0.721
we know that mas density of water [tex]{\gamma _{water}}[/tex] is 1000 kg/m3
specific gravity of water is given as S.G
[tex]S.G = \frac{\gamma _{fluid} }{\gamma _{water}}[/tex]
mass density of fluid is = 1000*0.721 = 721 kg/m3
A fluid is combined form used for gas and liquid.
it offer negligible resistance to stress (shear stress)