Answer:
1) Mass fraction of apple in the solid waste = 0.644
2) Total flow rate of the concentrate = 12.89 kg/min
3) Flow rate of pure water entering the overall system = 9.434 kg/min
Explanation:
Mass fraction of the slurry = 0.422
Rate at which the balance water is fed to a juicer, m = 27.3 kg/min
Stream fed to juice contains (0.422 * 27.3) = 11.5206 kg/min apple
Stream fed to juice contains (27.3 - 11.5206) = 15.7794 kg/min water
70% of the apples entering the juicer leave as part of the solid waste stream
Amount of Apples leaving the stream as part of the solid waste stream,
W = 0.7 * 11.5206
W = 8.06442 kg/min
Overall balance on juicer: m = p + W
p = m - W
p = 27.3 - 8.06442
p = 19.236 kg/min
p stream contains 32.9 wt% apples and 67.1 wt% water
1) Mass fraction of apple in the solid waste:
Let x = Mass fraction of apple in the solid waste
Apple in m stream = Apple in p stream + Apple in W stream
11.5206 = (0.329 * 19.236) + (x * 8.06442)
11.5206 = 6.33 + 8.06442x
8.06442x = 11.5206 - 6.33
x = 0.644
2) Total flow rate of the concentrate (kg/min)
Cider sent to evaporator where concentrate increases from 32.9 wt% to 49.1 wt%
Solid balance; 0.329 p = 0.491 c
0.329 * 19.236 = 0.491 c
c = 12.89 kg/min
Total flow rate of the concentrate = 12.89 kg/min
3 Flow rate of pure water entering the overall system (kg/min)
Overall balance for evaporator: p = v + c
Where P is the pure water from evaporator
v = p - c
v = 19.236 - 12.89
v = 6.346 kg/min
Rate of Water entering into the juicer = (1 - 0.422) *27.3 = 15.78 kg/min
Pure water from evaporator, v = 6.346 kg/min
Flow rate of pure water entering the overall system = 15.78 - 6.346
Flow rate of pure water entering the overall system = 9.434 kg/min
Based on the ELR, the stability classification of the atmosphere at the airport between the surface and 3000 m is
Answer:
The answer is "conditionally unstable"
Explanation:
The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.
Classification dependent on ELR:
Larger than 10 [tex]\frac{^{\circ}C}{1000}[/tex] m Around 10 and 6 [tex]\frac{^{\circ}C}{1000}[/tex] m or less 6 [tex]\frac{^{\circ}C}{1000}[/tex] m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 [tex]\frac{^{\circ}C}{1000}[/tex] m, that's why it is "conditionally unstable".
Find I in the circuit
Answer:
I = 6.364∠3.8° A
Explanation:
You can use KCL or KVL to write node or mesh equations for the voltages and currents in the network. Those require a matrix equation solver capable of working with complex numbers. Some calculators can do that. However, we're going to try a different approach here.
Starting from upper left (j4) and working clockwise around the outside, label the impedances Z1 .. Z4. Label the horizontal branch across the middle Z5. We're going to transform the Δ of Z1, Z2, Z5 into a Y of ZA, ZB, ZC that will facilitate computing the effective impedance of the bridge to the source voltage.
The Δ-Y transformation is symmetrical. The numerator of the equivalent impedance connected to each node is the product of the values currently connected to that node; the denominator is the sum of the values around the loop of the Δ.
So, If we transform the Δ of Z1, Z2, Z5 to a Y of ZA, ZB. ZC with ZA connected where Z1 and Z2 are now connected, ZB connected to Z4, and ZC connected to Z3, the network becomes a series-parallel network with an effective impedance of ...
Z = ZA + ((ZB +Z4) ║ (ZC +Z3))
__
For starters, we have ...
ZA = Z1·Z2/(Z1 +Z2 +Z5) = (j4)(-j3)/(j4-j3+8+j5) = 12/(8+j6) = 0.96-j0.72
ZB = (j4)(8+j5)/(8+j6) = (20+j32)/(8+j6) = 0.32+j3.76
ZC = (-j3)(8+j5)/(8+j6) = (15-j24)/(8+j6) = -0.24-j2.82
So, the left branch of the parallel combination is ...
ZB +Z4 = (0.32+j3.76) +(5-j2) = 5.32+j1.76
And the right branch is ...
ZC +Z3 = (-0.24-j2.82) +10 = 9.76-j2.82
Then the series-parallel combination we want is ...
ZA + (ZB+Z4)(ZC+Z3)/(ZB+Z4+ZC+Z3) ≈ 4.703671 -j0.3126067
That is, the impedance of the bridge circuit to the source voltage is about ...
4.7140478∠-3.802°
Dividing the source voltage by this impedance gives the source current, ...
I = (30∠0°)/(4.7140478∠-3.802°)
I ≈ 6.363958∠3.802° . . . amperes
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 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
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.
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.
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
enclosed by a well-fitted casing covering almost the whole length, which reduces the lateral expansion by half.
Determine the ratio between the axial strain when the casing is fitted and that when it is free to expand in diameter.
Take v = 0.3.
Answer:
[tex]\frac{e'_z}{e_z} = 0.87142[/tex]
Explanation:
Given:-
- The diameter of the cylinder, d = 50 mm.
- The compressive load, F = 80 KN.
Solution:-
- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.
- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).
εx = - [ σx - ν( σy + σz ) ] / E
εy = - [ σy - ν( σx + σz ) ] / E
εz = - [ σz - ν( σy + σx ) ] / E
- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:
εx = [ ν*σz ] / E
εy = [ ν*σz ] / E
εz = - [ σz ] / E .... Eq 1
- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.
- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:
εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx
εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E
εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy
εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E
- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.
- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).
Note: σx' = σy', The cylinder is radially enclosed around the entire surface.
Therefore,
- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2
σx' ( 1 - v ) = [ ν*σz ] / 2
σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]
- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):
εz' = - [ σz - ν( σy' + σx' ) ] / E
εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E
εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2
- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:
[tex]\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142[/tex]... Answer
Shaft AB has a 30-mm diameter and is made of a steel with an allowable shearing stress of 100 MPa; shaft BC has a 50-mm diameter and is made of an aluminum alloy with an allowable shearing stress of 60 MPa. Neglecting the effect of stress concentrations, determine the largest torque T that can be applied at A.
Answer: The largest torque that can be applied at A is the smaller of both
Tᵇc and Tᵃᵇ
which is (Tᵃᵇ = 530.14376 Nm)
Explanation:
First we take a look at Torque formula
T = M × Shear stress
T = π/2 × (d/2)³ × shear stress
{ where M = π/2 × (d/2)³ = π/2 × d³/8
For Shaft AB
d = 30mm = 30 × 10⁻³m
shaft stress = 100Mpa = 100 × 10⁶ N/m²
now Torque at A due to AB
Tᵃᵇ = π/2 × (d/2)³ × shear stress
Tᵃᵇ = π/2 × (30 × 10⁻³)³/2³ × 100 × 10⁶
Tᵃᵇ = π/2 × (0.000027 / 8) × 100 × 10⁶
Tᵃᵇ = 530.14376 Nm
For Shaft BC
the value for M is changed
M = π/2 × { (d₂/2)⁴ - (d₁/2)⁴}
d₁ = 30mm = 30 × 10⁻³m
d₂ = 50mm = 50 × 10⁻³m
M = π/2 × { (50 × 10⁻³/2)⁴ - (30 × 10⁻³/2)⁴}
M = 5.34 × 10⁻⁷
Torque formula is also changed which is
T = M × share stress / (d₂/2)
shear stress = 60Mpa = 60 × 10⁶ N/m²
so Torque A due to BC is
Tᵇc = (5.34 × 10⁻⁷ × 60 × 10⁶) / ( 50 × 10⁻³ / 2
Tᵇc = 1281.6 Nm
Therefore the largest torque that can be applied at A is the smaller of both
Tᵇc and Tᵃᵇ
which is (Tᵃᵇ = 530.14376 Nm)
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
Determine the voltage input to the inverting terminal of an op amp when −40 μV is applied to the non-inverting terminal and the output through an open-loop gain of 150,000 is 15 V.
Answer:
The voltage input to the inverting terminal is 60μV
Explanation:
Given;
open-loop gain, A = 150,000
output voltage, V₀ = 15 V
voltage at the inverting input, [tex]V_n[/tex] = −40 μV = 40 x 10⁻⁶ V
The relationship between output voltage and voltage at the inverting input is given as;
[tex]V_o = A(V_p -V_n)\\\\15 = 150,000(V_p -(-40*10^{-6}))\\\\15 = 150,000 (V_p +40*10^{-6}) \\\\V_p +40*10^{-6} = \frac{15}{150,000} \\\\V_p + 40*10^{-6} = 1 *10^{-4}\\\\V_p + 40*10^{-6} = 100 *10^{-6}\\\\V_p = 100 *10^{-6} - 40*10^{-6}\\\\V_p = 60 *10^{-6}\\\\V_p = 60 \ \mu V[/tex]
Therefore, the voltage input to the inverting terminal is 60μV
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:
For flow of a liquid metal through a circular tube, the velocity and temperature profiles at a particular axial location may be approximated as being uniform and parabolic, respectively. That is, u(r) = C, and TO-T, = C11-(r/ro)2], where Ci and C2 are constants. what is the value of the Nusselt number Nup at this location?
Answer:
Nup = 8
Explanation:
u(r) = C1 and T(r) = T(s) = C2 [ 1 - r/ro)^2 ]
The temperature and velocity particles at a particular axial location is Uniform and parabolic
The Nusset Number Nup at this location can be obtained by first determining/calculating for the convection coefficient = [tex]h = \frac{q^I s}{TsTn}[/tex]
Therefore the mean temperature is calculated as follows
attached is the detailed solution and free body diagram
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.
If these components have weights WA = 50000 lb , WB=8000lb, and WC=6000lb, determine the normal reactions of the wheels D, E, and F on the ground.
Answer:
F(F) = 15037 lb
F(E) = 24481.5 lb
F(D) = 24481.5 lb
Explanation:
(The diagram of the figure and Free Body Diagram is attached)
Data given:W(A) = 50,000 lb
W(B) = 8000 lb
W(C) = 6000 lb
∑F = 0F(F) + F(E) + F(D) - W(A) - W(B) - W(C) = 0
F(F) + F(E) + F(D) = W(A) + W(B) + W(C)
F(F) + F(E) + F(D) = 50000 + 8000 + 6000
F(F) + F(E) + F(D) = 64000 lb
∑M(o)∑M(o) = M(F) + M(E) + M(D) + M(A) + M(B) + M(C)
Where
M(F) = 27i × F(F)k = -27F(F)j
M(E) = 14j × F(E)k = 14F(E)i
M(D) = -14j × F(D)k = -14F(D)i
M(A) = 7i × -50000k = 350,000j
M(B) = (4i - 6j) × -8000k = 48000i + 32000j
M(C) = (4i + 8j) × -6000k = -48000i + 24000j
∑M(x) = ∑M(i) = 0∑M(i) = 14F(E) - 14F(D) = 0
F(E) = F(D)
∑M(y) = ∑M(j) = 0∑M(j) = -27F(F) + 350,000 + 32,000 + 24,000 = 0
27F(F) = 406,000
F(F) = 15037 lb
F(F) + F(E) + F(D) = 64000 lbF(E) = F(D)
F(F) + 2F(E) = 64000
2F(E) = 64000 - 15037
2F(E) = 48963
F(E) = 24481.5 lb
F(D) = 24481.5 lb
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.
Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]
I2 + KOH = KIO3 + KI + H2O Marque la(s) respuesta(s) falsas: La suma de coeficientes mínimos del agua y el agente reductor es 6 El KI es la forma reducida El KOH es el agente reductor La suma de electrones transferidos más el coeficiente mínimo del agua suman 16 La relación entre el agente oxidante y el agente reductor es 1
Answer:
Las declaraciones falsas incluyen
- El KOH es el agente reductor.
- La suma de electrones transferidos más el coeficiente mínimo de agua suman 16.
Todas las otras declaraciones son ciertas.
The false statements include
- The KOH is the reducing agent.
- The sum of transferred electrons plus the minimum coefficient of water add up to 16.
All the other statements are true.
Explanation:
Es evidente que esta es una reacción redox en presencia de medio básico. Entonces, equilibraremos esta reacción redox en pasos. I₂ + KOH → KIO₃ + KI + H₂O
Paso 1 Eliminar los iones espectadores; Estos son los iones que aparecen en ambos lados de la reacción. Es evidente que solo el ion de potasio (K⁺) es el ion espectador de esta reacción.
I₂ + OH⁻ → IO₃⁻ + I⁻ + H₂O
Paso 2
Separamos la reacción en las medias reacciones de oxidación y reductina. La oxidación es la pérdida de electrones que conduce a un aumento del número de oxidación del ion, mientras que la reducción es la ganancia de elecrones que conduce a una disminución en el número de oxidación del ion. También es evidente que es el gas de yodo el que se reduce y oxida para esta reacción.
El gas de yodo se reduce a I⁻ (el número de oxidación se reduce de 0 a -1) y el gas de yodo se oxida a IO₃⁻ (el número de oxidación de yodo aumenta de 0 en gas de yodo a +5 en IO₃⁻)
Reducción media reacción
I₂ → I⁻
Media reacción de oxidación
I₂ + OH⁻ → IO₃⁻ + H₂O
Paso 3
Equilibramos las medias reacciones y agregamos los respectivos electrones transferidos
Reducción media reacción
I₂ → 2I⁻
I₂ + 2e⁻ → 2I⁻
Media reacción de oxidación
I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O
I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻
Paso 4
Balancee el número de electrones en las dos medias reacciones
[I₂ + 2e⁻ → 2I⁻] × 5
[I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻] × 1
5I₂ + 10e⁻ → 10I⁻
I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻
Paso 5
Agregue las dos medias reacciones y elimine cualquier especie que aparezca en ambos lados
5I₂ + 10e⁻ + I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O + 10e⁻
Entonces, eliminamos los 10 electrones que fueron transferidos en la reacción balanceada
6I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O
Paso 6
Reintroducimos la especie eliminada desde el principio (el ion potasio)
6I₂ + 12KOH → 10KI + 2KIO₃ + 6H₂O
Los coeficientes mínimos son entonces
3I₂ + 6KOH → 5KI + KIO₃ + 3H₂O
Luego verificamos cada una de las declaraciones proporcionadas para elegir las falsas.
- La suma de los coeficientes mínimos del agua y el agente reductor es 6.
El gas yodo es el agente reductor y oxidante. Coeficiente mínimo de agua y gas de yodo = 3 + 3 = 6 Esta afirmación es cierta.
- El KI es la forma reductora KI resulta de la semirreacción de reducción.
Por lo tanto, es la forma reducida del gas de yodo. Esta afirmación es cierta. - El KOH es el agente reductor. KOH no es el agente reductor. Esta afirmación es falsa.
- La suma de los electrones transferidos más el coeficiente mínimo de agua suman 16.
Electrones transferidos = 10
Coeficiente mínimo de agua = 3
Suma = 13 y no 16.
Esta afirmación es falsa.
- La proporción del agente oxidante y el agente reductor es 1.
Dado que el gas yodo es el agente reductor y oxidante, la proporción de estos dos es verdaderamente 1. Esta afirmación es cierta.
¡¡¡Espero que esto ayude!!!
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]
Learn more about Air compressors:
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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.
Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of the nozzle is 50 cm2, and heat is being lost at a rateof 120 kJ/s. Determine the following:
a) the mass flow rateof the steam.
b) the exit velocity of the steam.
c) the exitarea of the nozzle.
Answer:
a) the mass flow rate of the steam is [tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam is [tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle is [tex]A_2[/tex] = 0.0015435 m²
Explanation:
Given that:
A steam with 5 MPa and 400° C enters a nozzle steadily
So;
Inlet:
[tex]P_1 =[/tex] 5 MPa
[tex]T_1[/tex] = 400° C
Velocity V = 80 m/s
Exit:
[tex]P_2 =[/tex] 2 MPa
[tex]T_2[/tex] = 300° C
From the properties of steam tables at [tex]P_1 =[/tex] 5 MPa and [tex]T_1[/tex] = 400° C we obtain the following properties for enthalpy h and the speed v
[tex]h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg[/tex]
From the properties of steam tables at [tex]P_2 =[/tex] 2 MPa and [tex]T_1[/tex] = 300° C we obtain the following properties for enthalpy h and the speed v
[tex]h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg[/tex]
Inlet Area of the nozzle = 50 cm²
Heat lost Q = 120 kJ/s
We are to determine the following:
a) the mass flow rate of the steam.
From the system in a steady flow state;
[tex]m_1=m_2=m_3[/tex]
Thus
[tex]m_1 =\dfrac{V_1 \times A_1}{v_1}[/tex]
[tex]m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}[/tex]
[tex]m_1 =\dfrac{0.4 }{0.057838 }[/tex]
[tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam.
Using Energy Balance equation:
[tex]\Delta E _{system} = E_{in}-E_{out}[/tex]
In a steady flow process;
[tex]\Delta E _{system} = 0[/tex]
[tex]E_{in} = E_{out}[/tex]
[tex]m(h_1 + \dfrac{V_1^2}{2})[/tex] [tex]= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})[/tex]
[tex]- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]V_2^2 = 316631.29 \ m/s[/tex]
[tex]V_2 = \sqrt{316631.29 \ m/s[/tex]
[tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle.
The exit of the nozzle can be determined by using the expression:
[tex]m = \dfrac{V_2A_2}{v_2}[/tex]
making [tex]A_2[/tex] the subject of the formula ; we have:
[tex]A_2 = \dfrac{ m \times v_2}{V_2}[/tex]
[tex]A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}[/tex]
[tex]A_2[/tex] = 0.0015435 m²
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.
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.
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
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.
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.
As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr
Answer:
α = kt
ω = [tex]\frac{kt^2}{2}[/tex]
θ = [tex]\frac{kt^3}{6}[/tex]
Explanation:
given data
Initial velocity ω = 0
time t = 10 s
Number of revolutions = 20
solution
we will take here first α = kt .....................1
and as α = [tex]\frac{d\omega}{dt}[/tex]
so that
[tex]\frac{d\omega}{dt}[/tex] = kt ..................2
now we will integrate it then we will get
∫dω = [tex]\int_{0}^{t} kt\ dt[/tex] .......................3
so
ω = [tex]\frac{kt^2}{2}[/tex]
and
ω = [tex]\frac{d\theta}{dt}[/tex] ..............4
so that
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{kt^2}{2}[/tex]
now we will integrate it then we will get
∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex] ...............5
solve it and we get
θ = [tex]\frac{kt^3}{6}[/tex]
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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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
Liquidated damages are intended to represent anticipated losses to the owner based upon circumstances existing at the time the contract was made. List at least five types of potential losses to the owner that would qualify for determination of such potential losses.
Answer:
1. Loss of income.
2. Rental costs.
3. Utility bills.
4. Loss of rent.
5. Storage costs.
Explanation:
Liquidated damages can be defined as pre-determined damages or clauses that are highlighted or indicated at the time of entering into a contract between a contractor and a client which is mainly based on evaluation of the actual loss the client may incur should the contractor fail to meet the agreed completion date.
Generally, liquidated damages are meant to be fair rather than being a penalty or punitive to the defaulter. It is usually calculated on a daily basis for the loss.
When entering into a contract with another, liquidated damages are intended to represent anticipated losses to the owner based upon circumstances existing at the time the contract was made.
Listed below are five (5) types of potential losses to the owner that would qualify for determination of such potential losses;
1. Loss of income.
2. Rental costs.
3. Utility bills.
4. Loss of rent.
5. Storage costs.