Answer is given below
Explanation:
Agreement is a official contract. It is written form or even in orally form. The Agreement can be written in formal or informal terms or we can use purely verbal language.
Agreement made between two or more party that allow the court to decide.
The main 6 elements are:
1. Offer
2. acceptance
3. consideration
4. intention to create legal relation
5. certainty
6. capacity
1. The first elements of the contract herein are ABC Sanctions Inc., without the offer, as it is not valid under the Contract Act 1950, Contract Act.
2. Once the offer is made in the contract, acceptance must take place. The agreement must be approved by Northern Inc. When Northern Inc is clear with the offer, it will accept it once the terms and conditions of the agreement are clear.
3. Contrarification is the most important aspect of a contract, when considering a contract, the other person will give something in return. It is considered an exchange between ABC Construction Inc and Northern Inc.
4. It is necessary to have these elements in the contract. Contract law 1950 is one of the requirements of a valid contract, although there is silence about the need for a legal relationship.
5. Another important aspect of the contract is of course. The Contract Agreement sets out the terms and conditions that must be clearly understood by both ABC Contracts Inc and Northern Inc.
6. The ability of a contract to have a legal capacity on either side of the contract is more than eighteen years, since the age of 18 years is specified as the age at which the contract is entered into.
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects calculate the viscosities of the oils.
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)
Answer:
Exit temperature = 32°C
Explanation:
We are given;
Initial Pressure;P1 = 100 KPa
Cp =1000 J/kg.K = 1 KJ/kg.k
R = 500 J/kg.K = 0.5 Kj/Kg.k
Initial temperature;T1 = 27°C = 273 + 27K = 300 K
volume flow rate;V' = 15 m³/s
W = 130 Kw
Q = 80 Kw
Using ideal gas equation,
PV' = m'RT
Where m' is mass flow rate.
Thus;making m' the subject, we have;
m' = PV'/RT
So at inlet,
m' = P1•V1'/(R•T1)
m' = (100 × 15)/(0.5 × 300)
m' = 10 kg/s
From steady flow energy equation, we know that;
m'•h1 + Q = m'h2 + W
Dividing through by m', we have;
h1 + Q/m' = h2 + W/m'
h = Cp•T
Thus,
Cp•T1 + Q/m' = Cp•T2 + W/m'
Plugging in the relevant values, we have;
(1*300) - (80/10) = (1*T2) - (130/10)
Q and M negative because heat is being lost.
300 - 8 + 13 = T2
T2 = 305 K = 305 - 273 °C = 32 °C
Decompose the signal (1+0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequency, and phase of each component. Hint: use the identity for cosacosb.
Answer:
amplitudes : 1 , 0.05, 0.05
frequencies : 50/[tex]\pi[/tex], 105/[tex]2\pi[/tex], 95/2[tex]\pi[/tex]
phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]
Explanation:
signal s(t) = ( 1 + 0.1 cos 5t )cos 100t
signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb
s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]
s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t
= cos100t + 0.05cos(105)t + 0.05cos 95t
= cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]
= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex] + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )
attached is the remaining part of the solution
A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.
Answer:
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Explanation:
A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:
[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]
Output power is cleared:
[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]
If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:
[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\dot W_{out} = 1752.825\,kW[/tex]
The power that fluid supplies to the turbine is 1752.825 kilowatts.
An undersea research chamber is spherical with an external diameter of 3.50 mm . The mass of the chamber, when occupied, is 21700 kg. It is anchored to the sea bottom by a cable. Find the followings
Required:
a. The buoyant force on the chamber.
b. The tension in the cable?
Answer:
a. The buoyant force on the chamber is 220029.6 N
b. The tension in the cable is 7369.6 N
Explanation:
The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg
Given;
diameter of the sphere, d = 3.50 m
radius of the sphere, r = 1.75 mm = 1.75 m
mass of the chamber, m = 21700 kg
density of water, ρ = 1000 kg/m³
(a)
Buoyant force is the weight of water displaced, which is calculated as;
Fb = ρvg
where;
v is the volume of sphere, calculated as;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]
Fb = 1000 x 22.452 x 9.8
Fb = 220029.6 N
(b)
The tension in the cable will be calculated as;
T = Fb - mg
T = 220029.6 N - (21700 x 9.8)
T = 220029.6 N - 212660 N
T = 7369.6 N
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:
a. The oil mean temperature.
b. The centerline temperature.
c. The axial gradient of the mean temperature.
d. The heat transfer coefficient.
Answer:
(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K
Explanation:
Solution
Given that:
The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s
Diameter of the tube, D = 1cm =0.01 m
Electrical heat rate, q =76 W/m
Wall Temperature, Ts = 370 K
Now,
From the properties table of engine oil we can deduce as follows:
thermal conductivity, k =0.139 W/m .K
Density, ρ = 854 kg/m³
Specific heat, cp = 2120 J/kg.K
(a) Thus
The wall heat flux is given as follows:
qs = q/πD
=76/π *0.01
= 2419.16 W/m²
Now
The oil mean temperature is given as follows:
Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)
Tb =370 - 11/24 * (2419.16 * 0.005/0.139)
Tb = 330.12 K
(b) The center line temperature is given below:
Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)
Tc =304.73 K
(c) The flow velocity is given as follows:
V = m/ρ (πR²)
Now,
The The axial gradient of the mean temperature is given below:
dTb/dx = 2 *qs/ρ *V*cp * R
=2 *qs/ρ*[m/ρ (πR²) *cp * R
=2 *qs/[m/(πR)*cp
dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120
dTb/dx = 19.81 K/m
(d) The heat transfer coefficient is given below:
h =48/11 (k/D)
=48/11 (0.139/0.01)
h =60.65 W/m². K
Waste cooking oil is to be stored for processing by pouring it into tank A, which is connected by a manometer to tank B. The manometer is completely filled with water. Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa. To what height h can waste oil be poured into tank A? If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height?
KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.
==================================
Answer:
(1). 1.2 metres.
(2). There is going to be the same pressure.
Explanation:
From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;
" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."
=> Also, the density of oil = 930
That is if Pressure, P in B > 18kpa there will surely be a burst.
The height, h the can waste oil be poured into tank A is;
The maximum pressure = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).
18 × 10^3 = (height, h × 10 × 930) + 10 × (2 - 1.25) × 1000.
When we make height, h the Subject of the formula then;
Approximately, Height, h = 1.2 metres.
(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.
Consider a refrigerator that consumes 400 W of electric power when it is running. If the refrigerator runs only one-quarter of the time and the unit cost of electricity is $0.13/kWh, what is the electricity cost of this refrigerator per month (30 days)
Answer:
Electricity cost = $9.36
Explanation:
Given:
Electric power = 400 W = 0.4 KW
Unit cost of electricity = $0.13/kWh
Overall time = 1/4 (30 days) (24 hours) = 180 hours
Find:
Electricity cost
Computation:
Electricity cost = Electric power x Unit cost of electricity x Overall time
Electricity cost = 0.4 x $0.13 x 180
Electricity cost = $9.36
Given:
Electric power = 400 W = 0.4 KW
Over all Time = 30(1/4) = 7.5 days
Unit cost of electricity = $0.13/kWh
Find:
Electricity cost.
Computation:
Electricity cost = Electric power x Unit cost of electricity x Over all Time
Electricity cost = 0.4 x 0.13 x 7.5
Electricity cost = $
Q#1: Provide an example of a software project that would be amenable to the following models. Be specific. a. Waterfall b. Prototype c. Extreme Programming
Answer:
Waterfall model
Explanation:
The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.
The waterfall model is very easy to use. This is the earliest approach of the SDLC.
There are different phase of the waterfall:
Requirement analysisSystem DesignImplementationTestingDeploymentMaintenanceSteam is contained in a closed rigid container which has a volume of 2 initially the the pressure and the temperature is the remeraturedrops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m3, occupied by saturated liquid at the final state?
The given question is incomplete. The complete question is as follows.
Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],
(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?
Explanation:
Using the property tables
[tex]T_{1} = 500^{o}C[/tex], [tex]P_{1}[/tex] = 10 bar
[tex]v_{1} = 0.354 m^{3}/kg[/tex]
(a) During the process, specific volume remains constant.
[tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]
T = [tex](150 - 160)^{o}C[/tex]
Using inter-polation we get,
T = [tex]154.71^{o}C[/tex]
The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].
(b) When the system will reach at state 3 according to the table at 0.5 bar then
[tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]
[tex]v_{g} = 3.24 m^{3} kg[/tex]
Let us assume "x" be the gravity if stream
[tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]
[tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]
= [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]
= 0.109
At state 3, the fraction of total mass condensed is as follows.
[tex](1 - x_{5})[/tex] = 1 - 0.109
= 0.891
The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.
(c) Hence, total mass of the system is calculated as follows.
m = [tex]\frac{v}{v_{1}}[/tex]
= [tex]\frac{1}{0.354}[/tex]
= 2.825 kg
Therefore, at final state the total volume occupied by saturated liquid is as follows.
[tex]v_{ws} = m \times v_{f}[/tex]
= [tex]2.825 \times 0.00103[/tex]
= [tex]2.9 \times 10^{-3} m^{3}[/tex]
The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].
The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T is that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa (50 ksi ) is exposed to a stress of 2023 MPa (293400 psi). Assume that the parameter Y has a value of 1.14. (a) If the largest surface crack is 0.2 mm (0.007874 in.) long, determine the critical stress .
Answer:
Explanation:
The formula for critical stress is
[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]
[tex]\sigma_c =\texttt{critical stress}[/tex]
K is the plane strain fracture toughness
Y is dimensionless parameters
We are to Determine the Critical stress
Now replacing the critical stress with 54.8
a with 0.2mm = 0.2 x 10⁻³
Y with 1
[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]
The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa
Therefore, the specimen does not failure for surface crack of 0.2mm
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, ν = 0.30). Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon, (c) the thickness of portion AB, (d) the cross- sectional area of portion AB.
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
PART (a)For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
PART (b)
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
PART(c)
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
PART (d)
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²
What is the stress and strain in the plate?Let us first find the cross sectional area of AB from the image attached;
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
We are given;
Tensile Load; F = 2.75 kN = 2.75 × 10³ N
Horizontal Stress is calculated from the formula;
σₓ = F/A
σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m
σₓ = 143.23 × 10⁶ Pa
Horizontal Strain is calculated from;
εₓ = σₓ/E
We are given E = 200 GPa = 200 × 10⁹ Pa
Thus;
εₓ = (143.23 × 10⁶)/(200 × 10⁹)
εₓ = 716.15 × 10⁻⁶
Formula for Vertical Strain is;
ε_y = -ν * εₓ
We are given ν = 0.30. Thus;
ε_y = -(0.3) * (716.15 × 10⁻⁶)
ε_y = -214.84 × 10⁻⁶
A) We are given;
Gauge Length; L = 0.05m
Change in gauge length is gotten from;
Δx = L * εₓ
Δx = 0.05 × 716.15 × 10⁻⁶
Δx = 35.808 × 10⁻⁶ m
B) From the attached diagram, the width is;
W = 0.012m
Change in width is;
Δy = W * ε_y
Δy = 0.012 * -214.84 × 10⁻⁶
Δy = -2.5781 × 10⁻⁶m
C) We are given;
Thickness of plate; t = 1.6 mm = 0.0016m
Change in thickness;
Δ_z = t * ε_z
where;
ε_z = ε_y
Thus;
Δ_z = t * ε_y
Δ_z = 0.0016 * -214.84 × 10⁻⁶
Δ_z = -343.74 × 10⁻⁹m
D) The change in cross sectional area is gotten from;
ΔA = A_final - A_initial
From calculating the areas, we have;
A = -8.25 × 10⁻⁹ m²
Read more about stress and strain in steel plates at; https://brainly.com/question/1591712
An 60-m long wire of 5-mm diameter is made of steel with E = 200 GPa and ultimate tensile strength of 400 MPa. If a factor of safety of 3.2 is desired, determine (a) the allowable tension in the wire (b) the corresponding elongation of the wire
Answer:
a) 2.45 KN
b) 0.0375 m
Explanation:
[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]
[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]
[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]
(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]
As an engineer who has just finished taking engineering materials course, your first task is to investigate the causes of an automobile accident. Your findings show that the right rear wheel has broken off at the axle. The axle is bent. The fracture surface reveals a Chevron pattern pointing toward the surface of the axle. Suggest a possible cause for the fracture and why?
Answer is given below
Explanation:
Evidence shows that the axle was not broken before the accident, while the clumsy axle meant that the wheel was still attached when the load was applied. This indicates that the Chevron prototype wheel suffered a severe impact shock, which caused the failure of the transmission to the axle. Preliminary evidence suggests that the driver lost control and crashed. Further examination of the surface, microstructure and structure and characteristics of the fracture can be modified if the axle is properly preparedCathy works in a welding shop. While working one day, a pipe falls from scaffolding above and lands on her head, injuring her. Cathy complains to OSHA, but the company argues that because it has a "watch out for falling pipe" sign in the workplace that it gave fair warning. It also says that if Cathy wasn’t wearing a hardhat that she is responsible for her own injury. Which of the following is true?1. Common law rules could hold Cathy responsible for her own injury.2. Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.3. OSHA rules can hold Cathy’s employer responsible for not maintaining a hazard-free workplace.4. More than one answer is correct.
Answer:1 common law
Explanation:
It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.
What are OSHA rules?In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.
According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.
You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.
Therefore, more than one answer is correct.
Learn more about OSHA, here:
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2. The block is released from rest at the position shown, figure 1. The coefficient of
kinetic friction over length ab is 0.22, and over length bc is 0.16. Using the
principle of work and energy, find the velocity with which the block passes
position c.
Answer:
Velocity = 4.73 m/s.
Explanation:
Work done by friction is;
W_f = frictional force × displacement
So; W_f = Ff * Δs = (μF_n)*Δs
where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24
Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;
mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).
Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;
ΔKE2 + (0.16)(mg cos 24)(2).
Plugging in the relevant values, we have;
1.22mg = ΔKE1 + 0.603mg
ΔKE1 = 1.22mg - 0.603mg
ΔKE1 = 0.617mg
Also,
0.813mg = ΔKE2 + 0.292mg
ΔKE2 = 0.813mg - 0.292mg
ΔKE2 = 0.521mg
Now total increase in Kinetic Energy is ΔKE1 + ΔKE2
Thus,
Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg
Putting 9.81 for g to give;
Total increase in kinetic energy = 11.164m
Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m
m cancels out to give; ½v² = 11.164
v² = 2 × 11.164
v² = 22.328
v = √22.328
v = 4.73 m/s.
9. A Co has 500,000 total shares outstanding and each share is priced at 20$. B Co has 300,000 total shares outstanding and each share is priced at 40$. You have 100 shares in A Co and 200 shares in B Cos. After consolidation how many new shares you will own in consolidated AB Co?
Answer:
In consolidated AB Co 300 shares.
Explanation:
Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.
If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respectively, determine the resultant moment produced by all the weights about point A.
Answer:
hello the required diagram is missing attached to the answer is the required diagram
7.9954 kip.ft
Explanation:
AB = 1550-Ib ( weight acting on AB )
BCD = 190 - Ib ( weight of cage )
169-Ib = weight of man inside cage
Attached is the free hand diagram of the question
calculate distance [tex]x![/tex]
= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]
[tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft
calculate distance x
= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]
x = 30 * cos 75⁰ = 7.765 ft
The resultant moment produced by all the weights about point A
∑ Ma = 0
Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )
Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )
= 4014.5 + 1950.35 + 2030.535
= 7995.385 ft. Ib ≈ 7.9954 kip.ft
5. Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm3, respectively. Atomic weight of Ge is 72.64 g/mol
Answer:
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
Explanation:
The masses of silicon and germanium contained in a cubic centimeter of the germanium-silicon alloy by apply the concepts of mass ([tex]m[/tex]), density ([tex]\rho[/tex]) and volume ([tex]V[/tex]), as well as the mass-mass proportion of Germanium ([tex]x[/tex]):
[tex]m_{Ge} = x \cdot \rho_{Ge}\cdot V_{sample}[/tex]
[tex]m_{Ge} = 0.15\cdot \left(5.32\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{3})[/tex]
[tex]m_{Ge} = 0.798\,g[/tex]
The amount of moles of Germanium is obtained after dividing previous outcome by its atomic weight. That is to say:
[tex]n = \frac{m_{Ge}}{M_{Ge}}[/tex]
[tex]n = \frac{0.798\,g}{72.64\,\frac{g}{mol} }[/tex]
[tex]n = 0.011\,mol[/tex]
There are 0.011 moles in a cubic centimeter of the germanium-silicon alloy. According to the Law of Avogadro, there are [tex]6.022 \times 10^{23}\,atoms[/tex] in a mole of Germanium. The quantity of atoms in a cubic centimeter is therefore found by simple rule of three:
[tex]y = \frac{0.011\,mol}{1\,mol}\times \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)[/tex]
[tex]y = 6.624 \times 10^{21}\,atoms[/tex]
There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.
An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.
Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?
Answer:
A.) 7.13 degree north east
B.) 806.23 km/h
C.) 7.44 hours
D.) 0.06 hours
Explanation:
Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction
Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.
A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula
Tan Ø = y/x
Substitute y = 100 km/h and x = 800km/h
Tan Ø = 100/800
Tan Ø = 0.125
Ø = Tan^-1(0. 125)
Ø = 7.13 degrees north east.
Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east
B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem
Speed = sqrt ( 800^2 + 100^2)
Speed = sqrt (650000)
Speed = 806.23 km/h
C.) Let us use the speed formula
Speed = distance / time
Substitute the speed and distance into the formula
806.23 = 6000/ time
Make Time the subject of formula
Time = 6000/806.23
Time = 7.44 hours
D.) If there is no cross wind,
Time = 6000/800
Time = 7.5 hour
Time difference = 7.5 - 7.44
Time difference = 0.06 hours
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = [tex]\frac{A*L}{Vo}[/tex] Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s
since the reaction is in first order
X = 1 - [tex]e^{-kt}[/tex]
[tex]e^{-kt}[/tex] = 1 - X
kt = In [tex]\frac{1}{1-X}[/tex]
k = In [tex]\frac{1}{1-X}[/tex] / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156 [tex]s^{-1}[/tex]
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]
therefore
[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
A piston-cylinder device initially at 0.45-m3 contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.2 m3. The work done on the gas during this compression process is _____ kJ.
Answer:
219kJ
Explanation:
The work done (W) on a gas in an isothermal process is given by;
W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex] -----------------(i)
Where;
P₁ = initial pressure of the gas
V₁ = initial volume of the gas
V₂ = final volume of the gas
From the question;
P₁ = 600kPa = 6 x 10⁵Pa
V₁ = 0.45m³
V₂ = 0.2m³
Substitute these values into equation (i) as follows;
W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]
W = -6 x 10⁵ x 0.45 x ln (0.444)
W = -6 x 10⁵ x 0.45 x -0.811
W = 2.19 x 10⁵
W = 219 x 10³
W = 219kJ
Therefore, the work done on the gas during the compression process is 219kJ
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You close the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
The velocity field of a flow is given by V = 2x2 ti +[4y(t - 1) + 2x2 t]j m/s, where x and y are in meters and t is in seconds. For fluid particles on the x-axis, determine the speed and direction of flow
Answer:
Explanation:
The value of a will be zero as it is provided that the particle is on the x-axis.
Calculate the velocity of particles along x-axis.
[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]
Substitute 0 for y.
[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]
Here,
[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]
Calculate the magnitude of vector V .
[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]
Substitute
[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]
[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]
The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]
Calculate the direction of flow.
[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]
Here, θ is the flow from positive x-axis in a counterclockwise direction.
Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.
[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]
The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K
Answer:
The exit temperature of the gas = 32° C
Explanation:
Solution
Given that:
Inlet temperature T₁ = 27°C ≈ 300.15 K
Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa
Volume flow rate , V = 15 m/s³
Diameter of the deduct, D = 500 mm = 0.5 m
Electric heater power, W heater = 130 kW = 130 * 10^3 W
The heat lost Q = 80 kW = 80 * 10^3 W
Now,
From the ideal gas law, density of the air at the inlet is given as :
ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300
=0.6667 kg/m³
The mass flow rate through the duct is computed below:
m = ρ₁ V = 0.6667 * 15 = 10 kg/s
Thus
Applying the first law of thermodynamics to the process is shown below:
Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)
So,
If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:
Q + m (h₁) = m (h₂) + W
or
Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)
Thus
h₂ - h₁ = Cp T₂ - T₁
Now by method of substitution the known values are:
(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)
Note: The heat transfer is taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas
So,
Solving for T₂,
T₂ = 32° C
Therefore the exit temperature of the gas = 32° C
When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus _____ percent of the areas of smaller flue collar outlets
Answer:
Fifty (50) percent. [50%]
Explanation:
Water heater is a home appliance that comprises of an electric or gas heating unit as well as a water-tank where water is heated and stored for use.
When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus fifty (50) percent of the areas of smaller flue collar outlets.
A water heater is primarily vented with an approved and standardized plastic or metallic pipe such as flue or chimney, which allows gas to flow out of the water heater into the surrounding environment.
For a draft hood-equipped water heater, both the water heater and the barometric draft regulators must be installed in the same room. Also, the technician should ensure that the vent is through a concealed space such as conduit and should be labeled as Type L or Type B.
The minimum capacity of a water heater should be calculated based on the number of bathrooms, bedrooms and its first hour rating.
It is proposed to use water instead of refrigerant-134a as the working fluid in air-conditioning applications where the minimum temperature never falls below the freezing point. Can water be used as the working fluid in air-conditioning applications?
Answer:
No, water can't be used.
Explanation:
No, water cannot be used as the working fluid in air-conditioning applications.
This is because, if we assume the water is maintained at 10°C in the evaporator, the evaporator pressure will now be the saturation pressure that corresponds to this pressure, which in this case would be 1.2 kPa.
So we can conclude that for the refrigerants in the evaporator the temperature of a saturated pressure would be very low and so it's not practical to maintain it with water
Thus, it's is not practical to design refrigeration or air conditioning devices with water as the working fluid because it will involve extremely low pressures.
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile. (2 marks) d) Find the speed of the automobile at the time it was overtaken by the police car. (2 marks)
Answer:
A.) Time = 13.75 seconds
B.) Total distance = 339 m
C.) V = 11.18 m/s
D.) V = 10.2 m/s
Explanation: Given that the automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone.
Then,
Initial velocity U of the motorist = 15.65m/s
acceleration a = - 3.05 m/s^2
Initial velocity u of the police man = 11.18 m/s
Acceleration a = 1.96 m/s^2
The police will overtake at distance S as the motorist decelerate and come to rest.
Where V = 0 and a = negative
While the police accelerate.
Using 2nd equation of motion for the motorist and the police
S = ut + 1/2at^2
Since the distance S covered will be the same, so
15.65t - 1/2×3.05t^2 = 11.18t +1/2×1.96t^2
Solve for t by collecting the like terms
15.56t - 1.525t^2 = 11.18t + 0.98t^2
15.56t - 11.18t = 0.98t^2 + 1.525t^2
4.38t = 2.505t^2
t = 4.38/2.505
t = 1.75 seconds approximately
But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit.
Therefore, the total time required for the police car to overtake the automobile will be:
12 + 1.75 = 13.75 seconds
B.) Using the same formula
S = ut + 1/2at^2
Where S = total distance travelled
Substitutes t into the formula
S = 11.18(13.75) + 1/2 × 1.96 (13.75)^2
S = 153.725 + 185.28
S = 339 m approximately
C.) The speed of the police car at the time it overtakes the automobile will be constant = 11.18 m/s
D.) Using first equation of motion
V = U - at
Since the motorist is decelerating
V = 15.65 - 3.05 × 1.75
V = 15.65 - 5.338
V = 10.22 m/s
Therefore, the speed of the automobile at the time it was overtaken by the police car is 10.2 m/ s approximately