Answer:
a) 4160 V
b) 12 kW and 81 kVAR
c) 54 kW and 477 kVAR
Explanation:
1) The phase voltage is given as:
[tex]V_p=\frac{3810.5}{\sqrt{3} }=2200 V[/tex]
The complex power S is given as:
[tex]S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA[/tex]
[tex]where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA[/tex]
The line current I is given as:
[tex]I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A[/tex]
The phase voltage at the sending end is:
[tex]V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV[/tex]
The magnitude of the line voltage at the source end of the line ([tex]V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V[/tex]
b) The Total real and reactive power loss in the line is:
[tex]S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000[/tex]
The real power loss is 12000 W = 12 kW
The reactive power loss is 81000 kVAR = 81 kVAR
c) The sending power is:
[tex]S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000[/tex]
The Real power delivered by the supply = 54000 W = 54 kW
The Reactive power delivered by the supply = 477000 VAR = 477 kVAR
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P
Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]
Where:
[tex]n[/tex] - Safety factor, dimensionless.
[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])
[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]
[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]
[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]
Where:
[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.
[tex]V[/tex] - Shear force, measured in kilonewtons.
[tex]A[/tex] - Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
[tex]V = \frac{P}{5}[/tex]
[tex]V = \frac{450\,kN}{5}[/tex]
[tex]V = 90\,kN[/tex]
The minimum allowable cross section area is cleared in the shearing stress equation:
[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]
If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:
[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]
[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]
The diameter is now cleared and computed:
[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]
[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]
[tex]D = 0.0457\,m[/tex]
[tex]D = 45.7\,mm[/tex]
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
We have that the minimum allowable bolt diameter is mathematically given as
d = 26.65mmFrom the question we are told
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of Assuming P to be P = 425 kN.DiameterGenerally the equation for the stress is mathematically given as
[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]
Therefore
Force = Stress * area
Force = P/2
F= 425,000 N / 2 = 212,500 N
Hence area of each bolt is given as
212,500 = 76.190*( 5* area of each bolt)
area of each bolt = 557.815
Since
area of each bolt=\pi*d^2/4
\pi*d^2/4 = 557.815
d = 26.65mmFor more information on diameter visit
https://brainly.com/question/8552546
A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of 62 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a design factor of 1.5, what dimension should the square cross section have
Answer:
The dimension of the square cross section is = 30mm * 30mm
Explanation:
Before proceeding with the calculations convert MPa to Kpsi
Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi
the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut
at 10^6 cycles" for 111.65 Kpsi = f ( fatigue strength factor) = 0.83
To calculate the endurance limit use Se' = 0.5 Sut since Sut < 1400 MPa
Se'( endurance limit ) = 0.5 * 770 = 385 Mpa
The surface condition modification factor
Ka = 57.7 ( Sut )^-0.718
Ka = 57.7 ( 770 ) ^-0.718
Ka = 0.488
Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one
The endurance limit at the critical location of a machine part can be expressed as :
Se = Ka*Kb*Se'
Se = 0.488 * 0.85 * 385 = 160 MPa
Next:
Calculating the constants to find the number of cycles
α = [tex]\frac{(fSut)^2}{Se}[/tex]
α =[tex]\frac{(0.83*770)^2}{160}[/tex] = 2553 MPa
b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]
b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex] = -0.2005
Next :
calculating the fatigue strength using the relation
Sf = αN^b
N = number of cycles
Sf = 2553 ( 10^4) ^ -0.2005
Sf = 403 MPa
Calculate the maximum moment of the beam
M = 2000 * 0.6 = 1200 N-m
calculating the maximum stress in the beam
∝ = ∝max = [tex]\frac{Mc}{I}[/tex]
Where c = b/2 and I = b(b^3) / 12
hence ∝max = [tex]\frac{6M}{b^3}[/tex] = 6(1200) / b^3 = 7200/ b^3 Pa
note: b is in (meters)
The expression for the factor of safety is written as
n = [tex]\frac{Sf}{\alpha max }[/tex]
Sf = 403, n = 1.5 and ∝max = 7200 / b^3
= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex] making b subject of the formula in other to get the value of b
b = 0.0299 m * 10^3 mm/m
b = 29.9 mm therefore b ≈ 30 mm
since the size factor assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2
the dimension = 30 mm by 30 mm
Use a delta-star conversion to simplify the delta BCD (40 , 16 , and 8 ) in the
bridge network in Fig. f and find the equivalent resistance that replaces the network
between terminals A and B, and hence find the current I if the source voltage is 52 V.
Answer:
Current, I = 4A
Explanation:
Since the connection is in delta, let's convert to star.
Simplify BCD:
[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]
[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]
[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]
From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.
Simplify:
[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]
Req = 10 ohms + 3 ohms
Req = 13 ohms
To find the current, use ohms law.
V = IR
Where, V = 52volts and I = 13 ohms
Solve for I,
[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]
Current, I = 4 A
Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in a) shaft AB b) shaft BC c) shaft CD (25 points) Given that the torque at B
Answer:
Explanation:
The image attached to the question is shown in the first diagram below.
From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.
IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :
From the diagram; we can determine the length of BC by using pyhtagoras theorem;
SO;
[tex]L_{BC}^2 = L_{AB}^2 + L_{AC}^2[/tex]
[tex]L_{BC}^2 = (3.5+2.5)^2+ 4^2[/tex]
[tex]L_{BC}= \sqrt{(6)^2+ 4^2}[/tex]
[tex]L_{BC}= \sqrt{36+ 16}[/tex]
[tex]L_{BC}= \sqrt{52}[/tex]
[tex]L_{BC}= 7.2111 \ m[/tex]
The cross -sectional of the cable is calculated by the formula :
[tex]A = \dfrac{\pi}{4}d^2[/tex]
where d = 4mm
[tex]A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2[/tex]
A = 1.26 × 10⁻⁵ m²
However, looking at the maximum deflection in length [tex]\delta[/tex] ; we can calculate for the force [tex]F_{BC[/tex] by using the formula:
[tex]\delta = \dfrac{F_{BC}L_{BC}}{AE}[/tex]
[tex]F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}[/tex]
where ;
E = modulus elasticity
[tex]L_{BC}[/tex] = length of the cable
Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for [tex]L_{BC}[/tex] and 0.006 m for [tex]\delta[/tex] ; we have:
[tex]F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}[/tex]
[tex]F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN[/tex] ---- (1)
Similarly; we can determine the force [tex]F_{BC}[/tex] using the allowable maximum stress; we have the following relation,
[tex]\sigma = \dfrac{F_{BC}}{A}[/tex]
[tex]{F_{BC}}= {A}*\sigma[/tex]
where;
[tex]\sigma =[/tex] maximum allowable stress
Replacing 190 × 10⁶ Pa for [tex]\sigma[/tex] ; we have :
[tex]{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN[/tex] ------ (2)
Comparing (1) and (2)
The magnitude of the force [tex]F_{BC} = 2.09676 \ kN[/tex] since the elongation of the cable should not exceed 6mm
Finally applying the moment equilibrium condition about point A
[tex]\sum M_A = 0[/tex]
[tex]3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0[/tex]
[tex]3.5 P - 3.328 F_{BC} = 0[/tex]
[tex]3.5 P = 3.328 F_{BC}[/tex]
[tex]3.5 P = 3.328 *2.09676 \ kN[/tex]
[tex]P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }[/tex]
P = 1.9937 kN
Hence; the maximum load P that can be applied is 1.9937 kN
g A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answer:
the overall elongation δ of the bar is 1.2337 mm
Explanation:
From the information given :
According to the principle of superposition being applied to the axial load P of the system; we have:
[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]
where;
δ = overall elongation
[tex]\delta _{AB}[/tex] = elongation of bar AB
[tex]\delta _{BC}[/tex] = elongation of bar BC
[tex]\delta _{CD} =[/tex] elongation of bar CD]
If we replace; [tex]\dfrac{PL}{AE}[/tex] for δ and bt for area;
we have:
[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]
where ;
P = load
L = length of the bar
A = area of the cross-section
E = young modulus of elasticity
Let once again replace:
P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex] (since load in all member of AB, BC and CD will remain the same )
[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],
[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and
[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]
[tex]2\dfrac{b}{3}[/tex] for [tex]b_{BC}[/tex]
b for [tex]b_{CD}[/tex]
[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]
[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]
[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]
The stress in the central portion can be calculated as:
[tex]\sigma = \dfrac{P}{A}[/tex]
[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]
[tex]\sigma = \dfrac{3P}{2bt}[/tex]
So; Now:
[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]
[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]
δ = 1.2337 mm
Therefore, the overall elongation δ of the bar is 1.2337 mm
Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most transmissions today use compound (multiple) planetary gear sets. Which technician is correct?
Answer:
Both technician A and technician B are correct
Explanation:
A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.
One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.
So, both technician A and technician B are correct.
Given in the following v(t) signal.
a. Find the first 7 harmonics of the Fourier series function in cosine form.
b. Plot one side spectrum
c. Find the first 7 harmonics of the Fourier series function in exponential form.
d. Plot two side spectrum Given in the following v(t) signal.
Answer:
Check the v(t) signal referred to in the question and the solution to each part in the files attached
Explanation:
The detailed solutions of parts a to d are clearly expressed in the second file attached.
Sensors are used to monitor the pressure and the temperature of a chemical solution stored in a vat. The circuitry for each sensor produces a HIGH voltage when a specified maximum value is exceeded. An alarm requiring a LOW voltage input must be activated when either the pressure or the temperature is excessive. Design a circuit for this application
4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, respectively. a. Find the load and shrinkage factors for this soil. b. How many trucks loads with a capacity of 5 lcy/truck would be required to haul 750,000 ccy of this soil to a project
Answer:
a. load factor = 0.893
shrinkage factor = 0.848
b. Number of Trucks loads = 113,585 Trucks loads
Explanation:
Here, we start by identifying the factors as given in the question.
γn = 2800 lb/cy
γloose = 2500 lb/cy
and γcompacted = 3300 lb/cy
a. Mathematically,
Load factor = γloose/γn = 2500/2800 = 0.893
Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848
b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;
ρlcy = 5
Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy
0.893 × 0.848 = 5/ρccy
ρccy =5/(0.893 × 0.848) = 6.603
The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads
2) Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Answer:
Explanation:
Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Strict schedule:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.
S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.
S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))
but T3 commits after T1. In a strict schedule T3 must read X after C1.
Cascadeless schedule:
Cascadeless schedule follows the below condition:
Tj reads X only? after Ti has written to X and terminated means aborted or committed.
S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.
But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,
T3 does not have to roll back if T1 is rolled back. The problem occurs because these
schedules are not serializable.
Recoverable schedule:
Schedule that follows the below condition:
-----Tj commits after Ti if Tj has?read any data item written by Ti.
Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is
recoverable one has to include abort operations. Thus in testing the recoverability abort
operations will have to used in place of commit one at a time. Also the strictest condition is
------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.
If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and
T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.
Strictest condition of schedule S3 is C3>C2.
If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
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
13000 + 300 - 8000 = T2
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core temperature is 37.0°C, and it has a surface area of 2.23 m2. What is the average thickness of its blubber? The conductivity of fatty tissue without blood is 0.20 (J/s · m · °C).
Answer:
The average thickness of the blubber is 0.077 m
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Describe with an example how corroded structures can lead to environment pollution?
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical shape. Initially, the balloon is filled up with air at the pressure and temperature of 100 kPa and 27°C respectively and the initial diameter (D) of the balloon is 10 m. Then the balloon is heated up to the point that the volume is 1.2 times greater than the original volume (V2 =1.2V1 ). Due to elastic material used in this balloon, the inside pressure ( P ) is proportional to balloonâs diameter, i.e. P = ð¼D, where ð¼ is a constant.
Required:
a. Show that the process is polytropic (i.e. PV" = Constant) and find the exponent n and the constant.
b. Find the temperature at the end of the process by assuming air to be ideal gas.
c. Find the total amount of work that is done by the balloon's boundaries and the fraction of this work that is done on the surrounding atmospheric air at the pressure of 100 kPa.
Answer:
a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]
[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]
We have;
[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]
[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]
[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]
p₂ = 100000/0.941 = 106.265 kPa
[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
When you do a vehicle check, what do you NOT need to keep an eye on?
A. Proper tire inflation
B. Cleanliness of windows and mirrors
C. Functioning indicator lights and headlights
D. Blindspot locations
Answer:
Blindspot Location
Explanation:
Just took the quiz
When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.
What is Blind spot location?A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.
While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.
Blind spot locations does not need to be checked.
Thus, the correct option is D.
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Caulking is recommended around the edges of partitions between apartments to... Group of answer choices reduce the need for trim. reduce sound transmission. reduce heat loss. increase the fire rating of the partition
Answer:
Reduce sound transmission.
Explanation:
A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.
Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.
Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.
The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.
A cylinder of metal that is originally 450 mm tall and 50 mm in diameter is to be open-die upset forged to a final height of 100 mm. The strength coefficient is 230 MPa and the work hardening exponent is 0.15 while the coefficient of friction of the metal against the tool is 0.1. If the maximum force that the forging hammer can deliver is 3 MN, can the forging be completed
Answer:
Yes, the forging can be completed
Explanation:
Given h = 100 mm, ε = ㏑(450/100) = 1.504
[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]
V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³
At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²
D = √(4·A/π) = 106.07 mm
[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042
F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN
Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.
Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]
Carrier frequency 100kHz
Upper side band
[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]
A) The frequency Components of the Detector Output are;
80 Hz, 120 Hz and 380 Hz
B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz
Message SignalsA) We are given the frequency components in the message signal as;
f1 = 100Hzf2 = 200Hzf3 = 400HzWe are told that the carrier signal has a frequency; fc = 100kHz
Thus, the frequency components generated are;
Lower side band:
100 kHz - 100 Hz = 99.9 kHz100 kHz - 200 Hz = 99.8 kHz100 kHz - 400 Hz = 99.6 kHzUpper side band:
100 kHz + 100 Hz = 100.1 kHz100 kHz + 200 Hz = 100.2 kHz100 kHz + 400 Hz = 100.4 kHzWe are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.
Thus, the output frequencies are;
100.02 kHz - 100.1 kHz = 80 Hz
100.02 kHz - 100.2 kHz = 180 Hz
100.02 kHz - 100.4 kHz = 380 Hz
B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;
100.02 kHz - 99.9 kHz = 120 Hz
100.02 kHz - 99.8 kHz = 220 Hz
100.02 kHz - 99.6 kHz = 420 Hz
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Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.
Answer:
(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0
Explanation:
Solution
Given that:
C = 6 µ which is = 6 * 10^ ⁻6
R = 2 MΩ, which is = 2 * 10^ 6
ε = 20 V
(a) When it is at the time constant we have the following:
λ = CR
= 6 * 10^ ⁻6 * 2 * 10^ 6
λ =12 seconds
(b) We solve for the half life of the circuit which is given below:
d₀ = d₀ [ 1- e ^ ⁺t/CR
d = decay mode]
d₀/2 = d₀ 1- e ^ ⁺t/12
2^⁻1 = e ^ ⁺t/12
Thus
t/12 ln 2
t = 12 * ln 2
t = 12 * 0.693
t = 8.31 seconds
(c) We find the current at t = 0
So,
I = d₀/dt
I = d₀/dt e ^ ⁺t/CR
= CE/CR e ^ ⁺t/CR
E/R e ^ ⁺t/CR
Thus,
at t = 0
I E/R = 20/ 2 * 10^ 6
= 10µ A
(d) We find the voltage across the capacitor at t = 0 which is shown below:
V = IR
= 10 * 10^ ⁻6 * 2 * 10^ 6
V = 20 V
(e) We solve for he voltage across the resistor.
At t = 0
I = 0
V =0
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
Answer:
[tex] T_A_C = 6.296 kN [/tex]
[tex] R = 10.06 kN [/tex]
Explanation:
Given:
[tex] T_A_B = 8.7 kN[/tex]
Required:
Find the tension TAC and magnitude R of this downward force.
First calculate [tex] \alpha, \beta, \gamma [/tex]
[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]
[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]
[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]
To Find tension in AC and magnitude R, use sine rule.
[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]
Substitute values:
[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]
Solve for T_A_C:
[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]
[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]
Solve for R.
[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]
[tex] R = 8.7 * 1.156 [/tex]
R = 10.06 kN
Tension AC = 6.296kN
Magnitude,R = 10.06 kN
Two blocks of rubber (B) with a modulus of rigidity G = 14 MPa are bonded to rigid supports and to a rigid metal plate A. Knowing that c = 80 mm and P = 46 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 7 mm.
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]
Making a the subject formula;
[tex]a = \frac{Gd}{r}[/tex]
Substituting into the above formula;
[tex]a = \frac{14000000*0.007}{1400000}[/tex]
[tex]a = \frac{98000}{1400000}[/tex]
[tex]a = 0.07m or 70mm[/tex]
a = 0.07m or 70mm.
Also, shearing stress;
[tex]r = \frac{P}{2bc}[/tex]
Making b the subject formula;
[tex]b = \frac{P}{2cr}[/tex]
Substituting into the above equation;
[tex]b = \frac{46000}{2*0.08*1400000}[/tex]
[tex]b = \frac{46000}{224000}[/tex]
[tex]b = 0.205m or 205mm[/tex]
b = 0.205m or 205mm
Sometimes, steel studs may not be used on outside walls because they are?
Answer:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Explanation:
We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...
Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have four arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in computer applications. For example, an app that determines the winner of a sales contest would input the number of units sold by each salesperson. The sales person who sells the most units wins the contest. Write pseudocode, then a C# app that inputs a series of 10 integers, then determines and displays the largest integer. Your app should use at least the following three variables:
Counter: Acounter to count to 10 (i.e., to keep track of how many nimbers have been input and to determine when all 10 numbers have been processed).
Number: The integer most recently input by the user.
Largest: The largest number found so far.
Answer:
See Explanation
Explanation:
Required
- Pseudocode to determine the largest of 10 numbers
- C# program to determine the largest of 10 numbers
The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;
The largest of the 10 numbers is then saved in variable Largest and printed afterwards.
Pseudocode (Number lines are used for indentation to illustrate the program flow)
1. Start:
2. Declare Number as 1 dimensional array of 10 integers
3. Initialize: counter = 0
4. Do:
4.1 Display “Enter Number ”+(counter + 1)
4.2 Accept input for Number[counter]
4.3 While counter < 10
5. Initialize: Largest = Number[0]
6. Loop: i = 0 to 10
6.1 if Largest < Number[i] Then
6.2 Largest = Number[i]
6.3 End Loop:
7. Display “The largest input is “+Largest
8. Stop
C# Program (Console)
Comments are used for explanatory purpose
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int[] Number = new int[10]; // Declare array of 10 elements
//Accept Input
int counter = 0;
while(counter<10)
{
Console.WriteLine("Enter Number " + (counter + 1)+": ");
string var = Console.ReadLine();
Number[counter] = Convert.ToInt32(var);
counter++;
}
//Initialize largest to first element of the array
int Largest = Number[0];
//Determine Largest
for(int i=0;i<10;i++)
{
if(Largest < Number[i])
{
Largest = Number[i];
}
}
//Print Largest
Console.WriteLine("The largest input is "+ Largest);
Console.ReadLine();
}
}
}
A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg was used. Calculate the Brinell hardness (in HB) of this material. Enter your answer in accordance to the question statement HB
Answer:
HB = 3.22
Explanation:
The formula to calculate the Brinell Hardness is given as follows:
[tex]HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }[/tex]
where,
HB = Brinell Hardness = ?
P = Applied Load in kg = 500 kg
D = Diameter of Indenter in mm = 10 mm
d = Diameter of the indentation in mm = 1.55 mm
Therefore, using these values, we get:
[tex]HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }[/tex]
HB = 3.22
A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the material is 240 MPa. A factor of safety of 2.5 is desired. The maximum permissible internal pressure is most nearly: 90 kPa 230 kPa 430 kPa D. 570 kPa csauteol psotolem here Pcr 8. A structural steel tube with a 203 mm x 203 mm square cross section has an average wall thickness of 6.35 mm. The tube resists a torque of 8 N m. The average shear flow is most nearly
A. 100 N/m
B. 200 N/m
C. 400 N/m
D. 800 N/m
Answer:
1) 2304 kPa
2) B. 200 N/m
Explanation:
The internal pressure of the of the tank can be found from the following relations;
Resisting wall force F = p×(1/4·π·D²)
σ×A = p×(1/4·π·D²)
Where:
σ = Allowable stress of the tank
A = Area of the wall of the tank = π·D·t
t = Thickness of the tank = 15 mm. = 0.015 m
D = Diameter of the tank = 25 m
p = Maximum permissible internal pressure pressure
∴ σ×π·D·t = p×(1/4·π·D²)
p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa
With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa
2) The formula for average shear flow is given as follows;
[tex]q = \dfrac{T}{2 \times A_m}[/tex]
Where:
q = Average shear flow
T = Torque = 8 N·m
[tex]A_m[/tex] = Average area enclosed within tube
t = Thickness of tube = 6.35 mm = 0.00635 m
Side length of the square cross sectioned tube, s = 203 mm = 0.203 m
Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²
[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]
Hence the average shear flow is most nearly 200 N/m.
Following are the solution to the given question:
Calculating the allowable stress:
[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]
[tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]
Calculating the Thickness:
[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]
The stress in a spherical tank is defined as
[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]
[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]
Calculating the shear flow:
[tex]\to q=\frac{T}{2A}[/tex]
[tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]
[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]
Therefore, the final answer is "".
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The lower half of a 7-m-high cylindrical container is filled with water (rho = 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and the bottom of the cylinder. (Round the final answer to one decimal place.)
Answer:
Pressure difference (ΔP) = 63,519.75 kpa
Explanation:
Given:
ρ = 1,000 kg/m³
Height of cylindrical container used (h) = 7m / 2 = 3.5m
Specific gravity (sg) = 0.85
Find:
Pressure difference (ΔP).
Computation:
⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ] ∵ [ g = 9.81]
⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]
⇒ Pressure difference (ΔP) = 34.335 [1,850]
⇒ Pressure difference (ΔP) = 63,519.75 kpa
A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected in series in order that the same current shall be supplied from 240 V, 50 Hz mains. Ignore the resistance of the inductor and calculate: i. the inductance of the inductor; ii. the impedance of the circuit; iii. the phase difference between the current and the applied voltage.
Answer:
(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°
Explanation:
Solution
Given that:
I= 5 A
V = 125V
Resistance R= Not known yet
Thus
To find the resistance we have the following formula which is shown below:
R = V/I
=125/15
R =8.333Ω
Now,
Voltage = 240
Frequency = 50Hz
Current (I) remain at = 15A
Z= not known (impedance)
so,
To find the impedance we have the formula which is shown below:
Z = V/I =240/15
Z= 16Ω⇒ Z = R + jXL
Z = 8.333 + jXL = 16
Thus
√8.333² + XL² = 16²
8.333² + XL² = 16²
XL² = 186.561
XL = 13.658Ω
Now
We find the inductance of the Inductor and the impedance of the circuit.
(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:
L = XL/w
=13.658/ 2π * 50
=13.658/314.15 = 0.043 = 43.43 mH
Note: w= 2πf
(ii) For the impedance of the circuit we have the following:
z = 8.333 + j 13.658
z = 16∠58.61° Ω
(iii) The next step is to find the phase difference between the applied voltage and current.
Q = this is the voltage across the inductor in a series of resonant circuit.
Q can also be called the applied voltage
Thus,
Q is described as an Impedance angle
Therefore, Q = 58.81°
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]
Weight transfer is given as follows;
[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]
[tex]F_R[/tex] + [tex]F_F[/tex] = 18000
[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]
[tex]F_R[/tex] = 18000*17/30 = 10200 N
[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel [tex]F_R[/tex] = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel [tex]F_F[/tex] = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]
The least coefficient of friction, μ = 0.625.
