Answer:
The concentration in mg/L equals 0.01 mg/L
Explanation:
Since the concentration is given as 10 parts per billion
We know that
1 part per billion equals 0.001 mg/L
Thus 10 parts per billion concentration equals
[tex]10\times 0.001mg/L\\\\=0.01mg/L[/tex]
What is a quasi-equilibrium process? What is its importance in engineering?
Answer:
Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.
Explanation:
Step1
Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.
Step2
All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.
Name the point of intersection, where the axis meet.
Answer:
origin
Explanation:
The point of intersection of axis is called origin.
In 2D origin is the intersection point of x-axis and y-axis if we go right to the origin then it is positive x axis, if we go left side of origin then it is negative x- axis
Similarly when we go above the origin then it positive y axis , and if we go bellow the origin then it is negative x axis
In 3D origin is the intersection of x-axis, y-axis and z-axis
NOTE- For defining i take here x axis as horizontal axis and y-axis as vertical axis
A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft is 0.7R. a) If both the shafts are subjected to the same torque, compare their shear stresses, angle of twist and mass. b) Find the strength to weight ratio for both the shafts.
Answer with Explanation:
By the equation or Torque we have
[tex]\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}[/tex]
where
T is the torque applied on the shaft
[tex]I_{p}[/tex] is the polar moment of inertia of the shaft
[tex]\tau [/tex] is the shear stress developed at a distance 'r' from the center of the shaft
[tex]\theta [/tex] is the angle of twist of the shaft
'G' is the modulus of rigidity of the shaft
We know that for solid shaft [tex]I_{p}=\frac{\pi R^4}{2}[/tex]
For a hollow shaft [tex]I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}[/tex]
Since the two shafts are subjected to same torque from the relation of Torque we have
1) For solid shaft
[tex]\frac{2T}{\pi R^4}\times r=\tau _{solid} [/tex]
2) For hollow shaft we have
[tex]\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4} [/tex]
Comparing the above 2 relations we see
[tex]\frac{\tau _{solid}}{\tau _{hollow}}=0.76[/tex]
Similarly for angle of twist we can see
[tex]\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316[/tex]
Part b)
Strength of solid shaft = [tex]\tau _{max}=\frac{T\times R}{I_{solid}}[/tex]
Weight of solid shaft =[tex]\rho \times \pi R^2\times L[/tex]
Strength per unit weight of solid shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}[/tex]
Strength of hollow shaft = [tex]\tau '_{max}=\frac{T\times R}{I_{hollow}}[/tex]
Weight of hollow shaft =[tex]\rho \times \pi (R^2-0.7R^2)\times L[/tex]
Strength per unit weight of hollow shaft = [tex]\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}[/tex]
Thus [tex]\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16[/tex]
A(n)______topology is a mixture of more than one type of topology.
Answer:
Hybrid topology is the connection of one or more than one topology.
Explanation:
Topology:
Topology is the arrangement of network.These network connects by line and nodes.
Type of topology:
1.Bus topology
2.Star topology
3.Ring topology
4.Mesh topology
Along with given above topology one topology is also used is known as hybrid topology.Hybrid topology is the connection of one or more than two one above given topology.
What is the difference between point-to-point and continuous path control in a motion control system?
Answer:
Point to point control motion system:
In point to point control motion system tool perform specific task at a particular location.Point to point control motion system is also called positioning system.It perform intermittent operation.
Ex: Drilling operation is a point to point motion control system.
Continuous path control system:
Continuous path control system is continuous operation that perform by tool.The program used in continuous path control system is more complex than point to point motion control system.
Ex :Milling operation is a continuous path control system.
The "Crawler" developed to transport the Saturn V launch vehicle from the assembly building to the launch pad is the largest land 6 vehicle ever built, weighing 4.9 x 10 -Ibs at sea level. a- What is its mass in slugs ? b- What is its mass in kilograms ?
Answer:
a) 152000 slugs
b) 2220000 kg or 2220 metric tons
Explanation:
A body with a weight of 4.9*10^6 lbf has a mass of
4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm
This mass value can then be converted to other mass values.
1 slug is 32.17 lbm
Therefore:
4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs
1 lb is 0.453 kg
Therefore:
4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg
A steel rectangular tube has outside dimensions of 150 mm x 50 mm and a wall thickness of 4 mm. State the inside dimensions, the area of its cross section, and the weight of a piece 1.22 m long.
Answer:
inside dimension [tex]= 142 mm \times 42 mm[/tex]
cross section area [tex]= 7.5\times 10^{-3} m^2[/tex]
mass of 1.2 meter log steel [tex] = 1.843\times 10^{-3} \rho[/tex]
Explanation:
given data:
Outside dimension of steel rectangular [tex]= 150 mm\times 50mm[/tex]
Thickness = 4 mm
Long = 1.22 m
inside dimension will be [tex]= (150- 8)mm \times ( 50-8)mm[/tex]
[tex]= 142 mm \times 42 mm[/tex]
cross section area [tex]= 150\times 50 mm^2[/tex]
[tex]= 7500\times 10^{-6} m^2[/tex]
[tex]= 7.5\times 10^{-3} m^2[/tex]
let the density be assumed as \rho
mass of 1.2 meter log steel will be [tex]= 1.2 \times (7.5\times 10^{-3} - 0.142\times 0.048)\times \rho[/tex]
[tex] = 1.843\times 10^{-3} \rho[/tex]
The acceleration of a particle as it moves along a straight line is given by a = (2t – 1) m/s2. If s = 1 m and v = 2 m/s when t = 0, determine the particle’s velocity and position when t = 6 s. Also determine the total distance the particle travels during this time period.
Answer:
1) Velocity at t =6 =32m/s
2) Position of particle at t = 6 secs = 67 meters
3) Distance covered in 6 seconds equals 72 meters.
Explanation:
By definition of acceleration we have
[tex]a=\frac{dv}{dt}\\\\dv=a(t)dt\\\\\int dv=\int a(t)dt\\\\v(t)=\int (2t-1)dt\\\\v(t)=t^{2}-t+c_{1}[/tex]
Now at t = 0 v = 2 m/s thus the value of constant is obtained as
[tex]2=0-0+c_{1}\\\\\therefore c_{1}=2[/tex]
thus velocity as a function of time is given by
[tex]v(t)=t^{2}-t+2[/tex]
Similarly position can be found by
[tex]x(t)=\int v(t)dt\\\\x(t)=\int (t^{2}-t+2)dt\\\\x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+c_{2}[/tex]
The value of constant can be obtained by noting that at time t = 0 x = 1.
Thus we get
[tex]1=0+0+0+c_{2}\\\\\therefore c_{2}=1[/tex]
thus position as a function of time is given by
[tex]x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+1[/tex]
Thus at t = 6 seconds we have
[tex]v(6)=6^{2}-6+2=32m/s[/tex]
[tex]x(6)=\frac{6^{3}}{3}-\frac{6^{2}}{2}+2\times 6 +1=67m[/tex]
The path length can be obtained by evaluating the integral
[tex]s=\int_{0}^{6}\sqrt{1+(t^{2}-t+2)^{2}}\cdot dt\\\\s=72meters[/tex]
What is the atmospheric temperature on Venus if the density is 67 kg/m^3 and the pressure is 9.3 mPa, absolute? Express in °C and °F. The atmosphere is composed of CO2.
Answer:
461 C
862 F
Explanation:
The specific gas constant for CO2 is
R = 189 J/(kg*K)
Using the gas state equation:
p * v = R * T
T = p * v / R
v = 1/δ
T = p / (R * δ)
T = 9.3*10^6 / (189 * 67) = 734 K
734 - 273 = 461 C
461 C = 862 F
How many joules are required to raise the temperature of a cubic meter of water by 10K?
Answer:
4.186 × 10⁷ J
Explanation:
Heat gain by water = Q
Thus,
[tex]m_{water}\times C_{water}\times \Delta T=Q[/tex]
For water:
Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)
Density of water= 1 kg/L
So, mass of the water:
[tex]Mass\ of\ water=Density \times {Volume\ of\ water}[/tex]
[tex]Mass\ of\ water=1 kg/L \times {1000\ L}[/tex]
Mass of water = 1000 kg
Specific heat of water = 4.186 kJ/kg K
ΔT = 10 K
So,
[tex]1000\times 4.186\times 10=Q[/tex]
Q = 41860 kJ
Also, 1 kJ = 1000 J
So, Q = 4.186 × 10⁷ J
Which renewable sources are growing at the fastest rate? Which renewable source is used to produce most electricity?
Answer:
Wind
Hydro electric
Explanation:
Energy are of two types
1.Renewable energy
These have unlimited source of energy.
Ex: Solar energy,wind energy,geothermal energy,hydro power ,biomass etc
2.Non renewable energy
These have limited source of energy.
Ex: Petroleum,Coal
Wind is the fastest renewable source of energy.This energy is produce by using the natural velocity of air.
Hydro electric power plant is the mostly used renewable source of energy to produce electricity.
Answer with Explanation:
The growth of renewable sources of energy depend on various factor's and their grown is also a regional dependent process. Many different countries use different renewable sources of energy and the growth of the renewable sources of energy depend on the location of the place. As an example in India the solar energy is the most widely growing source of energy due to the location of the place. while as in European union Bio energy is the source of renewable energy that has shown the most growth.
Water is the renewable source which is used to produce the most electricity in the world accounting for 16.3% of the total global electricity production.
Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is 0.9 cm. The water density is 1.0 kg/L. The empty weight of the metal pipe is 2500 N/m. In kN, what is the total weight (pipe plus water)?
Answer:
1113kN
Explanation:
The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:
Inside Diameter = Outside Diameter - Thickness
Inside Diameter = 61cm - 0.9cm = 60.1cm
Converting this diameter to meters, we have:
[tex]60.1cm*\frac{1m}{100cm}=0.601m[/tex]
This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:
[tex]V_{water}=\pi r^{2}h[/tex]
[tex]V_{water}=\pi (\frac{0.601m}{2})^{2}*120m[/tex]
[tex]V_{water}=113.28m^{3}[/tex]
The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:
[tex]d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}[/tex]
[tex]d_{water}=1000\frac{Kg}{m^{3}}[/tex]
Now, water density is given by the equation [tex]d=\frac{m}{V}[/tex], where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:
[tex]m_{water}=d_{water}.V_{water}[/tex]
[tex]m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}[/tex]
[tex]m_{water}=113280Kg[/tex]
With the water mass we can find the weight of water:
[tex]w_{water}=m_{water} *g[/tex]
[tex]w_{water}=113280kg*9.8\frac{m}{s^{2}}[/tex]
[tex]w_{water}=1110144N[/tex]
Finally we find the total weight add up the weight of the water and the weight of the pipe,so:
[tex]w_{total}=w_{water}+w_{pipe}[/tex]
[tex]w_{total}=1110144N+2500N[/tex]
[tex]w_{total}=1112644N[/tex]
Converting this total weight to kN, we have:
[tex]1112644N*\frac{0.001kN}{1N}=1113kN[/tex]
What is the mass in both slugs and kilograms of a 1000-lb beam?
Answer:
31.1 slug, 453.4 Kg
Explanation:
given,
mass of the beam is 1000 lb
to convert mass of beam into slugs and kilograms.
1 lb is equal to 0.0311 slug
1000 lb = 1000 × 0.0311
= 31.1 slug
now, conversion of lb into kg
1 lb is equal to 0.4534 kg
so,
1000 lb = 1000 × 0.4534
= 453.4 Kg
hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.
Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.
Answer:
The gravitational force between the masses is [tex]1.0\times 10^{-8}Newtons[/tex]
Explanation:
For 2 masses 'm' and 'M' separated by a distance 'd' the gravitational force between them is given by Newton as
[tex]F=G\cdot \frac{mM}{d^{2}}[/tex]
where
'G' is universal gravitational constant whose value is [tex]6.67\times 10^{-11}m^3kg^{-1}s^{-2}[/tex]
Applying the values in the above relation we get
[tex]F=6.67\times 10^{-11}\times \frac{8\times 12}{(800\times 10^{-3})^{2}}=1.0\times 10^{-8}Newtons[/tex]
Weight of 8 kg mass =[tex]8\times 9.81=78.45Newtons[/tex]
Weight of 12 kg mass =[tex]12\times 9.81=117.72Newtons[/tex]
thus we see that gravitational force between the masses is completely negligible as compared to the weight of the masses.
What is the density of an alloy formed by 10 cm^3 of copper (density = 8.9g / cm^3) and 10 cm^3 of silver (density= 10.5 g / cm^3)?
Answer:
9.7g / cm^3
Explanation:
To calculate a conbined density we must find the ratio between the sum of the masses and the sum of the volumes remembering that the equation to find the density is α=m/v, taking into account the above the following equation is inferred.
αc=copper density
αs=silver density
Vs=volume of silver
Vc=volume of copper
α= density of alloy
[tex]\alpha =\frac{({\alpha c}{Vc} +{\alpha s }{Vs} )}{Vs+Vc} \\\alpha =\frac{(8.9)(10) +(10.5)(10) }{10+10} \\\\\alpha =9.7g / cm^3[/tex]
the density of the alloy is 9.7g / cm^3
Define Plastic vs elastic deformation.
Answer:
Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.
Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.
An object is supported by a crane through a steel cable of 0.02m diameter. If the natural swinging of the equivalent pendulum is 0.95 rad/s and the natural time period of the axial vibration is found to be 0.35 sec. What is the mass of the object.
Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T= [tex]2\pi\sqrt{\frac{L}{g}}[/tex]
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35= [tex]2\pi\sqrt{\frac{L}{9.81}}[/tex]
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K = [tex]\frac{\textup{AE}}{\textup{L}}[/tex]
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K = [tex]\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}[/tex]
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ = [tex]\sqrt{\frac{K}{m}}[/tex]
or
mass, m = [tex]\sqrt{\frac{K}{\omega_n^2}}[/tex]
or
mass, m = [tex]\sqrt{\frac{20.66\times10^8}{0.95^2}}[/tex]
mass, m = 22.90 × 10⁸ kg
The engine is mounted on a foundation block which is spring - supported. Describe the steady - state vibration of the system if the block and engine have a total weight of 7500 N ( 750 kg) and the engine, when running, creates an impressed force F = (250 sin 2/) N, where t is in seconds. Assume that the system vibrates only in the vertical direction, with the positive displacement measured downward, and that the total stiffness of the springs can be represented as k = 30 kN/m. Determine the rotational speed omega of the engine which will cause resonance.
Answer:
wr = 6.32 rad/s
Explanation:
m = 750 kg
k = 30 kN/m
This system has no dampening, therefore the resonance frequency will simply be the natural frequency of the system.
[tex]wr = w0 = \sqrt{\frac{k}{m}}[/tex]
[tex]wr = \sqrt{\frac{30000}{750}} = 6.32 rad/s[/tex]
In this case the force applied doesn't matter. because we are calculating the resonance frequency.
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newtons do they experience when their separation is 0.7 m?
Answer:
The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons
Explanation:
We know that for two point charges of magnitude [tex]q_{1},q_{2}[/tex] the magnitude of force between them is given by
[tex]F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}[/tex]
where
[tex]k_{e}[/tex] is constant
[tex]r[/tex] is the separation between the charges
Initially when the charges are separated by 2.4 meters the force can be calculated as
[tex]F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208[/tex]
Now when the separation is reduced to 0.7 meters the force is similarly calculated as
[tex]F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}[/tex]
Applying value of the constant we get
[tex]F_{1}=\frac{62.208}{0.7^{2}}[/tex]
Thus [tex]F_{2}=126.955Newtons[/tex]
Find the diameter of the test cylinder in which 6660 N force is acting on it with a modulus of elasticity 110 x 103 Pa. The initial length of the rod is 380 mm and elongation is 0.50 mm.
Answer:
The diameter of the test cylinder should be 7.65 meters.
Explanation:
The Hooke's law relation between stress and strain is mathematically represented as
[tex]Stress=E\times strain\\\\\sigma =e\times \epsilon[/tex]
Where 'E' is modulus of elasticity of the material
Now by definition of strain we have
[tex]\epsilon =\frac{\Delta L}{L_{o}}[/tex]
Applying values to obtain strain we get
[tex]\epsilon =\frac{0.5}{380}=0.001316[/tex]
Thus the stress developed in the material equals
[tex]\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}[/tex]
Now by definition of stress we have
[tex]\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}[/tex]
Solving for 'D' we get
[tex]D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters[/tex]
Draw a flowchart to represent the logic of a program that allows the user to enter values for the current year and the user’s birth year. The program outputs the age of the user this year.
Answer:
Please, see the attachment.
Explanation:
First, we have to create two input boxes that allows the user to write the current year in one of them and his/her birth year in the another one. Also, we have to create a label that will show the result of the desired variable. We can write a message "Your age is:" and it will be attached to the result.
For the algorithm, let's call the variables as follows:
CY = Current Year
BY = Birth Year
X = Age of user
When the user inserts the current year and his/her birth year, the program will do the following operation:
X = CY - BY; this operation will give us the age of the user
After this the user will see something like "Your age is:" X.
A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?
Answer:43.70 MPa
Explanation:
Given
mass of engine [tex] 700 lb \approx 317.515 kg[/tex]
diameter of cable [tex]0.375 in.\approx 9.525 mm[/tex]
[tex]A=\frac{\pi d^2}{4}=71.26 mm^2[/tex]
we know stress([tex]\sigma [/tex])[tex]=\frac{load\ applied}{area\ of\ cross-section}[/tex]
[tex]\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa[/tex]
What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
Answer:
a) zero b) zero
Explanation:
Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.
Define volume flow rate of air flowing in a duct of area A with average velocity V.
Explanation:
Step1
Volume flow rate is the rate of change of volume of fluid that is flowing in the duct of pipe per unit time. It is measure in m³/s or l/s. Volume flow rate is very important parameter in fluid analysis.
Step2
For the given duct, the volume flow rate is the product of average velocity to the cross section area of duct.
Expression for volume flow rate is given as follows:
Q=AV
Here, Q is the flow rate, A is area of the duct and V is the average velocity of flowing fluid.
When a fluid flows through a sharp bend, low pressures may
developin localized regions of the bend. Estimate the
minimumabsolute pressure (in psi) that can develop without
causingcavitation if the fluid is water at 160 oF.
To avoid cavitation for water at 160 °F flowing through a sharp bend, the minimum absolute pressure should be slightly above the vapor pressure of water at this temperature, which is approximately 0.363 psi.
Explanation:When water flows through a sharp bend, cavitation can occur if the local pressure falls to or below the fluid's vapor pressure. To estimate the minimum absolute pressure without causing cavitation for water at 160 °F, we must consider water's vapor pressure at this temperature. At 160 °F (about 71 °C), the vapor pressure of water is approximately 0.363 psi. Since fluids cannot have a negative absolute pressure, and to avoid cavitation, the absolute pressure must stay above this vapor pressure. Therefore, considering atmospheric pressure to be approximately 14.7 psi, to avoid cavitation, the minimum absolute pressure in the system should be slightly above 0.363 psi to ensure no cavitation occurs.
Why do overhung rotors need to be balanced on or near resonance?
Explanation:
Balancing:
Generally balancing are of two types
1.Static balancing:In this only force balancing is done.
2.Dynamic balancing:in this force as well as moment balancing is done.
Balancing become compulsory for over hanging rotor because unbalance force produce lots of vibration and lots of sound due to this rotor or the whole system in which rotor is attached can be damage.
What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R
Answer:
-273.16 °C
-459.677 °F
0 °K
0 °R
Explanation:
The lowest temperature is the absolute zero.
Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.
Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.
Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.
The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.
Determine the weight of a 2,838 kg car in lbs. Round to the nearest 1lb.
Answer:
6243.6 lbs
Explanation:
We have given weight = 2838 kg
We have to convert this weight into lbs
Both kg and lbs are unit of measuring the weight of the body so they are changeable
We know that 1 lbs = 2.20 kg
So for converting kg into lbs we have to multiply with 2.20
So 2838 kg = 2838×2.2 = 6243.6 pound
So weight of 2838 kg will be equivalent to 6243.6 lbs
For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per hour c. 2000 km per hour d. 200 km per hour
Answer:d
Explanation:
Given
Temperature[tex]=200^{\circ}\approc 473 K[/tex]
Also [tex]\gamma for air=1.4[/tex]
R=287 J/kg
Flow will be In-compressible when Mach no.<0.32
Mach no.[tex]=\frac{V}{\sqrt{\gamma RT}}[/tex]
(a)[tex]1000 km/h\approx 277.78 m/s[/tex]
Mach no.[tex]=\frac{277.78}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.63
(b)[tex]500 km/h\approx 138.89 m/s[/tex]
Mach no.[tex]=\frac{138.89}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.31
(c)[tex]2000 km/h\approx 555.55 m/s[/tex]
Mach no.[tex]=\frac{555.55}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=1.27
(d)[tex]200 km/h\approx 55.55 m/s[/tex]
Mach no.[tex]=\frac{55.55}{\sqrt{1.4\times 287\times 473}}[/tex]
Mach no.=0.127
From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.
A football player can sprint 40 meters in 4 sec. He has constant acceleration until he reaches his top speed at 10 meters and has constant speed after that. What is his acceleration? What is his constant speed?
Answer:
1) Acceleration of player is 7.8125 [tex]m/s^{2}[/tex].
2) Constant speed of player is 12.5 m/s.
Explanation:
Let the acceleration of the player by 'a' and let he complete the initial 10 meters in [tex]t_1[/tex] time
The initial 10 meters are case of uniformly accelerated motion and hence we can relate the above quantities using second equation of kinematics as
[tex]s=ut+\frac{1}{2}at^{2}\\\\[/tex]
now since the player starts from rest hence u = 0 thus the equation can be written as
[tex]10=\frac{1}{2}at_1^2\\\\at_{1}^{2}=20...........(i)[/tex]
The speed the player reaches after [tex]t_{1}[/tex] time is obtained using first equation of kinematics as
[tex]v=u+at\\v=at_{1}[/tex]
Since the total distance traveled by the player is 40 meters hence the total time of trip is 4 seconds hence we infer that he covers 30 meters of distance at a constant speed in time of [tex]4-t_{1}[/tex] seconds
Hence we can write
[tex](4-t_{1})\cdot at_{1}=30.......(ii)\\\\[/tex]
Solving equation i and ii we get
from equation 'i' we obtain [tex]a=\frac{10}{t_{1}^{2}}[/tex]
Using this in equation 'ii' we get
[tex](4-t_{1})\cdot \frac{20}{t_{1}^{2}}\cdot t_{1}=30\\\\30t_{1}=80-20t_{1}\\\\\therefore t_{1}=\frac{80}{50}=1.6seconds\\\\\therefore a=\frac{20}{1.6^2}=7.8125m/s^{2}[/tex]
Thus constant speed equals [tex]v=7.8125\times 1.6=12.5m/s[/tex]