A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heated uniformly to115 oC, determine the length and the diameter of this rod to the nearest micron at the new temperature if the linear coefficient of thermal expansion of steel is 12.5 x 10-6 m/m/oC. What is the stress on the rod at 115oC.
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
a basketball player pushes down with a force of 50N on a basketball that is indlated to a gage pressure of 8.0x10^4 Pa. What is the diameter of comtact between the ball nad the floor
Answer:
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Explanation:
The basketball experiments a normal stress ([tex]\sigma[/tex]), in pascals, due to normal force from the floor ([tex]N[/tex]). By definition of normal stress, we have the following equation:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex] (1)
Where [tex]D[/tex] is the diameter of the contact area between the ball and the floor, in meters.
Please notice that magnitude of the normal force equals the magnitude of external force given by the basketball player and weight is negligible in comparison with normal and external forces.
If we know that [tex]N = 50\,N[/tex] and [tex]\sigma = 8.0\times 10^{4}\,Pa[/tex], then the diameter of the contact area is:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex]
[tex]D^{2} = \frac{4\cdot N}{\pi\cdot \sigma}[/tex]
[tex]D = 2\cdot \sqrt{\frac{N}{\pi\cdot \sigma} }[/tex]
[tex]D = 2\cdot \sqrt{\frac{50\,N}{\pi\cdot (8\times 10^{4}\,Pa)} }[/tex]
[tex]D\approx 0.028\,m[/tex] [tex](28\,mm)[/tex]
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Elliptic curve cryptography is considered as the latest and probably the one with a future. Having seen RSA in earlier modules, in which ways do YOU think elliptic cryptography is more advanced than RSA. You may read other material or get this information from the internet to answer this question. But make sure to provide necessary references when you do cite others.
Answer:
The answer is below.
Explanation:
Some of the ways, how I think elliptic cryptography is more advanced than RSA are the following:
1. ECC - Elliptic Curve Cryptography uses smaller keys for the same level of security, particularly at greater levels of security.
2. ECC can work well and at a faster rate on a small-capacity device compared to RSA
3. It uses offer speedier SSL handshakes that enhance security
4. It offers fast signatures
5. It allows signatures to be computed in two stages, which enables lower latency than inverse throughput.
6. Relatively quick encryption and decryption
A pinion and gear pair is used to transmit a power of 5000 W. The teeth numbers of pinion
and gear are 20 and 50. The module is 5 mm, the pressure angle is 20o
and the face width is 45 mm. The
rotational speed of pinion is 300 rev/min. Both the pinion and the gear material are Nitralloy 135 Grade2 with a hardness of 277 Brinell. The quality standard number Qv is 5 and installation is open gearing
quality. Find the AGMA bending and contact stresses and the corresponding factors of safety for a
pinion life of 109
cycles and a reliability of 0.98
Answer:
mark me as a brainleast
Explanation:
209781
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at ft/s^2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 18 ft/s^2.
Required:
Determine the minimum distance d between the cars so as to avoid a collision.
Answer:
Explanation:
Using the kinematics equation [tex]v = v_o + a_ct[/tex] to determine the velocity of car B.
where;
[tex]v_o =[/tex] initial velocity
[tex]a_c[/tex] = constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
[tex]S = d + v_ot + \dfrac{1}{2}at^2[/tex]
Then:
[tex]v_B = 60-12t[/tex]
The distance traveled by car B in the given time (t) is expressed as:
[tex]S_B = d + 60 t - \dfrac{1}{2}(12t^2)[/tex]
For car A, the needed time (t) to come to rest is:
[tex]v_A = 60 - 18(t-0.75)[/tex]
Also, the distance traveled by car A in the given time (t) is expressed as:
[tex]S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
Relating both velocities:
[tex]v_B = v_A[/tex]
[tex]60-12t = 60 - 18(t-0.75)[/tex]
[tex]60-12t =73.5 - 18t[/tex]
[tex]60- 73.5 = - 18t+ 12t[/tex]
[tex]-13.5 =-6t[/tex]
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
[tex]S_B = S_A[/tex]
[tex]d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
[tex]d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2[/tex]
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
In heavy traffic areas you should wave pedestrians across the street if there is no crosswalk
In heavy traffic areas, you should wave pedestrians across the street if there is no crosswalk: False.
What is a crosswalk?A crosswalk can be defined as the marked or specially paved part of a road that is characterized by heavy traffic, so as to enable pedestrians have right of way to cross the street because drivers are required by traffic law to stop for them.
However, a driver or other road users in heavy traffic areas shouldn't wave pedestrians across the street if there is no crosswalk
Read more on traffic laws here: https://brainly.com/question/22768531
Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b. DSSS and CDMA are fundamental technologies used with WiMAX BWA c. OFDM is implemented to increase spectral efficiency and to improve noise performance d. all of the statements are correct
Answer:
d. all of the statements are correct.
Explanation:
WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.
A network has three independent file servers, each with 90 percent reliability. The probability that the network will be functioning correctly (at least one server is working) at a given time is:
Answer:
The correct answer is "99.9%".
Explanation:
According to the information given in the question,
[tex]P(1 \ fail) = 0.1[/tex]
The probability of all fail will be:
[tex]P(all \ fail) = (0.1)^3[/tex]
[tex]=0.001[/tex]
hence,
[tex]P(not \ all \ fail)= 1-P(all \ fail)[/tex]
[tex]=0.999[/tex]
[tex]=99.9[/tex] (%)
Thus the above is the right answer.
Given : x² + 200x = 166400 The current park is a square, and the addition will increase the width by 200 meters to give the expanded park a total area of 166,400 square meters To Find : the side length of the current square park. Solution: x² + 200x = 166400 => x(x + 200) = 166400 166400 = 320 * 520 => (320)(320 + 200) = 166400 => x = 320 side length of the current square park. = 320 m Learn More: Which expression is a possible leading term for the polynomial ... brainly.In/question/13233517
Answer:
320 m
Explanation:
To find the side length of the current park, x, we solve the quadratic equation for the area of the park
x² + 200x = 166400
x² + 200x - 166400 = 0
We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x
So, we re-write the expression as
x² + 200x - 166400 = 0
x² + 520x - 320x - 166400 = 0
We write out the factors of the expression,
x² + 520x - 320x - 320 × 520 = 0
Factorizing the expression, we have
x(x + 520) - 320(x + 520) = 0
(x + 520)(x - 320) = 0
x + 520 = 0 or x - 320 = 0
x = -520 or x = 320
Since x is not negative, we take the positive answer.
So, x = 320 m
Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.
True
False
Answer:
True ❤️
-Solid by solid can make Cylindrical wire doubles Strengths in tension
If an elevator repairer observes that cables begin to fray after 15 years, what process might he or she use to create a maintenance schedule for their replacement? fallacious reasoning reductive reasoning inductive reasoning deductive reasoning
Answer:
inductive reasoning
Explanation:
Inductive reasoning is one of the type of reasoning method in which generalized consequences are derived from limited observations. By observing few data, general conclusions are drawn. The conclusions drawn are false in inductive reasoning. In the given situation, the conclusion drawn by the elevator repairer has been drawn by inductive reasoning. His observation of some cables led him to draw the conclusion about all the cables. The result of the reasoning is false.
A investor will invest in mutual fundwith a probability of 0.6, will invest in government fundwith a probability of 0.3, and will invest in both fundswith a probability of 0.15. Find the probability that the investor will invest in either mutual fundor government fund.
Answer:
0.75
Explanation:
From this question above we have the following information
A = probability of investment in mutual fund= 0.6
B = probability of investment in government fund = 0.3
C = probability of investing in both the mutual fund and the government= 0.15
Where to find the probability of this investor investing in either of these two
= Prob(a) + prob(b) - prob(c)
= 0.6 + 0.3 - 0.15
= 0.9 - 0.15
= 0.75
Explain ROLAP, and list the reasons you would recommend its use in the relational database environment.
Answer:
ROLAP is a branch of OLAP that is used to contain Relational database ( RDB ). which is a very fast database ( quick process of queries )
Very fast to access and also fast in processing queries provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesExplanation:
ROLAP ( Relational On-line Analytical processing ) is a branch of OLAP that is used to contain Relational database ( RDB ).
Advantages of ROLAP ( reasons for the use of ROLAP )
Very fast to access ( fast in processing queries )provides multidimensional view of data / supports multidimensional database schema with RDBMssupports large databasesThe value of universal gas constant is same for all gases?
a) yes
b)No
Answer:
The answer of these questions is
Explanation:
b) NO
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?
Answer:
[tex]V_{pp}=2V[/tex]
Explanation:
Source Voltage [tex]V= 120V[/tex]
Frequency [tex]f=60Hz[/tex]
Peak output voltage [tex]Vp=15V[/tex]
Peak Output Voltage with filter [tex]V_p'=14V[/tex]
Generally the equation for Peak to peak voltage is mathematically given by
[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]
Therefore
[tex]V_{pp}=2(V_p-v_p')[/tex]
[tex]V_{pp}=2(15-14)[/tex]
[tex]V_{pp}=2V[/tex]
An ideal gas within a piston-cylinder assembly undergoes a Carnot refrigeration cycle. The isothermal compression occurs at 325 K from 2 bar to 4 bar. The isothermal expansion occurs at 250 K. Determine:
a. the coefficient of performance
b. the heat transfer to the gas during the isothermal expansion, in kj per kmol of gas
c. the magintude of the net work input, in kj per kmol of gas.
Answer:
a) [tex]\mu=3.3[/tex]
b) [tex]Q=1440.7KJ/Kmol[/tex]
c) [tex]W=1872.9KJ/Kmol[/tex]
Explanation:
From the question we are told that:
Initial Temperature [tex]T_1=325k[/tex]
initial Pressure [tex]P_1=2 bar[/tex]
Final Pressure [tex]P_2=4 bar[/tex]
iso-thermal expansion [tex]T_2=250k[/tex]
a)
Generally the equation for Coefficient of performance is mathematically given by
[tex]\mu=\frac{T_2}{T_1-T_2}[/tex]
[tex]\mu=\frac{250}{325-250}[/tex]
[tex]\mu=3.3[/tex]
b)
Generally the equation for Heat Expansion is mathematically given by
[tex]Q=RT_2 In(\frac{P_2}{P_1})[/tex]
Where
R=Gas constant
[tex]R=8.314462618[/tex]
Therefore
[tex]Q=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]Q=1440.7KJ/Kmol[/tex]
c)
Generally the equation for work input is mathematically given by
[tex]W=RT_1 In(\frac{P_2}{P_1})[/tex]
[tex]W=8.314462618*250 In(\frac{4}{2})[/tex]
[tex]W=1872.9KJ/Kmol[/tex]
The coefficient of performance is 3.33, the heat transfer in the isothermal expansion is 1440.71kJ/K.mol and the work input is calculated as 1872.92kJ/K.mol
Given Data:
T1 = 325KP1 = 2 barP2 = 4 barT2 = 250KIsothermal expansion occurs at 250K.
a) The coefficient of performanceThis is calculated as
COP =[tex]\frac{T_2}{T_1-T_2}=\frac{250}{325-250} =3.33[/tex]
b) Heat Transfer in isothermal expansion[tex]Q = RT_2In(\frac{p_2}{p_1})[/tex]
Therefore; In isothermal process du = 0
R = 8.314 AkJ/K.mol
Q = 8.314 * 250 In(4/2)
Q = 1440.71kJ/K.mol
c) Work InputW[tex]_i_n[/tex]=[tex]RT_1In(\frac{p_2}{p_1})\\W_i_n=8.314*325In(4/2)\\W_i_n=1872.92kJ/K.mol[/tex]
The work input is 1872.92kJ/K.mol
Learn more on Carnot cycle here:
https://brainly.com/question/14983940
https://brainly.com/question/13170743
In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2 and a Y value of 1.0. An in-service stress level of 200 MPa has been calculated. What is the length of a surface crack (in mm) that will lead to fracture
Answer:
[tex]l=24mm[/tex]
Explanation:
From the question we are told that:
Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]
Y value [tex]Y=1.0[/tex]
Stress level of[tex]\sigma =200 MPa[/tex]
Generally the equation for length of a surface crack is mathematically given by
[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]
[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]
[tex]l=0.024m[/tex]
Therefore
in mm
[tex]l=24mm[/tex]
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more ________________ than bypass and absorb the energy of the transient.
Answer:
I think ( MOV Metal oxide varistors )
Transients (surges) on a line can cause spikes or surges of energy that can damage delicate electronic components. A SPD device contains one or more MOV Metal oxide varistors than bypass and absorb the energy of the transient.
Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at 2208C with a mass flow rate of 1.2 kg/s. Refrigerant exits at 7 bar, 708C. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in m3 /s, and (b) the power input to the compressor, in kW.'
Answer:
a)[tex]V_1=4.88m^2/s[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)[tex]P=-119.18kW[/tex]
Explanation:
From the question we are told that:
Steady State Saturated vapor [tex]T_1= -20C=>253k[/tex]
Mass Flow rate [tex]M=1.2kg/s[/tex]
Exit Pressure [tex]P_2=7bar[/tex]
Exit Temperature [tex]T_2=70C=>373k[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>P_1=1.399 bar[/tex]
Generally the equation for Volumetric Flow rate is mathematically given by
For Inlet
[tex]V_1=m\frac{RT_1}{P_1}[/tex]
[tex]V_1=m\frac{8314*253}{1.399*10^3}[/tex]
[tex]V_1=18.97m^2/s[/tex]
For outlet
[tex]V_2=m\frac{RT_2}{P_2}[/tex]
[tex]V_2=1.2*\frac{8314*343}{7*10^3}[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)
Generally the equation for Steady state mass and energy equation is mathematically given by
[tex]P=m(h_1-h_2)[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>h_1=24.76kJ/kg[/tex]
[tex]T_2= 70C =>h_2=124.08kJ/kg[/tex]
Therefore
[tex]P=1.2(12.76-124.08)[/tex]
[tex]P=-119.18kW[/tex]
Therefore
Power input into the compressor is
[tex]P=-119.18kW[/tex]
A force measuring instrument comes with a certificate of calibration that identifies two instrument errors and assigns each an uncertainty at 95% confidence over its range. Provide an estimate of the instrument design-stage uncertainty.
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Answer:
[tex]U=\pm 0.382N[/tex]
Explanation:
From the question we are told that:
Resolution: 0.25 N
Range: 0 to 100 N
Linearity error: within 0.20 N over range
Hysteresis error: within 0.30 N over range
Generally the equation for Stage Uncertainty is mathematically given by
[tex]U=\sqrt{u_0^2+u_T^2}[/tex]
Where
[tex]u_0=Zero\ order\ uncertainty[/tex]
[tex]u_0=\pm 0.5*0.25[/tex]
[tex]u_0=\pm=0.125[/tex]
And
u_T=Total instrumental Uncertainty
[tex]u_T=\sqrt{l_e^2+h_e^2}[/tex]
Where
l_e=Error of linearity
h_e=Error due to hysteresis
Hence
[tex]u_T=\sqrt{0.20^2+0.30^2}[/tex]
[tex]u_T=\pm 0.36[/tex]
Therefore
[tex]U=\sqrt{(0.125)^2+0.36^2}[/tex]
[tex]U=\pm 0.382N[/tex]
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
122 cm
182 cm
82 cm
22 cm
108.2 cm
Answer: (b)
Explanation:
Given
Original length of the rod is [tex]L=100\ cm[/tex]
Strain experienced is [tex]\epsilon=82\%=0.82[/tex]
Strain is the ratio of the change in length to the original length
[tex]\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm[/tex]
Therefore, new length is given by (Considering the load is tensile in nature)
[tex]\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm[/tex]
Thus, option (b) is correct.
Why water parameters of Buriganga river vary between wet and dry seasons?
Explain.
A(n) ____ combines two planetary gearsets to provide more gear ratio possibilities. A)compound planetary gearset B)orifice C)detent D)lock-up torque converter
Answer:
The answer is A. Compound Planetary Gearset.
Explanation:
The Compound Planetary Gear block represents a planetary gear train with composite planet gears. Each composite planet gear is a pair of rigidly connected and longitudinally arranged gears of different radii. One of the two gears engages the centrally located sun gear while the other engages the outer ring gear.
Compound planetary gear sets have at least two planet gears attached in line to the same shaft, rotating and orbiting at the same speed while meshing with different gears. Compounded planets can have different tooth numbers, as can the gears they mesh with.
a video inspection snake is use
Answer:
very good thx
Explanation:
4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape and the ground (assumed to be uniform) be if the temperature of the tape rises to 30 C. Neglect the expansion of the earth (the radius of the earth is 6.37 X 106 m)
Answer:
2102.1 m
Explanation:
Temperature at the equator = 0⁰
Radius of the earth = 6.37x10⁶
Required:
We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.
Final temperature = ∆T = 303-273 = 30
S = 11x10^-6
The clearance R = Ro*S*∆T
=6.37x10⁶x 11x10^-6x30
= 2102.1m
Or 2.102 kilometers
Thank you
What statement about the print() function is true?
print() has a variable number of parameters.
print() can have only one parameter.
print() can be used to obtain values from the keyboard.
print() does not automatically add a line break to the display.
Explanation:
print() has a variable number of parameters. this is the answer.
hope this helps you
have a nice day
Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.20 kg; the mass of the liquid is 1.85 kg. Balance D reads 3.10 kg and balance E reads 7.50 kg. The volume of block A is 4.15 × 10−3 m3.
a) What is the density of the liquid?
b) What will the balance D read if block A is pulled up out of the liquid?
c) What will the balance E read if block A is pulled up out of the liquid?
Answer:
a) [tex]m_e= 3.05 Kg[/tex]
b) [tex]\rho=1072.3kg/m^3[/tex]
c) [tex]m_e= 3.05 Kg[/tex]
Explanation:
From the question we are told that:
Beaker Mass [tex]m_b=1.20[/tex]
Liquid Mass [tex]m_l=1.85[/tex]
Balance D:
Mass [tex]m_d=3.10[/tex]
Balance E:
Mass [tex]m_e=7.50[/tex]
Volume [tex]v=4.15*10^{-3}m^3[/tex]
a)
Generally the equation for Liquid's density is mathematically given by
[tex]m_e=m_b+m_l+(\rho*v)[/tex]
[tex]\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}[/tex]
[tex]\rho=1072.3kg/m^3[/tex]
b)
Generally the equation for D's Reading at A pulled is mathematically given by
m_d = mass of block - mass of liquid displaced
[tex]m_d=m- (\rho *v )[/tex]
[tex]m=3.10+ (1072.30 *4.15*10^{-3}m^3 )[/tex]
[tex]m=18.10kg[/tex]
c)
Generally the equation for E's Reading at A pulled is mathematically given by
[tex]m_e=m_b+m_l[/tex]
[tex]m_e = 1.20 + 1.85[/tex]
[tex]m_e= 3.05 Kg[/tex]
A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage of 120 V rms. The load draws a total of 10 kW at a power factor of 0.85 (lagging). Calculate the rms value of the phase currents and the magnitude of the per-phase load impedance. Draw a phasor diagram showing all tlme voltages and currents.
Answer:
Following are the solution to the given question:
Explanation:
Line voltage:
[tex]V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v[/tex]
Power supplied to the load:
[tex]P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi[/tex]
[tex]10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A[/tex]
Check wye-connection, for the phase current:
[tex]I_{ph}=I_L= 32.68\ A[/tex]
Therefore,
Phasor currents: [tex]32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A[/tex]
Magnitude of the per-phase load impedance:
[tex]Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega[/tex]
Phase angle:
[tex]\phi = \cos^{-1} \ (0.85) =31.79^{\circ}[/tex]
Please find the phasor diagram in the attached file.
Compute the minimum length of vertical curve that will provide 220 m stopping sight distance for a design speed of 110 km/h at the intersection of a -3.50% grade and a +2.70% grade.
i have made notes and saved it as a pdf u can take it to answer question and make ur concept good
The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m
We are given;
Stopping sight distance; S = 220 m
Design Speed; V = 110 km/h
Intersection grade 1; G1 = +2.7
Intersection Grade 2; G2 = -3.5
From the AASHTO Table attached, we can trace the value of the radius of vertical curvature for the given stopping sight distance and design speed.From the table, at S = 220 m and V = 110 km/h, we can see that;
Radius of vertical curvature; K = 74
Now, the difference in grade given is;A = G1 - G2
A = 2.7 - (-3.5)
A = 2.7 + 3.5
A = 6.2
Formula for the minimum length of vertical curve is;L = KA
Thus;
L = 74 × 6.2
L = 458.8 m
Read more about stopping sight distance at; https://brainly.com/question/2087168
Which of the following is not a part sympathetic activation during the fight or flight response?
Answer:
Digestion functions become more active
Explanation:
I just took the text!
