Answer:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
A bridge has been constructed between the mainland and
anisland. The total cost (excluding toll) to travel across
thebridge is expressed as C=50+0.5V, where V is the number of
veh/dayand C is the cost/vehicle in cents. The demand for
travelacross the bridge is V=2500-10C.
(a)Determine the volume of traffic across the bridge.
(b)If a toll of 25 cents is added, what is the volume
acrossthe bridge?
Answer:
a) 333,33 b)291,67
Explanation:
you have both equations :
C=50+0.5V and V=2500-10C
and you want to know the variable V so, you can calculate V in function of C, and you have already clear the variable "C", then you replace for (a):
V=2500-10CV=2500-10(50+0.5V)V=2500-500-5VV+5V=2500-5006V=2000V=2000/6V=333,33for b) they tell you that you increase your constant of the equiation C=50+0.5V (remember that the constant is the one alone, wihout any variable, in this case "50") increase in 25 so, your equiation for this point of C is C=75+0.5V, then you do the same:
V=2500-10CV=2500-10(75+0.5V)V=2500-750-5VV+5V=2500-7506V=1750V=1750/6V=291,67It's a system of two equations and two variables, wich gives you a compatible define system, it gives you only one solution.
I hope it helps you.
To determine the volume of traffic across the bridge, we can solve an equation relating cost and traffic volume. The volume of traffic across the bridge is approximately 333 vehicles per day. If a toll of 25 cents is added, the volume of traffic across the bridge is approximately 292 vehicles per day.
Explanation:To determine the volume of traffic across the bridge, we need to solve the equation for V. The equation relating cost and volume of traffic is given by V = 2500 - 10C. Substituting the cost function C = 50 + 0.5V into the equation, we get V = 2500 - 10(50 + 0.5V). Simplifying the equation, we find V = 2500 - 500 - 5V. Combining like terms, we get 6V = 2000, which yields V = 333.33. Therefore, the volume of traffic across the bridge is approximately 333 vehicles per day.
To find the volume across the bridge if a toll of 25 cents is added, we can modify the cost function to be C = 50 + 0.5V + 25 cents. We again substitute this into the equation V = 2500 - 10C and solve for V. Substituting in the new cost function, we get V = 2500 - 10(50 + 0.5V + 25). Simplifying the equation, we find V = 2500 - 10(75 + 0.5V). Continuing to simplify, we have V = 2500 - 750 - 5V. Combining like terms, we get 6V = 1750, which yields V = 291.67. Therefore, the volume across the bridge with the toll of 25 cents is approximately 292 vehicles per day.
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]%
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]
Calculate the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, the centers of which are separated by a distance of 2.9 nm.
Calculating the force of attraction between a cation and an anion using Coulomb's Law involves converting the given distance into meters, using the elementary charge for valences, and then applying the formula.
Explanation:The question is about calculating the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, separated by a distance of 2.9 nm. To find this, we use Coulomb's Law, which is given by the equation F = k * |q1*q2| / r2, where F is the force between the charges, k is Coulomb's constant (8.9875517873681764 × 109 N m2 C-2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges. However, given the units of distance as nanometers (nm) and charge in terms of electron charge, we first convert the distance to meters (1 nm = 1 × 10-9 meters) and use the elementary charge (1.6022 x 10-19 C) for the charges. With a +2 valence for the cation and a -1 valence for the anion, the charges become +2e and -1e, respectively. The distance r is 2.9 nm or 2.9 × 10-9 meters. Plugging these values into Coulomb's formula, the attraction force can be calculated, taking into account that the force will be negative, signifying attraction.
Calculate the time taken to completely empty aswimming pool 15
m long and 9 m wide through an opening at thebottom as shown in the
fig. The swimming pool holds water to depth1.5 m.Opening area is
0.3 sq.m and Cd=0.62.
Answer:
Time needed to empty the pool is 401.35 seconds.
Explanation:
The exit velocity of the water from the orifice is obtained from the Torricelli's law as
[tex]V_{exit}=\sqrt{2gh}[/tex]
where
'h' is the head under which the flow of water occurs
Thus the theoretical discharge through the orifice equals
[tex]Q_{th}=A_{orifice}\times \sqrt{2gh}[/tex]
Now we know that
[tex]C_{d}=\frac{Q_{act}}{Q_{th}}[/tex]
Thus using this relation we obtain
[tex]Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}[/tex]
Now we know by definition of discharge
[tex]Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}[/tex]
Using the above relations we obtain
[tex]Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\[/tex]
The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0
Solving for 't' we get
[tex]\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.[/tex]
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.
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%
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]
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.
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]
Select the most accurate statement. A diffuser converts some of a fluid's _(i)____ to ___(ii)___. Select one: a. (i) pressure, (ii) workb. (i) work, (ii) pressurec. (i) kinetic energy, (ii) heat d. (i) heat, (ii) kinetic energye. (i) enthalpy, (ii) kinetic energy f. (i) kinetic energy, (ii) enthalpy
Answer:
1.Kinetic energy 2.Enthalpy
Explanation:
We know that diffuser is a device which convert kinetic energy of fluids into pressure energy of fluids,on the other hand nozzle is used to convert the pressure energy of fluids into kinetic energy of fluids. If we want exit velocity of fluid is too high then we use nozzle and if we wand exit velocity of fluid is low then we use diffuser.
We know that enthalpy is also related with pressure so we can say that ,diffuser covert kinetic energy into enthalpy.
Enthalpy h= u + p.v
Where u is the internal energy ,p is the pressure and v is volume.
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.
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 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]
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.
A shaft is to transmit 3.5 kW power while rotating at 350 rpm. If the shaft is made of plain carbon steel with 100 MPa yield strength, calculate: a) Torque being transmitted by the shaft b) Diameter of the shaft
Answer:
a)T=95.5414 N.m
b)d=21.35 mm
Explanation:
Given that
P=3.5 KW
Speed N=350 RPM
Yield strength= 100 MPa
So Shear strength = 0.5 x 100 =50 MPa
We know that
[tex]P=\dfrac{2\pi NT}{60}[/tex]
Where N is the speed ,T is the torque and P is the power.
Now by putting the values
[tex]P=\dfrac{2\pi NT}{60}[/tex]
[tex]3500=\dfrac{2\pi 350T}{60}[/tex]
T=95.5414 N.m
T=95,541.4 N.mm
We also know that
[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]
d is the diameter of shaft
[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]
[tex]50=\dfrac{16\times 95541.4}{\pi d^3}[/tex]
d=21.35 mm
Torque,T=95.5414 N.m
Diameter ,d=21.35 mm
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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.
An exercise room has six weight-lifting machines that have no motors and seven treadmills, each equipped with a 2.5-hp (shaft output) motor. The motors operate at an average load factor of 0.7, at which their efficiency is 0.77. During peak evening hours, all 13 pieces of exercising equipment are used continuously, and there are also two people doing light exercises while waiting in line for one piece of the equipment. Assuming the average rate of heat dissipation from people in an exercise room is 740 W, determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions.
Answer:
25.4 kW
Explanation:
There are 15 people doing exercises, each will dissipate 740 W of heat, so they will disspiate a total of p = 15*740 = 11100 W = 11.1 kW
There are 7 treadmills, each has a 2.5 hp motor (1.86 kW) running at a load factor of 0.7 with an efficiency of 0.77. So their total power would be p = 7*1.86*0.77/0.7 =14.3 kW
So the total heat dissipated would be 11.1 + 14.3 = 25.4 kW.
The rate of heat gain from people and equipment at peak load is approximately 14.955 kW, including treadmill and people's heat dissipation.
To calculate the rate of heat gain of the exercise room from people and equipment at peak load conditions, we need to consider the heat dissipated by both the exercising equipment and the people in the room.
1. Heat dissipation from exercising equipment:
- For the treadmills: Each treadmill has a 2.5-hp motor operating at 0.7 load factor and 0.77 efficiency. So, the power consumed by each treadmill is [tex]\( P_{\text{treadmill}} = 2.5 \, \text{hp} \times 0.7 \times 0.77 = 1.925 \, \text{kW} \).[/tex]
- For the weight-lifting machines: Since they have no motors, they do not contribute to heat dissipation.
2. Heat dissipation from people:
- There are two people doing light exercises. Assuming each person dissipates heat at a rate of 740 W, the total heat dissipated by people is [tex]\( P_{\text{people}} = 2 \times 740 \, \text{W} = 1480 \, \text{W} \).[/tex]
Now, we can calculate the total rate of heat gain of the exercise room:
[tex]\[ \text{Total heat gain} = P_{\text{treadmill}} \times \text{number of treadmills} + P_{\text{people}} \][/tex]
Given there are 7 treadmills, we have:
[tex]\[ \text{Total heat gain} = 1.925 \, \text{kW} \times 7 + 1480 \, \text{W} \]\[ \text{Total heat gain} = 13.475 \, \text{kW} + 1480 \, \text{W} \]\[ \text{Total heat gain} \approx 13.475 \, \text{kW} + 1.48 \, \text{kW} \]\[ \text{Total heat gain} \approx 14.955 \, \text{kW} \][/tex]
So, the rate of heat gain of the exercise room from people and equipment at peak load conditions is approximately 14.955 kW.
A small metal particle passes downward through a fluid medium while being subjected to the attraction of a magnetic field such that its position is observed to be s = (15t^3 - 3t) mm, where t is measured in seconds. Determine (a) the particle's displacement from t = 2 s to t = 4 s, and (b) the velocity and acceleration of the particle when t = 5 s.
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
[tex]a=450\ mm/s^2[/tex]
Explanation:
Given that
[tex]s = 15t^3 - 3t\ mm[/tex]
a)
When t= 2 s
[tex]s = 15t^3 - 3t\ mm[/tex]
[tex]s = 15\times 2^3 - 3\times 2\ mm[/tex]
s= 114 mm
At t= 4 s
[tex]s = 15t^3 - 3t\ mm[/tex]
[tex]s = 15\times 4^3- 3\times 4\ mm[/tex]
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
[tex]V=\dfrac{ds}{dt}[/tex]
[tex]\dfrac{ds}{dt}=45t^2-3[/tex]
At t= 5 s
[tex]V=45t^2-3[/tex]
[tex]V=45\times 5^2-3[/tex]
V=1122 mm/s
We know that acceleration a
[tex]a=\dfrac{d^2s}{dt^2}[/tex]
[tex]\dfrac{d^2s}{dt^2}=90t[/tex]
a= 90 t
a = 90 x 5
[tex]a=450\ mm/s^2[/tex]
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.
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
Which component stores energy as potential energy in a mechanical system ?
Answer:
Spring
Explanation:
We know that energy associated with motion is known as kinetic energy and the energy associated with position or elevation is known as potential energy.
A ball on floor which on the rest condition have zero potential energy but on the other hand a ball at a height h from the floor have mgh potential energy.
Spring is the mechanical component which stored potential energy.
The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in Appendix A to obtain a conversion
Answer:
The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)
Explanation:
Since the given volume flow rate is gallons per minute.
We know that 1 gallon = 3.785 liters and
1 minute = 60 seconds
Let the flow rate be [tex]Q\frac{gallons}{minute}[/tex]
Now replacing the gallon and the minute by the above values we get
[tex]Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}[/tex]
Thus [tex]Q'=0.631Q\frac{liters}{second}[/tex]
Now since we know that 1 liter = [tex]0.0353ft^{3}[/tex]
Using this in above relation we get
[tex]Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q[/tex]
From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.
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 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]
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.
What is the lehr and what purpose does it serve?
Explanation:
Step1
Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.
Step2
Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.
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
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.