Answer:
Specific weight of the pop, [tex]w_{s} = 8619.45 N/m^{3}[/tex]
Density of the pop, [tex]\rho_{p} = 8790.76 kg/m^{3}[/tex]
[tex]g_{s} = 8.79076[/tex]
[tex]w_{w} = 9782.36 N/m^{3}[/tex]
Given:
Volume of pop, V = 360 mL = 0.36 L = [tex]0.36\times 10^{-3} m^{3}[/tex]
Mass of a can of pop , m = 0.369 kg
Weight of an empty can, W = 0.153 N
Solution:
Now, weight of a full can pop, W
W' = mg = [tex]0.369\times 9.8 = 3.616 N[/tex]
Now weight of the pop in can is given by:
w = W' - W = 3.616 - 0.513 = 3.103 N
Now,
The specific weight of the pop, [tex]w_{s} = \frac{weight of pop}{volume of pop}[/tex]
[tex]w_{s} = \frac{3.103}{0.36\times 10^{- 3}} = 8619.45 N/m^{3}[/tex]
Now, density of the pop:
[tex]\rho_{p} = \frac{w_{s}}{g}[/tex]
[tex]\rho_{p} = \frac{86149.45}{9.8} = 8790.76 kg/m^{3}[/tex]
Now,
Specific gravity, [tex]g_{s} = \frac{\rho_{p}}{density of water, \rho_{w}}[/tex]
where
[tex]g_{s} = \frac{8790.76}{1000} = 8.79076[/tex]
Now, for water at [tex]20^{\circ}c[/tex]:
Specific density of water = [tex]998.2 kg/m^{3}[/tex]
Specific gravity of water = [tex]0.998 kg/m^{3}[/tex]
Specific weight of water at [tex]20^{\circ}c[/tex]:
[tex]w_{w} = \rho_{20^{\circ}}\times g = 998.2\times 9.8 = 9782.36 N/m^{3}[/tex]
A 200 m3 swimming pool has been overly chlorinated by accidently
dumping an entire
container of chlorine, which amounted to 500 grams. This creates a
chlorine
concentration well in excess of the 0.20 mg/L that was the
objective at the time. With
which volumetric rate (in L/min) of freshwater should the pool be
flushed to decrease the
concentration to the desired level in 36 hours? You may assume the
chlorine does not
degrade chemically during those hours.
Answer:
Water needs to be added into the pool at the rate of 1064.814 Liters/min.
Explanation:
The concentration in mg/L of the chlorine in the pool that is induced by dumping the whole container of chlorine into the pool equals
[tex]\frac{500\times 10^{3}}{200\times 10^{3}}=2.5mg/l[/tex]
Since this is in excess to the required concentration of 0.20 mg/L
Let the amount of pure water we need to add be equal to 'V' liters now since pure water does not have any chlorine thus
By the conservation of mass principle we have
[tex]0\times V+2.5\times 200\times 10^{3}=0.20\times (V+200\times 10^{3})[/tex]
Solving for V we get
[tex]V=\frac{2.5\times 200\times 10^{3}-0.20\times 200\times 10^{3}}{0.20}=2300000Liters[/tex]
Thus 2300000 Liters of water needs to be further added in 36 hours
Thus the rate of flow in L/min equals
[tex]Q=\frac{2300000}{36\times 60}=1064.814Liters/min[/tex]
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.
The principal component in glass manufacturing is_____
Answer:
Silica is the principal component in glass.
Explanation:
Step1
Glass is the amorphous solid and transparent. Glass products have many of the shapes and design that are available in market.
Step2
Natural quartz is the primary source of glass in sand. Silica is the principal component in approximately all glass. Lime stone, soda ash and aluminum oxide are added in the galas for desired properties depending upon application. So, silica is the principal component in glass.
The acceleration due to gravity at sea level is g=9.81 m/s^2. The radius of the earth is 6370 km. The universal gravitational constant is G = 6.67 × 10−11 N-m^2/kg^2. Use this information to determine the mass of the earth.
Answer:
Mass of earth will be [tex]M=5.96\times 10^{24}kg[/tex]
Explanation:
We have given acceleration due to gravity [tex]g=9.81m/sec^2[/tex]
Radius of earth = 6370 km =[tex]6370\times 10^3m[/tex]
Gravitational constant [tex]G=6.67\times 10^{-11}Nm^2/kg^2[/tex]
We know that acceleration due to gravity is given by
[tex]g=\frac{GM}{R^2}[/tex], here G is gravitational constant, M is mass of earth and R is radius of earth
So [tex]9.81=\frac{6.67\times 10^{-11}\times M}{(6370\times 10^3)^2}[/tex]
[tex]M=5.96\times 10^{24}kg[/tex]
So mass of earth will be [tex]M=5.96\times 10^{24}kg[/tex]
A metallic material with yield stress of 140 MPa and cross section of 300 mm x 100 mm, is subjected to a tensile force of 8.00 MN. Will the sample experience plastic deformation? You must justify your answer.
Answer:Yes,266.66 MPa
Explanation:
Given
Yield stress of material =140 MPa
Cross-section of [tex]300\times 100 mm^2[/tex]
Force(F)=8 MN
Therefore stress due to this Force([tex]\sigma [/tex])
[tex]\sigma =\frac{F}{A}=\frac{8\times 10^6}{300\times 100\times 10^{-6}}[/tex]
[tex]\sigma =266.66 \times 10^{6} Pa[/tex]
[tex]\sigma =266.66 MPa[/tex]
Since induced stress is greater than Yield stress therefore Plastic deformation occurs
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]
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
a single-cylinder pump feeds a boiler through a delivery
pipeof 25mm internal diameter. the plunger speed is such as togive
a vlocity of 2 m/s. how many m3 ofwater can
be pumped into the boiler in 1 hour?
Answer:
Net discharge per hour will be 3.5325 [tex]m^3/hr[/tex]
Explanation:
We have given internal diameter d = 25 mm
Time = 1 hour = 3600 sec
So radius [tex]r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m[/tex]
We know that area is given by
[tex]A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2[/tex]
We know that discharge is given by [tex]Q=AV[/tex], here A is area and V is velocity
So [tex]Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec[/tex]
So net discharge in 1 hour = [tex]981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour[/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.
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.
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]
Estimate the magnitude of the force, in lbf, exerted on a 12-lb goose in a collision of duration 10^-3s with an airplane taking off at 150 miles/h.
Answer:
The collision force is 81987.577 lbf
Explanation:
Apply Newton’s second law of motion for required collision force exerted on the goose.
Given:
Mass of goose is 12 lb.
Time of collision is [tex]10^{-3}[/tex] s.
Taking off speed is 150 miles/h.
Calculation:
Step1
Convert take off speed in ft/s as follows:
[tex]v=(150mi/h)(\frac{\frac{5280}{3600}ft/s}{1mi/h})[/tex]
v = 220 ft/s
Step2
Collision force is calculated as follows:
[tex]F=\frac{mv}{t}[/tex]
[tex]F=\frac{(\frac{12}{32.2})220}{10^{-3}}[/tex]
F = 81987.577 lbf.
Thus, the collision force is 81987.577 lbf.
A pump collects water (rho = 1000 kg/m^3) from the top of one reservoir and pumps it uphill to the top of another reservoir with an elevation change of Δz = 800 m. The work per unit mass delivered by an electric motor to the shaft of the pump is Wp = 8,200 J/kg. Determine the percent irreversibility associated with the pump.
Answer:
4.29%
Explanation:
Given:
Density of water, ρ = 1000 kg/m³
elevation change of Δz = 800 m
work per unit mass delivered, Wp = 8,200 J/kg
Now,
The percent irreversibility = [tex](1-n_p)\times100[/tex]
where,
[tex]n_p\frac{W_{actual}}{W_{theoretical}}[/tex]
also,
[tex]W_{actual}=\frac{g\Delta z}{1000}[/tex]
Where, g is the acceleration due to the gravity
on substituting the values, we get
[tex]W_{actual}=\frac{9.81\times800}{1000}[/tex]
or
[tex]W_{actual}=7.848\ KJ/kg[/tex]
or
[tex]W_{actual}=7848\ J/kg[/tex]
Therefore,
The percent irreversibility = [tex](1-n_p)\times100[/tex]
on substituting the values, we get
The percent irreversibility = [tex](1-\frac{7848}{8,200})\times100[/tex]
or
The percent irreversibility = 4.29%
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.
If you measured a pressure difference of 50 mm of mercury across a pitot tube placed in a wind tunnel with 200 mm diameter, what is the velocity of air in the wind tunnel? What is the Reynolds number of the air flowing in the wind tunnel? Is the flow laminar or turbulent? Assume air temperature is 25°C.
Answer:
V=33.66 m/s
[tex]Re=448.8\times 10^6[/tex]
Re>4000, The flow is turbulent flow.
Explanation:
Given that
Pressure difference = 50 mm of Hg
We know that density of Hg=136000[tex]Kg/m^3[/tex]
ΔP= 13.6 x 1000 x 0.05 Pa
ΔP=680 Pa
Diameter of tunnel = 200 mm
Property of air at 25°C
ρ=1.2[tex]Kg/m^3[/tex]
Dynamic viscosity
[tex]\mu =1.8\times 10^{-8}\ Pa.s[/tex]
Velocity of fluid given as
[tex]V=\sqrt{\dfrac{2\Delta P}{\rho_{air}}}[/tex]
[tex]V=\sqrt{\dfrac{2\times 680}{1.2}}[/tex]
V=33.66 m/s
Reynolds number
[tex]Re=\dfrac{\rho _{air}Vd}{\mu }[/tex]
[tex]Re=\dfrac{1.2\times 33.66\times 0.2}{1.8\times 10^{-8}}[/tex]
[tex]Re=448.8\times 10^6[/tex]
Re>4000,So the flow is turbulent flow.
Which undesirable emissions are produced by the combustion of fossil fuels? What adverse effects are produced by these emissions?
Answer:
Emissions produce by combustion of fossil fuel:
1.Sulfur di oxide
2.Carbon mono oxide
3.Nitrogen oxide
4.Particular matter
5.Lead
6.Hydrocarbon
7.Photo chemical smog
Effects of emissions produce by combustion of fossil fuel:
1.Sulfur and nitrogen oxide leads to acid rain which corrodes the monuments.
2.Carbon mono oxide affects the central nervous system.
3.Particular matter produce asthma diseases.
4.Lead damage red blood cells.
5.Free silica produce cough and chest pain.
Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotational. The velocity varied with the radius across the flow as V = 1/r m/s where r is in meters. Find the difference in depth of the liquid from the inside to the outside radius. The inside radius of the bend is 1 m and the outside radius is 3 m.
Answer:
9 cm
Explanation:
The liquid on the bend will be affected by two accelerations: gravity and centripetal force.
Gravity will be of 9.81 m/s^2 pointing down at all points.
The centripetal acceleration will be of
ac = v^2/r
Pointing to the center of the bend (perpendicular to gravity).
The velocity will depend on the radius
v = (1 m^2/s) / r
Replacing:
ac = (1/r)^2 / r
ac = (1 m^4/s^2) / r^3
If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be
a = (-1/r^3 * i - 9.81 * j)
The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.
The potential energy of the gravity field is:
pg = g * h
The potential energy of the centripetal force is:
pc = ac * r
Then the potential field is:
p = -1/r^2 * - 9.81*h
Points on the surface at r = 1 m and r = 3 m have the same potential.
-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2
-1 - 9.81*h1 = -1/9 - 9.81*h2
-1 + 1/9 = 9.81 * (h1 - h2)
h1 - h2 = (-8/9) / 9.81
h2 - h1 = 0.09 m
The outer part will be 9 cm higher than the inner part.
What's the difference between accuracy and percision in measuring and gaging?
Answer:
The term Accuracy means that how close our result to the original result.
Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.
And the term Precision means how likely we get result like this.
Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.
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]
Please explain what is the difference between engineering stress and true stress in a tensile test?
Answer:
Explanation:
Engineering Stress is defined as Load applied to the original cross-sectional area which we have taken in the start.
True stress is defined as the load divided by area of cross-section of specimen at that instant.
Engineering stress and true stress can be expressed by relation
[tex]\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )[/tex]
Where
[tex]\sigma _T=True\ stress[/tex]
[tex]\sigma _E=Engineering\ stress[/tex]
[tex]\epsilon _E=Engineering\ Strain[/tex]
Two closed systems A (4000 kJ of thermal energy at 30°C) and B (3000 kJ of thermal energy at 40°C) are brought into contact with each other. Determine the direction of any heat transfer between the two systems and explain why.
Answer:
The direction of any heat transfer between the systems will be from system B to System A.
Explanation:
The direction of any heat transfer between the systems will be from system B to System A.
Even if the system A has a higher thermal Energy it will be from B to A because the system B has a higher temperature than A. The energy will be transferred from B to A until the temperature of the two systems equals and the temperature is in equilibrium.
This comes from the fundamental law of thermodynamics.
Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hours.
Answer:
The availability of system will be 0.9
Explanation:
We have given mean time of failure = 900 hours
Mean time [to repair = 100 hour
We have to find availability of system
Availability of system is given by [tex]\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}[/tex]
So availability of system [tex]=\frac{900}{900+100}=\frac{900}{1000}=0.9[/tex]
So the availability of system will be 0.9
A heat pump has a work input of 2 kW and provides 7 kW of net heat transfer to heat a house. The system is steady, and there are no other work or heat interactions. Is this a violation of the first law of thermodynamics? Select one: a. Yes b. No
Answer:
No, is not a violation of the first law of thermodynamics.
Explanation:
The first law of thermodynamics states that energy is neither created nor destroyed in an isolated system (a system without mass or energy transfer with ambient).
In a heat pump work is used to transfer heat from a cold body to a hot body (see figure). In a free system heat would go from the heat source to the cold one, that is why you need work. Work and heat are energy in transit.
W + Qin = Qout
In your case
2 kW + Qin = 7 kW
Qin =5 kW
It would be a violation to the first law of thermodynamics if Qout is less than 2 kW.
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 small pad subjected to a shearing force is deformed at the top of the pad 0.12 in. The heigfit of the pad is 1.15 in. What is the shearing strain (rad) on the pad?
Answer:
Shearing strain will be 0.1039 radian
Explanation:
We have given change in length [tex]\Delta L=0.12inch[/tex]
Length of the pad L = 1.15 inch
We have to find the shearing strain
Shearing strain is given by
[tex]\alpha =tan^{-1}\frac{\Delta L}{L}=tan^{-1}\frac{0.12}{1.15}=5.9571^{\circ}[/tex]
Shearing strain is always in radian so we have to change angle in radian
So [tex]5.9571\times \frac{\pi }{180}=0.1039radian[/tex]
2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in thepump is measured to be 500 kPa and the motor efficiency is 90 percent, determine the mechanical efficiencyof the pump.
Answer:
[tex]\eta = 91.7[/tex]%
Explanation:
Determine the initial velocity
[tex]v_1 = \frac{\dot v}{A_1}[/tex]
[tex] = \frac{0.1}{\pi}{4} 0.08^2[/tex]
= 19.89 m/s
final velocity
[tex]v_2 =\frac{\dot v}{A_2}[/tex]
[tex]= \frac{0.1}{\frac{\pi}{4} 0.12^2}[/tex]
=8.84 m/s
total mechanical energy is given as
[tex]E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}[/tex]
[tex]\dot v = \dot m v[/tex] [tex]( v =v_1 =v_2)[/tex]
[tex]E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}[/tex]
[tex] = mv\Delta P + \dot m \frac{v_2^2 -v_1^2}{2}[/tex]
[tex]= \dot v \Delta P + \dot v \rho \frac{v_2^2 -v_1^2}{2}[/tex]
[tex] = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}[/tex]
[tex]E_{mech} = 36.34 W[/tex]
Shaft power
[tex]W = \eta_[motar} W_{elec}[/tex]
[tex]=0.9\times 44 =39.6[/tex]
mechanical efficiency
[tex]\eta{pump} =\frac{ E_{mech}}{W}[/tex]
[tex]=\frac{36.34}{39.6} = 0.917 = 91.7[/tex]%
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
A car accelerates with a = 0.01s m/s^2 with sin meters. The car starts at t = 0 at s = 100 m with v = 12 m/s. Determine the speed at s = 420 m and the time to get there.
Answer:
Part 1) Speed at s = 420 meters =12.26 m/s
Part 2) Time required to cover the distance = 26 seconds
Explanation:
This problem can be solved using third equation of kinematics as follows
[tex]v^2=u^2+2as[/tex]
where
'v' is the final velocity
'u' is the initial velocity
'a' is the acceleration of the car
's' is the distance covered between change in velocities
Now during the time at which the car moves it cover's a distance of 420 m-100 m=320 meters
Thus applying values in the above equation we get
[tex]v^2=12^{2}+2\times 0.01\times 320\\\\v^2=150.4\\\\\therefore v=12.26m/s[/tex]
The time to reach this velocity can be found using first equation of kinematics as
[tex]v=u+at\\\\\therefore t=\frac{v-u}{a}\\\\t=\frac{12.26-12}{0.01}=26seconds[/tex]
Given a reservoir watershed of 22 square miles, and assuming a
rainfall of 0.75
inches of rain, with 35% of the rainfall draining overland as
surface runoff and
entering the reservoir, what does this runoff contribution amount
to in acre-feet of
water?
Answer:
308 acre-ft of water
Explanation:
Given:
Area of the watershed = 22 square miles
Depth of Rainfall = 0.75 in =[tex]\frac{\textup{0.75}}{\textup{12}}[/tex]=0.0625 ft
Percentage rainfall falling in reservoir as runoff = 35%
Now,
1 square mile = 640 acre
Thus,
22 square miles = 22 × 640 = 14,080 acres
Thus,
The total volume of rainfall = Area of watershed × Depth of the rainfall
or
The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft
also,
only 35% of the total rainfall is contributing as runoff
thus,
Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water