Answer:
(a) The ball's initial speed is 8.4 m/s
(b) The height of the ball is 3.28 m
Explanation:
Given;
maximum height of the ball, H = 3.6 m
Apply conservation of energy;
¹/₂mu² + mgh = ¹/₂mv² + mgH
Where;
m is the mass of the ball
u is the initial velocity of the ball
v is the final velocity of the ball
h is the initial height of the ball
H is the maximum height of the ball
(a) The ball's initial speed, u;
¹/₂mu² + mgh = ¹/₂mv² + mgH
¹/₂u² + gh = ¹/₂v² + gH
make u the subject of the formula
[tex]u = \sqrt{v^2 +2gH -2gh} \\\\[/tex]
at maximum height, the final velocity = 0
maximum height H = 3.6 m
initial height, h = 0
g is acceleration due to gravity = 9.8 m/s²
[tex]u = \sqrt{0^2 +2*9.8*3.6 -2*9.8*0} \\\\u = \sqrt{2*9.8*3.6} \\\\u = 8.4 \ m/s[/tex]
(b) The height of the ball when it has a speed of 2.5 m/s
v = 2.5 m/s
[tex]u = \sqrt{v^2 +2gH -2gh} \\\\u^2 = v^2 +2gH -2gh \ (h = 0)\\\\u^2 = v^2 +2gH\\\\2gH = u^2 - v_2\\\\H = \frac{u^2 - v^2}{2g} \\\\H = \frac{8.4^2 - 2.5^2}{2*9.8}\\\\H = 3.28 \ m[/tex]
A) The ball's initial speed is : 8.4 m/s
B) The height of the ball when it has a speed of 2.5 m/s is : 3.28 m
Given data :
Maximum height reached by ball = 3.60 m
applying conservation of energy relation
¹/₂mu² + mgh = ¹/₂mv² + mgH -------- ( 1 )
where : u = initial velocity , v = final velocity , H = max height , h = initial height
A) Determine the ball initial speedGiven that the value of m is not given equation ( 1 ) becomes
¹/₂u² + gh = ¹/₂v² + gH --- ( 2 )
solve for u ( initial velocity )
u = [tex]\sqrt{v^2 +2gH -2gh}[/tex] ----- ( 3 )
where : g = 9.8 m/s², v = 0 , H = 3.60 m , h = 0
insert values into equation ( 3 )
u ( initial speed of ball ) = 8.4 m/s
B) Determine the height of the ball when it has a speed of 2.5m/sapplying equation ( 3 )
u = [tex]\sqrt{v^2 +2gH -2gh}[/tex]
where : u = 8.4 m/s , v = 2.5 m/s , H = ? , h = 0
solve for H
H = u² - v² / 2g
= ( 8.4² - 2.5² ) / ( 2 * 9.8 )
= 3.28 m
Hence we can conclude that The ball's initial speed is : 8.4 m/s and The height of the ball when it has a speed of 2.5 m/s is : 3.28 m.
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Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW.
Answer:
(a) 48.2 %
(b) 0.4137
(c) 2385.9 kW
Explanation:
The given values are:
Initial pressure,
p₁ = 100 kPa
Initial temperature,
T₁ = 300 K
Mass,
M = 6 kg/s
Pressure ration,
r = 10
Inlent temperature,
T₃ = 1400 K
Specific heat ratio,
k = 1.4
At T₁ and p₁,
⇒ [tex]c_{p}=1.005 \ KJ/Kg.K[/tex]
Process 1-2 in isentropic compression, we get
⇒ [tex]\frac{T_{2}}{T_{1}}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}[/tex]
[tex]T_{2}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}. T_{1}[/tex]
On putting the estimated values, we get
[tex]=(10)^{\frac{1.4-1}{1.4}}(300)[/tex]
[tex]=579.2 \ K[/tex]
Process 3-4,
⇒ [tex]\frac{T_{4}}{T_{3}}=(\frac{p_{4}}{p_{3}})^{\frac{k-1}{k}}[/tex]
[tex]T_{4}=(\frac{1}{10})^{\frac{1.4-1}{1.4}}(1400)[/tex]
[tex]=725.13 \ K[/tex]
(a)...
The thermal efficiency will be:
⇒ [tex]\eta =\frac{\dot{W_{t}}-\dot{W_{e}}}{\dot{Q_{in}}}[/tex]
[tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
⇒ [tex]\dot{Q_{in}}=\dot{m}(h_{1}-h_{2})[/tex]
[tex]=\dot{mc_{p}}(T_{3}-T_{2})[/tex]
[tex]=6\times 1005\times (1400-579.2)[/tex]
[tex]=4949.4 \ kJ/s[/tex]
⇒ [tex]\dot{Q_{out}}=\dot{m}(h_{4}-h_{1})[/tex]
[tex]=6\times 1.005\times (725.13-300)[/tex]
[tex]=2563.5 \ KJ/S[/tex]
As we know,
⇒ [tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]
On putting the values, we get
[tex]=1-\frac{2563.5}{4949.4}[/tex]
[tex]=0.482 \ i.e., \ 48.2 \ Percent[/tex]
(b)...
Back work ratio will be:
⇒ [tex]bwr=\frac{\dot{W_{e}}}{\dot{W_{t}}}[/tex]
Now,
⇒ [tex]\dot{W_{e}}=\dot{mc_{p}}(T_{2}-T_{1})[/tex]
On putting values, we get
[tex]=6\times 1.005\times (579.2-300)[/tex]
[tex]=1683.6 \ kJ/s[/tex]
⇒ [tex]\dot{W_{t}}=\dot{mc_{p}}(T_{3}-T_{4})[/tex]
[tex]=6\times 1.005\times (1400-725.13)[/tex]
[tex]=4069.5 \ kJ/s[/tex]
So that,
⇒ [tex]bwr=\frac{1683.6}{4069.5}=0.4137[/tex]
(c)...
Net power is equivalent to,
⇒ [tex]\dot{W}_{eyele}=\dot{W_{t}}-\dot{W_{e}}[/tex]
On substituting the values, we get
[tex]= 4069.5-1683.6[/tex]
[tex]=2385.9 \ kW[/tex]
Following are the solution to the given points:
Given :
Initial pressure [tex]p_1 = 100\ kPa \\\\[/tex]
Initial temperature [tex]T_1 = 300\ K \\\\[/tex]
Mass flow rate of air [tex]m= 6\ \frac{kg}{s}\\\\[/tex]
Compressor pressure ratio [tex]r =10\\\\[/tex]
Turbine inlet temperature [tex]T_3 = 1400\ K\\\\[/tex]
Specific heat ratio [tex]k=1.4\\\\[/tex]
Temperature [tex]\ T_1 = 300\ K[/tex]
pressure [tex]p_1 = 100\ kPa\\\\[/tex]
[tex]\to c_p=1.005\ \frac{kJ}{kg\cdot K}\\\\[/tex]
Process 1-2 is isen tropic compression
[tex]\to \frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \\\\[/tex]
[tex]\to T_2=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \ T_1 \\\\[/tex]
[tex]=(10)^{\frac{1.4-1}{1.4}} (300)\\\\ =(10)^{\frac{0.4}{1.4}} (300) \\\\[/tex]
[tex]\to T_2 = 579.2\ K \\\\[/tex]
Process 3-4 is isen tropic expansion
[tex]\to \frac{T_4}{T_3}=(\frac{P_4}{P_3})^{\frac{k-1}{k}}\\\\ \to T_4=(\frac{1}{10})^{\frac{1.4-1}{1.4}} (1400)\\\\\to T_4= 725.13\ K \\\\[/tex]
For point a:
The thermal efficiency of the cycle:
[tex]\to \eta = \frac{W_i-W_e}{Q_{in}} \\\\\to \eta = \frac{Q_{in}- Q_{out}}{Q_{in}}\\\\\to \eta =1 - \frac{Q_{out}}{Q_{in}} \\\\\to Q_{in}= m(h_3-h_1) = mc_p (T_4-T_1) =(6)(1.005)(725.13-300) = 2563 \ \frac{kJ}{S}\\\\\to \eta =1- \frac{Q_{out}}{Q_{in}}\\\\[/tex]
[tex]=1-\frac{2563.5}{4949.4}\\\\ = 0.482\\\\[/tex]
[tex]\eta = 48.2\%\\\\[/tex]
For point b:
The back work ratio
[tex]\to bwr =\frac{W_e}{W_t}[/tex]
Now
[tex]\to W_e =mc_p (T_2 -T_1)[/tex]
[tex]=(6) (1.005)(579.2 -300)\\\\ =1683.6 \ \frac{kJ}{S}\\\\[/tex]
[tex]\to W_t=mc_p(T_3-T_4)[/tex]
[tex]=(6)(1.005)(1400 - 725.13)\\\\ = 4069.5 \frac{KJ}{s}[/tex]
[tex]\to bwr =\frac{W_s}{W_t}= \frac{1683.6}{4069.5}=0.4137[/tex]
For point c:
The net power developed is equal to
[tex]\to W_{cycle} = W_t-W_e \\\\[/tex]
[tex]= ( 4069.5-1683.6)\\\\ = 2385.9 \ kW\\[/tex]
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The column is constructed from high-strength concrete and eight A992 steel reinforcing bars. If the column is subjected to an axial force of 200 kip.
a) Determine the average normal stress in the concrete and in each bar. Each bar has a diameter of 1 in.
b) Determine the required diameter of each bar so that 60% of the axial force is carried by concrete.
Answer:
d= 2.80inch
Explanation:
Given:
Axial force= 30kip
d= 1inch
CHECK THE ATTACHMENT FOR DETAILED EXPLANATION
A) The average normal stress in the concrete and in each bar are; σ_st = 15.52 kpi ; σ_con = 2.25 kpi
B) The required diameter of each bar so that 60% of the axial force is carried by concrete is; 0.94 inches
Concrete Column Design
We are told that;
Column has eight A992 steel reinforcing bars.
Column is subjected to an axial force of 200 kip.
A) Diameter of each bar is 1 inch.
Using equations of equilibrium, we have;
∑fy = 0;
8P_st + P_con = 200 ------(eq 1)
Using compatibility concept, we know from the image attached that;
δ_st = δ_con
where δ_st is change in length of steel and δ_con is change in length of concrete.
Thus;
δ_st = (P_st * L)/(A_st * E_st)
where;
P_st is tensile force of steel
L is length of steel = 3 ft = 36 inches
A_st is area of steel = π/4 * 1² = 0.7854 in²
E_st is young's modulus of steel = 29000 ksi
Similarly;
δ_con = (P_con * L)/(A_con * E_con)
where;
P_con is tensile force of concrete
L is length of concrete = 3 ft = 36 inches
E_con is young's modulus of concrete = 4200 ksi
A_con is area of concrete with diameter of 8 inches = (π/4 * 8²) - 6(π/4 * 1²) = 45.5531 in²
Thus;
From δ_st = δ_con;
(P_st * 36)/(0.7854 * 29000) = (P_con * 36)/(45.5531 * 4200)
Solving this gives;
P_st = 0.119P_con -----(eq 2)
Put 0.119P_con for P_st in eq 1 to get;
8(0.119P_con) + P_con = 200
1.952P_con = 200
P_con = 102.459 kip
Thus; P_st = 12.193 kip
Thus, average normal stress is;
Steel; σ_st = P_st/A_st
σ_st = 12.193/0.7854
σ_st = 15.52 kpi
Concrete; σ_con = P_con/A_con
σ_con = 102.459/45.5531
σ_con = 2.25 kpi
B) Since 60% of the axial force is carried by the concrete. Then it means that 40% will be carried by the steel.
Thus;
P_con = 60% * 200 = 120 kip
P_st = 40% * 200 = 80 kip
Using compatibility again;
δ_st = δ_con
Thus;
(P_st * L)/(A_st * E_st) = (P_con * L)/(A_con * E_con)
6(π/4 * d²)) = (80 * ((π/4 * 8²) - 6(π/4 * d²)) * 4200)/(120 * 29000)
⇒ 4.712d² = 0.09655(50.2655 - 4.712d²)
⇒ 4.712d²/0.09655 = 50.2655 - 4.712d²
⇒ 48.8037d² = 50.2655 - 4.712d²
Solving this gives;
d = 0.94 inches
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For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe , the velocity of fluid, the density of fluid and the viscosity of the fluid. Show that = ∅ ൬ ൰
Answer:
Explanation:
La vaca
El pato
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
After impact testing a sample at -100oC you realize that the fracture surface is very dull and fibrous. Is the sample behaving in a ductile of brittle manner at this temperature
Answer:
Ductile
Explanation:
So, from the question, we have the following information or parameters or data which is going to help us in solving this particular problem or question;
=> " impact testing a sample = -100oC shows that the fracture surface is very DULL AND FIBROUS"
TAKE NOTE: DULL AND FIBROUS.
IMPACT TESTING is used by engineers in the configuration of a sample or object.
In order to determine whether a specimen is ductile or brittle, it can be shown from its appearance for instance;
A DUCTILE SAMPLE will be DULL AND FIBROUS thus, our answer!
But a brittle sample will have a crystal shape.
Define Engineering Economy and explain the foundation of Engineering Economy in terms of seven basic principles.
Answer:
Check Explanation.
Explanation:
ENGINEERING ECONOMY:
In a simple way, Engineering Economy simply refers to the study of Economics which is related to engineers that is the study of Economic decisions by people in the engineering field. The study of Engineering Economy is very important because Engineering is a major manufacturing part in every country's economy.
With the study of Economics by Engineering that is Engineering Economy, engineers can make rational decisions after seeing alternatives.
The foundation of Engineering Economy in terms of seven basic principles:
(A). Creation of Alternatives: there will always be a problem and every problem had one or more solutions. When a problem has been seen as a problem alternative solutions come in.
(B). Differences in the Alternatives : this part is when engineers makes the best decision(choice) among alternates.
(C). Your viewpoint should be consistent: consistency is power. In order to make decisions in Engineering works or projects, viewpoint should be consistent.
(D). Develop Common Performance Measures: in order to make sure that the project is perfected there should be common performance measures.
(E). Considering Relevant Criteria: relevant Criteria will be met before the best choice is decided
(F). Risk making: Engineering projects should not be put under risk and thus is why this principle is very important.
(G). Decision retargeting: go back to the alternatives and recheck your choices.
Before you attempt to change a tire yourself, you should _____.
A. put on a pair of gloves
B. read your vehicle owner's manual for any special directions or warnings.
C. always call for emergency assistance first
D. let the remaining air out of the flat tire
Answer: read your vehicle owner's manual for any special directions or warnings.
Answer:
B. read your vehicle owner's manual for any special directions or warnings.
Explanation:
The closed feedwater heater of a regenerative Rankine cycle is to heat 7000 kPa feedwater from 2608C to a saturated liquid. The turbine supplies bleed steam at 6000 kPa and 3258C to this unit. This steam is condensed to a saturated liquid before entering the pump. Calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
Answer:
the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
Explanation:
Given that:
Pressure of the feed water = 7000 kPa
Temperature of the closed feedwater heater = 260 ° C
Pressure of of the turbine = 6000 kPa
Temperature of the turbine = 325 ° C
The objective is to calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.
From the table A-4 of saturated water temperature table at temperature 260° C at state 1 ;
Enthalpies:
[tex]h_1 = h_f = 1134.8 \ kJ/kg[/tex]
From table A-6 superheated water at state 3 ; the value of the enthalpy relating to the pressure of the turbine at 6000 kPa and temperature of 325° C is obtained by the interpolating the temperature between 300 ° C and 350 ° C
At 300° C; enthalpy = 2885.6 kJ/kg
At 325° C. enthalpy = 3043.9 kJ/kg
Thus;
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-{h_{300^0}}}{{h_{350^0}}- {h_{300^0}}}[/tex]
[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]\dfrac{25}{50}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]
[tex]h_{325^0} = 2885.6 + \dfrac{25}{50}({3043.9-2885.6 )[/tex]
[tex]h_{325^0} = 2885.6 + 0.5({3043.9-2885.6 )[/tex]
[tex]h_{325^0} =2964.75 \ kJ/kg[/tex]
At pressure of 7000 kPa at state 6; we obtain the enthalpies corresponding to the pressure at table A-5 of the saturated water pressure tables.
[tex]h_6 = h_f = 1267.5 \ kJ/kg[/tex]
From state 4 ;we obtain the specific volume corresponding to the pressure of 6000 kPa at table A-5 of the saturated water pressure tables.
[tex]v_4 = v_f = 0.001319\ m^3 /kg[/tex]
However; the specific work pump can be determined by using the formula;
[tex]W_p = v_4 (P_5-P_4)[/tex]
where;
[tex]P_4[/tex] = pressure at state 4
[tex]P_5[/tex] = pressure at state 5
[tex]W_p = 0.001319 (7000-6000)[/tex]
[tex]W_p = 0.001319 (1000)[/tex]
[tex]W_p =1.319 \ kJ/kg[/tex]
Using the energy balance equation of the closed feedwater heater to calculate the amount of bleed steam required to heat 1 kg of feed water ; we have:
[tex]E_{in} = E_{out} \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6[/tex]
where;
[tex]m_1 = 1 \ kg[/tex]
Replacing our other value as derived above into the energy balance equation ; we have:
[tex]1 \times 1134.8 +m_3 \times 2964.75 + m_3 \times 1.319 = (1+m_3)\times 1267.5[/tex]
[tex]1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3[/tex]
Collect like terms
[tex]2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8[/tex]
[tex]1698.569 \ m_3 =132.7[/tex]
[tex]\ m_3 = \dfrac{132.7}{1698.569}[/tex]
[tex]\mathbf{ m_3 = 0.078 \ kg/s}[/tex]
Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s
A 1.7 cm thick bar of soap is floating in water, with 1.1 cm of the bar underwater. Bath oil with a density of 890.0 kg/m{eq}^3 {/eq} is added and floats on top of the water. How high on the side of the bar will the oil reach when the soap is floating in only the oil?
Answer:
The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm
Explanation:
The water level on the bar soap = 1.1 m mark
Therefore, the proportion of the bar soap that is under the water is given by the relation;
Volume of bar soap = LW1.7
Volume under water = LW1.1
Volume floating = LW0.6
The relative density of the bar soap = Density of bar soap/(Density of water)
= m/LW1.7/(m/LW1.1) = 1.1/1.7
Given that the oil density = 890 kg/m³
Relative density of the oil to water = Density of the oil/(Density of water)
Relative density of the oil to water = 890/1000 = 0.89
Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89
Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7
Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7
Where:
X = The height of the oil on the side of the bar when the soap is floating in only the oil
Therefore;
X = 1.236 cm.
why is the peak value of the rectified output less than the peak value of the ac input and by how much g
Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
____ is based on the observation that the rate of increase in transistor density on microchips had increased steadily, roughly doubling every 18 to 24 months.
Answer:
Moore's Law
Explanation:
An observation that the number of transistors in a dense integrated circuit doubles about every two years (24 months), was made by Gordon E. Moore, the co-founder of Intel, and this observation became Moore's Law in 1965.
Therefore, Moore's Law is based on the observation that the rate of increase in transistor density on microchips had increased steadily, roughly doubling every 18 to 24 months.
is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction. What is Q
Answer:
Transverse shear stress formula
Explanation:
Transverse shear stress also known as the beam shear, is the shear stress due to bending of a beam.
Generally, when a beam is made to undergo a non-uniform bending, both bending moment (I) and a shear force (V) acts on its cross section or width (t).
Transverse shear stress formula is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction.
Mathematically, the transverse shear stress is given by the formula below;
[tex]T' = \frac{VQ}{It}[/tex]
Also note, T' is pronounced as tau.
Where;
V is the total shear force with the unit, Newton (N).
I is the Moment of Inertia of the entire cross sectional area with the unit, meters square (m²).
t is the thickness or width of cross sectional area of the material perpendicular to the shear with the unit centimeters (cm).
Q is the statical moment of area.
Mathematically, Q is given by the formula;
[tex]Q = y'P^{*} = ∑y'P^{*}[/tex]
Where [tex]P^{*}[/tex] is the section supporting a downward shear force in the y' direction.
You are tasked with designing an ICS/SCADA system. You must choose a type of ICS/SCADA system from the options listed below: Correctional facility Paint processing plant Water distribution facility Considering the type of system you chose from the list above, discuss the model you would use (time or event-based or a combination) and for what purposes. What considerations do you need to take into account in your design? How does the PLC fit into this system?
Answer:
The type of ICS/SCADA i choose is the Paint processing plant.
I will consider both the the time and event based as we need to mix various colors at different time interval and different quantities.
The PLC is used to process different tasks based on the commands assigned to it. In a paint processing plant, when a command from a computer is given to PLC for processing that tasks at that time whatever is the quantity is considered to mix it is carried out by PLC.
Explanation:
Solution
From the given question, i will select the Paint processing plant.
Here, the Supervisory Control and Data Acquisition System (SCADA) refers to a control system which uses computer network to control and manage the various processes from a single computer.
Since we consider the paint processing system for this we make use of both the time and event based as we need to mix various colors at different time interval and different quantities.
The programming Logic Controller (PLC): This is used to process different inputs based on the commands assigned to it.
In paint processing plant, when a command from a computer is assigned to PLC for processing that function at that time whatever is the quantity is required to mix it is carried out by PLC.
So, PLC is very useful device which is also the main processing device which carries out tasks assigned by the SCADA.
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lane is provided on the local road to allow vehicles from the ramp to turn right onto the local road without stopping. The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08 for superelevation.
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]
where;
R= radius
[tex]f_s[/tex] = coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
[tex]f_s[/tex] = 0.20
So;
[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]
[tex]R = \dfrac{900}{15(0.28)}[/tex]
[tex]R = \dfrac{900}{4.2}[/tex]
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.30 mm/rev and a depth of 4.0 mm. At a speed of 100 m/min, flank wear = 0.12 mm at 1 min, 0.27 mm at 5 min, 0.45 mm at 11 min, 0.58 mm at 15 min, 0.73 at 20 min, and 0.97 mm at 25 min. At a speed of 155 m/min, flank wear = 0.22 mm at 1 min, 0.47 mm at 5 min, 0.70 mm at 9 min, 0.80 mm at 11 min, and 0.99 mm at 13 min. The last value in each case is when final tool failure occurred.(a) On a single piece of linear graph paper, plot flank wear as a function of time for both speeds. You may use Excel to help yourself to plot the curve. Using 0.75 mm of flank wear as the criterion of tool failure, determine the tool lives for the two cutting speeds.(b) Calculate the values of n and C in the Taylor equation solving simultaneous equations.
Answer:
A) n = 0.6143, c ≈ 640m/min
B) n = 0.6143 , c = 637.53m/min
Explanation:
using the given data
A) A plot of flank wear as a function of time and also A plot for tool when
Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min
is attached below
calculate for the constant N from the second plot
note : the slope will be negative because cutting speed decreases as time of cutting increase
V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min
= - N = [tex]\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}[/tex]
therefore - N = [tex]\frac{5.043 - 4.605}{2.302 -3.015}[/tex]
= - 0.6143
THEREFORE ( N ) = 0.6143
Determine for the constant C from the second plot as well
note : C is the intercept on the cutting speed axis in 1 min tool life
connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point
hence C ≈ 640m/min
B) Calculate the values of N and C in the Taylor equation solving simultaneous equations
using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation
Taylor tool life equation = vT = C ------------- equation 1
cutting speed = v = 100m/min and 155m/min
tool life = T = 20.4 min and 10 min
also constant n and c are obtained from the previous plot
back to taylor tool life equation = 100 * 20.4 = C
therefore C = (100)(20.4)^n ---------------- equation 2
also using the second values of v and T
taylor tool life equation = 155 * 10 = C
therefore C = ( 155 )(10)^n ----------------- equation 3
Equate equation 2 and equation 3 and solve simultaneously
(100)(20.4)^n = (155)(10)^n
To find N
take natural log of both sides of the equation
= In ((100)(20.4)^n) = In((155)(10)^n)
= In (100) + nIn(20.4) = In(155) + nIn(10)^n
= n(3.0155) - n (2.3026) = 5.043 - 4.605
= 0.7129 n = 0.438
therefore n = 0.6143
To find C
substitute 0.6143 for n in equation 2
C = (100)(20.4) ^ 0.6143
C = 637.53 m/min
Attached are the two plots for solution A
For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:
a. . . . A B C A B C B A C B A . . .
b. . . . A B C A B C B C A B C . . .
Copy the stacking sequences and indicate the position(s) of planar defect(s) with a vertical dashed line.
Answer:
a) The planar defect that exists is twin boundary defect.
b) The planar defect that exists is the stacking fault.
Explanation:
I am using bold and underline instead of a vertical line.
a. A B C A B C B A C B A
In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.
b. A B C A B C B C A B C
In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 mL of the base. Assuming complete neutralization of the acid,
1) What was the normality of the acid solution?
2) What was the molarity of the acid solution?
Answer:
a. 0.4544 N
b. [tex]5.112 \times 10^{-5 M}[/tex]
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]
[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]
[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]
= 0.27264 g
[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]
[tex]= \frac{0.27264\ g}{40g/mol}[/tex]
= 0.006816 mol
Now
Moles of [tex]H_2SO_4[/tex] needed is
[tex]= \frac{0.006816}{2}[/tex]
= 0.003408 mol
[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]
[tex]= 0.003408\ mol \times 98g/mol[/tex]
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]
[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]
= 0.4544 N
b. And, the acid solution molarity is
[tex]= \frac{moles}{Volume}[/tex]
[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]
= 0.00005112
=[tex]5.112 \times 10^{-5 M}[/tex]
We simply applied the above formulas
The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is
21.30 mL, which gives the following acid solution approximate values;
1) Normality of the acid solution is 0.4544 N
2) The molarity of the acid is 0.2272
How can the normality, molarity of the solution be found?Molarity of the NaOH = 0.3200 M
Volume of NaOH required = 21.30 mL
1) The normality of the acid solution is found as follows;
The chemical reaction is presented as follows;
H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O
Number of moles of NaOH in the reaction is found as follows;
[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]
Therefore;
The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M
[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]
Which gives;
[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]
The normality of the acid solution, H₂SO₄(aq), N ≈ 0.45442) The molarity is found as follows;
[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]
The molarity of the acid solution is 0.2272 MLearn more about the normality and the molarity of a solution here:
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The value of an SMT capacitor is signified by a
Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time have a ductility of at least 12%EL. Justify your decision
Answer:
First we determine the tensile strength using the equation;
Tₓ (MPa) = 3.45 × HB
{ Tₓ is tensile strength, HB is Brinell hardness = 225 }
therefore
Tₓ = 3.45 × 225
Tₓ = 775 Mpa
From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10
When the percentage of cold work for steel is up to 10,the ductility is 16% EL.
And 16% EL is greater than 12% EL
Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL
Which of the following reduces friction in an engine A)wear B)drag C)motor oil D)defractionation
It is motor oil, as oil is used to reduce friction
A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate:
a. the thermal efficiency of the cycle
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answer:
a. [tex]\eta _{th}[/tex] = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)
[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
[tex]\eta _{th}[/tex] = 77.65%
b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]
Power developed is given by the relation;
[tex]\dot m \cdot w_{net, out}[/tex]
[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
A shaft made of aluminum is 40.0 mm in diameter at room temperature (21°C). Its coefficient of thermal expansion = 24.8 x 10-6 mm/mm per °C. If it must be reduced in size by 0.20 mm in order to be expansion fitted into a hole, determine the temperature to which the shaft must be cooled.
Answer:
Temperature to which the shaft must be cooled, [tex]\theta_2 = -180.61 ^0C[/tex]
Explanation:
Diameter of the shaft at room temperature, d₁ = 40 mm
Room temperature, θ₁ = 21°C
Coefficient of thermal expansion, [tex]\alpha = 24.8 * 10^{-6} / ^0 C[/tex]
The shaft is reduced in size by 0.20 mm:
Δd = - 0.20 mm
The temperature to which the shaft must be cooled, θ₂ = ?
The coefficient of thermal expansion is given by the equation:
[tex]\alpha = \frac{\triangle d}{d_1 * \triangle \theta}\\\\24.8 * 10^{-6} = \frac{-0.20}{40 * \triangle \theta}\\\\\triangle \theta = \frac{-0.20 }{24.8 * 10^{-6} * 40} \\\\\triangle \theta = - 201.61 ^0 C\\\triangle \theta = \theta_2 - \theta_1\\\\- 201.61 = \theta_2 - 21\\\\\theta_2 = -201.61 + 21\\\\\theta_2 = -180.61 ^0C[/tex]
A hot plate with a temperature of 60 C, 50 triangular profile needle wings of length (54 mm), diameter 10 mm (k = 204W / mK) will be added and cooled. Ambient temperature is 20 C and heat transfer coefficient is 20 Since it is W / m2K; a-) Wing efficiency, b-) Total heat transfer rate (W) from the wings, c-) Calculate the effectiveness of a wing.
Complete question is;
A hot plate with a temperature of 60 °C will be cooled by adding 50 triangular profile needle blades (k = 204 W/m.K) with a length of 54 mm and diameter 10 mm. According to the ambient temperature 20 °C and the heat transfer coefficient on the surface 20 W/m².K. Calculate,
a-) Wing efficiency
b-) Total heat transfer rate (W) from the wings,
c-) Calculate the effectiveness of a wing.
Answer:
A) Efficiency = 96.05 %
B) Total heat transfer rate = 166.68 W/m
C) Wing Effectiveness = 10.42
Explanation:
Please find attached explanation for all the answers given.
By saying that the electrostatic field is conservative, we do not mean that:_______ The potential difference between any two points is zero. It is the gradient of scalar potential. Its circulation is identically zero along any path. Its curl is identically zero everywhere. The work done in moving a charge along closed path inside the field is zero.
Answer:
(a) The potential difference between any two points is zero.
Explanation:
A conservative field is;
i. a vector field that is the gradient of some function. Electrostatic field is the gradient of scalar potential, hence it is conservative.
ii. a vector field where the integral along every closed path is zero. This means that the work done in a closed cycle is zero. For an electrostatic field, the charge along closed path inside the field is zero. Hence, electrostatic field is conservative.
iii. a vector field if curl of its potential(vector product of the del operator and the potential) is zero. The curl of electrostatic field is identically zero everywhere.
iv. a vector field whose circulation is zero along any path.
v. a vector field whose potential difference between two points is independent of the path taken. The potential difference between any two points is not necessarily zero.
Other examples of conservative fields are;
i. gravitational field.
ii. magnetic field.
When we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.
What is a conservative field?A conservative field refers to a form of force between the Earth and another mass whose work is determined only by the final displacement of the object acted upon.
What we mean by saying an electrostatic field is conservative includes:
It is the gradient of scalar potentialIts circulation is identically zero along any pathIts curl is identically zero everywhereThe work done in moving a charge along closed path inside the field is zero.Hence, when we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.
Therefore, the Option A is correct.
Read more about conservative field
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For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, after 114 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 87% completion
y = 1 - exp(-kt^n)
Answer:
total time = 304.21 s
Explanation:
given data
y = 50% = 0.5
n = 1.1
t = 114 s
y = 1 - exp(-kt^n)
solution
first we get here k value by given equation
y = 1 - [tex]e^{(-kt^n)}[/tex] ...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 × [tex]10^{4}[/tex]
so here
y = 1 - [tex]e^{(-kt^n)}[/tex]
1 - y = [tex]e^{(-kt^n)}[/tex]
take ln both side
ln(1-y) = -k × [tex]t^n[/tex]
so
t = [tex]\sqrt[n]{-\frac{ln(1-y)}{k}}[/tex] .............2
now we will put the value of y = 87% in equation with k and find out t
t = [tex]\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}[/tex]
total time = 304.21 s
A smooth, flat plate of length l = 6 m and width b = 4 m is placed in water with an upstream velocity of U = 0.5 m/s. Determine the boundary layer thickness and the wall shear stress at the center and the trailing edge of the plate. Assume a laminar boundary layer
Answer:
At x = 3m , Tw = 0.0176 N/m^2
At x = 6m , Tw = 0.056 N/m^2
Explanation:
Here in this question, we are concerned with calculating the boundary layer thickness and the wall shear stress at the center and the trailing edge of the plate , assuming a laminar boundary layer.
Please check attachment for complete solution
A thick casting with a thermal diffusivity of 5 x 10-6 m2/s is initially at a uniform temperature of 150oC. One surface of the casting is suddenly exposed to a high-speed water jet at 20oC, resulting in a very large convective heat transfer coefficient (Hint: assume imposed surface temperature). The thermal conductivity of the casting is 20 W/m-K. Determine the temperature 20 mm in from the surface after 40 seconds. Check your answer using an alternative technique. Your final answer should be about 101.3 oC.
Answer:
[tex]T_o = 141.81 ^0C[/tex]
Explanation:
Given that;
Thermal diffusivity [tex]\alpha = 5 \times 10 ^{-6} m^2/s[/tex]
Thermal conductivity [tex]k = 20 \ W/m.K[/tex]
Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K
Initial temperature = 150 ° C = (150+273) K = 423 K
Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K
Time = 40 seconds
Length = 20mm = 0.02 m
The objective is to determine the temperature at the surface at a depth of 20 mm after 40 seconds.
[tex]Bi = \dfrac{hL}{k}[/tex]
[tex]Bi = \dfrac{20*0.02}{20}[/tex]
Bi == 0.02
[tex]\tau = \dfrac{\alpha t}{L^2}[/tex]
[tex]\tau= \dfrac{5*10^{-6 }* 40}{0.020^2}[/tex]
[tex]\tau = 0.5[/tex]
For a wall at 0.2 Bi
[tex]A_1 = 1.0311[/tex]
[tex]\lambda _1 = 0.4328[/tex]
Therefore;
[tex]\dfrac{T_o - T_{\infty}}{T_i - T_{\infty}}= A_1 e ^{-( \lambda_1^2 \ \tau)[/tex]
[tex]\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )[/tex]
[tex]\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )[/tex]
[tex]\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085[/tex]
[tex]\dfrac{T_o - 293 }{130}= 0.937[/tex]
[tex]T_o - 293= 0.937 \times 130[/tex]
[tex]T_o - 293= 121.81[/tex]
[tex]T_o = 121.81+ 293[/tex]
[tex]T_o = 414.81 \ K[/tex]
[tex]T_o = (414.81 - 273)^0C[/tex]
[tex]T_o = 141.81 ^0C[/tex]
1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.
i. True
ii. False
2. The cosine rule is used when we are given either a) three sides or b) two sides and the included angle.
i. True
ii. False
Answer:
A. Yes
B. Yes
Explanation:
We want to evaluate the validity of the given assertions.
1. The first statement is true
The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.
Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side
Or
We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.
For notation purposes;
We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles
a/Sin A = b/Sin B = c/Sin C
2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate
So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A
That would be;
a^2 = b^2 + c^2 -2bcCosA
So yes, the cosine rule can be used for the scenario above