Answer:
[tex]v_2 = 160.23 m/s[/tex]
[tex]T_2 = 475.797 k[/tex]
Explanation:
given data:
Diameter =[tex] d_1 = 200mm[/tex]
[tex]t_1 =195 degree[/tex]
[tex]p_1 =500 kPa[/tex]
[tex]v_1 = 100m/s[/tex]
[tex]p_2 = 85kPa[/tex]
[tex]d_2 = 158mm[/tex]
from continuity equation
[tex]A_1v_1 = A_2v_2[/tex]
[tex]v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}[/tex]
[tex]v_2 = \frac{d_2v_1}{d_2^2}[/tex]
[tex]v_2 = [\frac{d_1}{d_2}]^2 v_1[/tex]
[tex]= [\frac{0.200}{0.158}]^2 \times 100[/tex]
[tex]v_2 = 160.23 m/s[/tex]
by energy flow equation
[tex]h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w[/tex]
[tex]z_1 =z_2[/tex] and q =0, w =0 for nozzle
therefore we have
[tex]h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2} [/tex]
[tex]dh = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
but we know dh = Cp dt
hence our equation become
[tex]Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)[/tex]
[tex]Cp (T_2 -T_1) = 7836.94[/tex]
[tex](T_2 -T_1) = \frac{7836.94}{1.005*10^3}[/tex]
[tex](T_2 -T_1) = 7.797 [/tex]
[tex]T_2 = 7.797 +468 = 475.797 k[/tex]
A convenient and cost-effective way to store biogas is to use light-weight, rigid gas containers. The pressure and temperature of biogas stored in a 10 m^3 container are 1.5 bar and 5°C, respectively, as measured at an early morning time (state 1). The temperature of the biogas is expected to increase to 40°C at noon on the same day, without any significant change in the volume of the container or amount of methane in the container (state 2). The biogas can be approximated as methane (CH4) modelled as an ideal gas with constant specific heat ratio, k = 1.4. The effects of gravity and motion are negligible. The reference environment pressure and temperature are, respectively, po = 100 kPa and To = 0 °C. The molecular weight of methane is M = 16.04 gmol-1. The universal gas constant is R= 8.31 J mol-1K-1. (a) Calculate the mass, in kg, and amount of substance, in mol, for the methane in the container.
Answer:
10.41 kg
Explanation:
The gas state equation is:
p * V = n * R * T
For this equation we need every value to be in consistent units
1.5 bar = 150 kPa
5 C = 278 K
n = p * V / (R * T)
n = 150000 * 10 / (8.31 * 278) = 649 mol
Multiplying the amount of moles by the molecular weight of the gas we obtain the mass:
m = M * mol
m = 16.04 * 649 = 10410 g = 10.41 kg
A freight train and a passenger train share the same rail track. The freight train leaves station A at 8:00 A.M. The train travels at a speed of 45 km/h for the first 10 minutes and then continues to travel at a speed of 60 km/h. At 8:35 A.M., the passenger train leaves station A. The pas¬senger train travels first at a speed of 75 km/h for 5 minutes and then continues to travel at a speed of 105 km/h. Determine the location of the siding where the freight train will have to be parked to allow the faster passenger train to pass through. As a safety precaution, it is determined that the time headway between the two trains should not be allowed to fall below 5 minutes.
Answer:
74.2 km from station A.
Explanation:
We set a frame of reference with the station at the origin and the X axis pointing in the direction the trains run.
The freight train leaves at 8 AM and travels 10 minutes at 45 km/h.
For this problem it is better to convert the speeds to km/min
45 km/h = 0.75 km/min
The equation for position under constant speed is:
X(t - t0) = X0 + v0 * (t - t0)
Since we know the time it will stop moving at this speed:
X(10 - 0) = 0 + 0.75 * (10 - 0) = 7.5 km
After it ran those 7.5 km it will keep running at 60 km/h.
60 km/h = 1 km/min
The position equation for it is now:
X(t - 10) = 7.5 + 1 * (t - 10)
The passenger train leaves the station at 8:35 AM. It travels at 75 km/h for 5 minutes.
75 km/h = 1.25 km/min
After those 5 minutes it will have traveled:
X(40 - 35) = 0 + 1.25 * (40 - 35) = 6.25 km
Then it travels at 105 km/h
105 km/h = 1.75 km/min
Its position equation is now:
X(t - 40) = 6.25 + 1.75 * (t - 40)
Equating both positions we find the time at which they would meet:
7.5 + 1 * (t - 10) = 6.25 + 1.75 * (t - 40)
7.5 + t - 10 = 6.25 + 1.75*t - 70
t - 1.75*t = 6.25 - 70 +10 - 7.5
-0.75*t = -61.25
t = 61.25 / 0.75
t = 81.7 minutes
The freight train will have to be parked 5 minutes before this at t = 76.7 minutes.
At that moment the freight train will be at:
X(76.7 - 10) = 7.5 + 1 * (76.7 - 10) = 74.2 km
why HF (hydrogen fluoride) has higher boiling temperature than HCl (hydrogen chloride), even thought HF has lower molecular weight?
Answer:
Boiling point of HF is higher as compared to HCl because of presence of hydrogen bonding in it.
Explanation:
In HF, intermolecular force of attraction is hydrogen bonding.
Hydrogen bonding is a type of electrostatic force of attraction existing between H atom and electronegative atom.
For a molecule to have hydrogen bonding, H atom must be bonded to electronegative atom, O, N and F.
Hydrogen bonding can be intermolecular and intramolecular.
So, in HF hydrogen bonding present.
In HCl, only van der Waals force exists. van der Waals forces are weak as compared to hydrogen bonding.
Because of presence of hydrogen bonding, HF molecules are held tightly and so requires more heat to boil.
Therefore, boiling point of HF is more as compared to HCl.
Six kilograms of nitrogen at 30 °C are cooled so that the internal energy decreases by 60 kJ. Find the final temperature of the gas. Assume that the specific heat of nitrogen is 0.745 kJ / kg °C.
Answer:
Final temperature will be 16.57°C
Explanation:
We have given mass of nitrogen m = 6 kg
Initial temperature [tex]T_1=30^{\circ}[/tex]
Decrease in internal energy [tex]\Delta U=-60KJ[/tex]
Specific heat of nitrogen [tex]c_v[/tex]= 0.745 KJ/kg
Let final temperature is [tex]T_2[/tex]
Change in internal energy is given by [tex]\Delta U=mc_v\Delta T=mc_v(T_2-T_1)[/tex]
[tex]-60=6\times 0.745(T_2-30)[/tex]
[tex]T_2=16.57^{\circ}C[/tex]
So final temperature will be 16.57°C
Thermal conductivity of AISI 316 Stainless Steel at 90ºC is 14.54 W/m K. Convert this value to IP system.
Answer:
the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]
Explanation:
Given that
Thermal conductivity K=14.54 W/m.K
This above given conductivity is in SI unit.
SI unit IP unit Conversion factor
m ft 0.3048
W Btu/hr 0.293
The unit of conductivity in IP is Btu./ft.hr.F.
Now convert into IP divided by 1.73 factor.
[tex]0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}[/tex]
So
[tex]0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]
[tex]8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}[/tex]
So the value of conductivity in IP is [tex]8.406\dfrac{Btu}{ft.hr.F}[/tex]
What is the ratio between the maximum elastic moment, MY, and the maximum plastic moment, MP, for a solid rectangular section made from a ductile, elastoplastic material? What is this ratio referred to?
Answer:
Shape factor
Explanation:
Shape factor is the ratio of maximum plastic moment to maximum elastic moment.Shape factor is denoted by K.
Shape factor can be given as
[tex]K=\dfrac{M_p}{M_y}[/tex]
[tex]K=\dfrac{\sigma _yZ_p}{\sigma _y Z}[/tex]
[tex]K=\dfrac{Z_p}{ Z}[/tex]
For a solid rectangular section made from ductile material shape factor is 1.5 .
How much would a 10.0 inch long, 0.25 inch diameter AISI 1020 Q&T bolt stretch when loaded with 2000 lbs?
Answer:
0.014 in
Explanation:
The Young's module of steel is of E = 210 GPa = 30*10^6 psi
The section of the bolt would be:
A = π/4 * D^2
The stiffness would be:
k = E * A / L
k = π * D^2 * E / (4 * L)
k = π * 0.25^2 * 30*10^6 / (4 * 10) = 147000 lb / in
If I apply a force of 2000 lb, I calculate the stretching with Hooke's law:
Δx = f / k
Δx = 2000 / 147000 = 0.014 in
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The system is heated until the volume is 0.2 m3 and the pressure is 180 kPa. What is the work done by the air? External pressure is 100 kPa.
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = [tex]\frac{60+180}{2}[/tex] = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated power density of the wind that is hitting the turbine. Calculate the mechanical power developed at the shaft connecting rotor and generator. Assume rotor diameter 100 m and air density 1.225 kg/m^3.
Answer:
Rated power = 1345.66 W/m²
Mechanical power developed = 3169035.1875 W
Explanation:
Wind speed, V = 13 m/s
Coefficient of performance of turbine, [tex]C_p[/tex] = 0.3
Rotor diameter, d = 100 m
or
Radius = 50 m
Air density, ρ = 1.225 kg/m³
Now,
Rated power = [tex]\frac{1}{2}\rho V^3[/tex]
or
Rated power = [tex]\frac{1}{2}\times1.225\times13^3[/tex]
or
Rated power = 1345.66 W/m²
b) Mechanical power developed = [tex]\frac{1}{2}\rho AV^3C_p[/tex]
Here, A is the area of the rotor
or
A = π × 50²
thus,
Mechanical power developed = [tex]\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3[/tex]
or
Mechanical power developed = 3169035.1875 W
Define drag and lift forces
Answer with Explanation:
Drag and lift are the forces that act on an object which moves in a fluid and are explained as under:
1)Drag: Drag is the force that opposes the motion of an object moving in a fluid and hence is analogous to fluid frictional force in a fluid. Power is needed to be spent by an object to overcome the frictional drag. The drag force is different from the classical friction as we know that the frictional force on a dry surface is independent of the velocity of the object but for drag the force is proportional to the square of the velocity of the object.
Mathematically
[tex]F_{Drag}=\frac{1}{2}C_{d}A\rho _{fluid}V^2[/tex]
where
[tex]C_d[/tex] is a constant known as coefficient of drag and A is the projected area of the object.
2) Lift: Lift force is a component of the force that a fluid exerts on an object moving in it that counters the weight of the object and hence has the tendency to lift the object and hence is known as lift force. The lift force is perpendicular to the incoming fluid. Lift force is useful as it allows the aeroplanes to fly as the lift force that is generated by the air on the wings of the plane is used to overcome the force of gravity on the plane.
Mathematically
[tex]F_{Lift}=\frac{1}{2}C_{L}A\rho _{fluid}V^2[/tex]
where
[tex]C_L[/tex] is a constant known as coefficient of lift and A is the projected area of the object moving with velocity 'v'.
Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance +/- 0.000005 m^3/kg)?
Answer:
[tex]specific\ volume=0.00097\ m^3/kg[/tex]
Explanation:
Given that
Specific gravity of sea water = 1.025
So density of sea water = 1.025 x 1000 [tex]kg/m^3[/tex]
Density of sea water = 1025 [tex]kg/m^3[/tex]
We know that
[tex]Density=\dfrac{mass}{Volume}[/tex] ---1
Specific volume
[tex]specific\ volume=\dfrac{Volume}{mass}[/tex] ---2
From equation 1 and 2
We can say that
[tex]specific\ volume=\dfrac{1}{density}\ m^3/kg[/tex]
[tex]specific\ volume=\dfrac{1}{1025}\ m^3/kg[/tex]
[tex]specific\ volume=0.00097\ m^3/kg[/tex]
Evaluate (204 mm)(0.004 57 kg) / (34.6 N) to three
significantfigures and express the answer in SI units using an
appropriateprefix.
Answer:
the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]
Explanation:
We have evaluate [tex]\frac{(204mm\times 0.00457kg)}{34.6N}[/tex]
We know that 1 mm [tex]=10^{-3}m[/tex]
So 240 mm [tex]=204\times 10^{-3}m[/tex]
Newton can be written as [tex]kgm/sec^2[/tex]
So [tex]\frac{(204\times 10^{-3}m)\times 0.00457kg}{34.6kgm/sec^2}=2.69\times 10^{-5}sec^{2}[/tex]
So the evaluation in SI unit will be [tex]2.69\times 10^{-5}sec^{2}[/tex]
Why does an object under forced convection reach a steady-state faster than an object subjected to free-convection?
Answer:
Free convection:
When heat transfer occurs due to density difference between fluid then this type of heat transfer is know as free convection.The velocity of fluid is zero or we can say that fluid is not moving.
Force convection:
When heat transfer occurs due to some external force then this type of heat transfer is know as force convection.The velocity of fluid is not zero or we can say that fluid is moving in force convection.
Heat transfer coefficient of force convection is high as compare to the natural convection.That is why heat force convection reach a steady-state faster than an object subjected to free-convection.
We know that convective heat transfer given as
q = h A ΔT
h=Heat transfer coefficient
A= Surface area
ΔT = Temperature difference
calculate the density of air(as a perfect gas) when the pressure
is1.01325*105 N/m2 and the temperature
is288.15 K.
Answer:
1.2253 kg/m3
Explanation:
from ideal gas equation we know that
[tex]PV = n\timesRT[/tex]
[tex]\frac{n}{V} = \frac{P}{(RT)}[/tex]
where,
pressure P = 1.01325 x 10^5 N/m2 = 101325 Pascals
gas constant R = 8.314 J/mol/K
T = 288.15 K
[tex]\frac{n}{V} = \frac{101325}{(8.314*288.15K)}[/tex]
[tex]\frac{n}{V} = 42.2948967 Pa*mol/J[/tex]
we know that 1 Pa = 1 J/m3
so[tex]\frac{n}{V} = 42.2948967\ mol/m3[/tex]
as we know, Air consist of 21% oxygen and 79% nitrogen
therefore
Molar mass of air:
[tex]0.21 *(32 g/mol) + 0.79 *(28 g/mol) = 28.97 g/mol[/tex]
So [tex](42.2948967\ mol/m3) \times 28.97\ g/mol[/tex]
= 1 225.27 g/m^3
= 1.22528316 kg/m^3 ~ 1.2253 kg/m3
A rocket developed by an amateur was traveling upwards at a velocity given by v = (11 + 0.2s) m/s, where s is in meters. Determine the time for the rocket to reach an altitude of s = 80 m. Initially, s = 0 when t = 0. [Hint: obtain initial velocity].
Answer:
Time taken to reach 80 meters equals 4.4897 seconds.
Explanation:
We know that velocity is related to position as
[tex]v=\frac{ds}{dt}[/tex]
Now it is given that [tex]v=(11+0.2v[/tex]
Using the given velocity function in the above relation we get
[tex]\frac{ds}{dt}=(11+0.2s)\\\\\frac{ds}{(11+0.2s)}=dt\\\\\int \frac{ds}{(11+0.2s)}=\int dt\\\\[/tex]
Now since the limits are given as
1) at t = 0 , s=0
Using the given limits we get
[tex]\int_{0}^{80} \frac{ds}{(11+0.2s)}=\int_{o}^{t} dt\\\\\frac{1}{0.2}[ln(11+0.2s)]_{0}^{80}=(t-0)\\\\5\times (ln(11+0.2\times 80)-ln(11))=t\\\\\therefore t=4.4897seconds[/tex]
Derive the following conversion factors: (a) Convert a viscosity of 1 m^2/s to ft^2/s. (b) Convert a power of 100 W to horsepower. (c) Convert a specific energy of 1 kJ/kg to Btu/lbm.
Answer:
(a) 10.76 [tex]ft^2/sec[/tex]
(b) 0.134 lbm
(c) 0.4308 Btu/lbm
Explanation:
We have to do conversion
(a) conversion of viscosity [tex]1m^2/sec\ to\ ft^2/sec[/tex]
We know that [tex]1m^2=10.76feet^2[/tex]
So [tex]1m^2/sec=10.76ft^2/sec[/tex]
(b) We have to convert power 100 W in hp
We know that 1 W = 0.00134 hp
So 100 W = 100×0.00134=0.134 hp
(c) We have to convert 1 KJ/kg to Btu/lbm
We know that 1 KJ = 0.94780 Btu
And 1 kg = 2.20 lbm
So [tex]1KJ/kg=\frac{0.9478Btu}{2.20lbm}=0.4308Btu/lbm[/tex]
Specific cutting energy increases with increasing the cutting speed. a) True b) False
Answer:
b)false
Explanation:
Specific cutting energy:
Energy required to remove unit volume of material is called specific energy.In other words we can say that the ratio of energy to the volume removal rate is called specific cutting energy.
When cutting speed is increases then specific energy goes to decrease.As well as when depth of cut and feed of tool is increase then specific cutting energy will decrease.
What is the governing ratio for thin walled cylinders?
Answer:
D/t>20
Explanation:
Lets take
D =Diameter of thin cylinder
t =Thickness of thin cylinder
So a cylinder is called thin cylinder if the ratio of diameter to the thickness is greater than 20 (D/t>20 ).
But on the other hand a cylinder is called thick cylinder is ratio of thickness to the diameter is greater than 20 (t/D>20 ).
So the governing ratio of thin walled cylinder is 20.
To 3 significant digits, what is the change of entropy of air in kJ/kgk if the pressure is decreased from 400 to 300 kPa and the temperature is increased from 300 to 900 K? DO NOT ASSUME constant specific heats.
Answer:
The change of entropy is 1.229 kJ/(kg K)
Explanation:
Data
[tex] T_1 = 300 K [/tex]
[tex] T_2 = 900 K[/tex]
[tex] p_1= 400 kPa[/tex]
[tex] p_2= 300 kPa[/tex]
[tex] R= 0.287 kJ/(kg K)[/tex] (Individual Gas Constant for air)
For variable specific heats
[tex]s(T_2, p_2) - s(T_1, p_1) = s^0(T_2) - s^0(T_1) - R \, ln \frac{p_2}{p_1}[/tex]
where [tex] s^0(T) [/tex] is evaluated from table attached
[tex] s^0(900 K) = 2.84856 kJ/(kg K)[/tex]
[tex] s^0(300 K) = 1.70203 kJ/(kg K)[/tex]
Replacing in equation
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 2.84856 kJ/(kg K) - 1.70203 kJ/(kg K) - 0.287 kJ/(kg K) \, ln \frac{300 kPa}{400 kPa}[/tex]
[tex]s(900 K, 300 kPa) - s(300 K, 400 kPa) = 1.229 kJ/(kg K)[/tex]
During an office party, an office worker claims that a can of cold beer on his table warmed up to 20oC by picking up energy from the surrounding air, which is 25oC. Is there any truth to his claim? Explain.
Answer:
Yes
Explanation:
As we know that heat transfer take place from high temperature body to low temperature body.
In the given problem ,the temperature of the air is high as compare to the temperature of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the both body will become at the same and this stage is called thermal equilibrium.
So an office worker claim is correct.
What is the weight in pounds of a gallon of oil that has a specific gravity of .86
Answer:
Mass of oil will be 7.176 pound
Explanation:
We have given specific gravity of oil = 0.86
We know that specific gravity is given by [tex]specific\ gravity=\frac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]0.86=\frac{density\ of\ oil}{1000}[/tex]
Density of oil = [tex]860kg/m^3[/tex]
We have given volume of oil = 1 gallon
We know that 1 gallon = 0.003785 [tex]m^3[/tex]
So mass of oil = volume ×density
mass = 0.003785×860 = 3.2551 kg
We know that 1 kg = 2.2046 pound
So 3.2551 kg = 3.2551×2.2046 = 7.176 pound
What are the two types of furnaces used in steel production?
Explanation:
The two types of furnaces used in steel production are:
Basic oxygen furnace
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
Electric arc furnace
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.
A material’s chemical property could be described as it’s reaction with other chemicals (gases, liquids and solid materials). a) True b) False
Answer:
The given statement is true.
Explanation:
A chemical property of any material be it solid liquid or gas is defined as how it interacts chemically with other substances after an interaction takes place between them. The interaction in language of chemistry is known as chemical reaction. Different materials in nature show different interactions with various other substances.The 2 substances can react with each other and form a different compound or may not react with each other and are termed as inert chemicals. Infact it is this interaction between the various chemicals that we can group different into classes based on their behavior with different chemicals. The interaction of different materials act as their signatures that help us in identifying them.
A piston-cylinder assembly contains air, initially at 2 bar, 300 K, and a volume of 2 m^3. The air undergoes a process to a state where the pressure is 1 bar, during which the pressure-volume relationship is pV = constant. Assuming ideal gas behavior for the air, determine the mass of the air, in kg, and the work and heat transfer, each in kJ.
Answer:
The mass of the air is 4.645 kg.
The work done or heat transfer is 277.25 kj.
Explanation:
In isothermal process PV=constant. Take air as an ideal gas. For air gas constant is 287 j/kgK.
Given:
Initial pressure of the gas is 2 bar.
Initial volume of the gas is 2 m³.
Initial temperature is 300 K.
Final pressure is 1 bar.
Calculation:
Step1
Apply ideal gas equation for air as follows:
PV=mRT
[tex]2\times10^{5}\times2=m\times 287\times300[/tex]
m = 4.645 kg.
Thus, the mass of the air is 4.645 kg.
Step2
For isothermal process work done is same as heat transfer.
Work done or the heat transfer is calculated as follows:
[tex]W=P_{i}V_{i}ln\frac{P_{i}}{P_{f}}[/tex]
[tex]W=2\times10^{5}\times2\times ln\frac{2}{1}[/tex]
[tex]W=2.7725\times10^{5}[/tex] j
Or,
W=277.25 kj.
Thus, the work done or heat transfer is 277.25 kj.
Yield and tensile strengths and modulus of elasticity with increasing temperature. (increase/decrease/independent)
Answer:
Decrease
Explanation:
Generally with increasing the temperature the mechanical properties of material decreases.But some materials have exceptions like tempered steel because when temperature increase then young modulus of elasticity of tempered steel increases.
So we can say that when with increasing temperature the properties of materials Yield and tensile strengths and modulus of elasticity decreases.
Decrease
Answer:
Decreases
Explanation:
Yield and tensile strength and modulus of elasticity decreases with increasing temperature. As the temperature is increased most materials decrease in their elasticity.
The modulus of elasticity is proportional to its tensile strength.
How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the same size carrying the same load?
Answer:
The volume of the extra water is [tex]2.195 ft^{3}[/tex]
Solution:
As per the question:
Mass of the canoe, [tex]m_{c} = 175 lb + w[/tex]
Height of the canoe, h = 21.5 ft
Mass of the kevlar canoe, [tex]m_{Kc} = 38 lb + w[/tex]
Now, we know that, bouyant force equals the weight of the fluid displaced:
Now,
[tex]V\rho g = mg[/tex]
[tex]V = \frac{m}{\rho}[/tex] (1)
where
V = volume
[tex]\rho = 62.41 lb/ft^{3}[/tex] = density
m = mass
Now, for the canoe,
Using eqn (1):
[tex]V_{c} = \frac{m_{c} + w}{\rho}[/tex]
[tex]V_{c} = \frac{175 + w}{62.41}[/tex]
Similarly, for Kevlar canoe:
[tex]V_{Kc} = \frac{38 + w}{62.41}[/tex]
Now, for the excess volume:
V = [tex]V_{c} - V_{Kc}[/tex]
V = [tex]\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}[/tex]
A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine the force R which the elevator floor exerts on her feet and the lifting force L which she exerts on the package during the acceleration in-travel. If the elevator supports cables suddenly and completely fail, what values would R and L acquire
Answer:
force R = 846.11 N
lifting force L = 110.36 N
if cable fail complete both R and L will be zero
Explanation:
given data
mass woman mw = 60 kg
mass package mp = 9 kg
accelerates rate a = g/4
to find out
force R and lifting force L and if cable fail than what values would R and L acquire
solution
we calculate here first reaction R force
we know elevator which accelerates upward
so now by direction of motion , balance the force that is express as
R - ( mw + mp ) × g = ( mw + mp ) × a
here put all these value and a = g/4 and use g = 9.81 m/s²
R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4
R = ( 69 ) × 9.81/4 + ( 69 ) 9.81
R = 69 ( 9.81 + 2.4525 )
force R = 846.11 N
and
lifting force is express as here
lifting force = mp ( g + a)
put here value
lifting force = 9 ( 9.81 + 9.81/4)
lifting force L = 110.36 N
and
we know if cable completely fail than body move free fall and experience no force
so both R and L will be zero
What is the relation between Poisson's ration, young's modulus, shear modulus for an material?
Answer:
Explanation:
Poisson's ration= is the ratio between the deformation that occurs in the material in the direction perpendicular to the force applied with the deformation suffered by the material in the same direction of the force.
young's modulus= is the ratio between the stress applied to a material with respect to its unit deformation when this material complies with the hooke's law.
shear modulus=change in the way an elastic material experiences when subjected to shear stresses
Why should a toolpath be verified on the screen of a CAM system prior to creating the program code?
Answer:
The tool's trajectory in a CAM program refers to the places where the tool will be during the work. It is important to review it before generating the program for the following reasons
1. analyze the machining strategy and identify which one is better for each piece.
2.Avoid the collision of the tool holder with the work piece.
3.Avoid the shock of the tool with the piece.
4. Prevent the collision of the tool with elements that are not displayed on the CAM such as clamping flanges or screws.
A 4,000-km^2 watershed receives 102cm of precipitation in one
year.The avg. flow of the river draining the watershed is 34.2
m^3/s.Infiltration is est. to be 5.5 x 10^(-7) cm/s
andevapotranspiration is est. to be 40 cm/y. Determine the change
instorage in the watershed over one year. The ratio of runoff
toprecipitation (both in cm) is termed the runoff
coefficient.Compute the runoff coefficient for this
watershed.
Answer:
1) The change in storage of the catchment is 707676800 cubic meters.
2) The runoff coefficient of the catchment is 0.83.
Explanation:
The water budget equation of the catchment can be written as
[tex]P+Q_{in}=ET+\Delta Storage+Q_{out}+I[/tex]
where
'P' is volume of precipitation in the catchment =[tex]Area\times Precipitation[/tex]
[tex]Q_{in}[/tex] Is the water inflow
ET is loss of water due to evapo-transpiration
[tex]\Delta Storage[/tex] is the change in storage of the catchment
[tex]Q_{out}[/tex] is the outflow from the catchment
I is losses due to infiltration
Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get
[tex]4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365[/tex]
[tex]\therefore \Delta Storage=707676800m^3[/tex]
Part b)
The runoff coefficient C is determined as
[tex]C=\frac{P-I}{P}[/tex]
where symbols have the usual meaning as explained earlier
[tex]\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83[/tex]