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Use the Node analysis to find the power of all resistors

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The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m

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 table, at S = 220 m and V = 110 km/h, we can see that;

Radius of vertical curvature; K = 74

A = G1 - G2

A = 2.7 - (-3.5)

A = 2.7 + 3.5

A = 6.2

L = KA

Thus;

L = 74 × 6.2

L = 458.8 m

Read more about stopping sight distance at; https://brainly.com/question/2087168

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

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.

Explanation:

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Answer:

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Answer:

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Answer:

what do you need help with

Explanation:

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]

Explanation:

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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.

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.

Answer:

For fluid power, a working fluid is a gas or liquid that primarily transfers force, motion, or mechanical energy. ... Examples without phase change include air or hydrogen in hot air engines such as the Stirling engine, air or gases in gas-cycle heat pumps, etc.

Explanation:

Answer:

A working fluid is a gas or liquid that primarily transfers force, motion or mechanical energy

Examples:Air, pentane, chlorofluorocarbons, butane, propane and ammonia

Explanation:

print() has a variable number of parameters. this is the answer.

hope this helps you

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Answer:

the minimum diameter of the bar is 1.634 in

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.

Answer:

New Zealand may use some of these solutions to prevent air pollution

Explanation:

Using public transports.

Recycle and Reuse

No to plastic bags

Reduction of forest fires and smoking

Use of fans instead of Air Conditioner

Use filters for chimneys

Avoid usage of crackers

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

Answer:

answer is C. his seat belt

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

Answer:

Explanation:

The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.

The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ),  BUT

When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )

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.

Answer:

[tex]D=0.41m[/tex]

Explanation:

From the question we are told that:

Discharge rate [tex]V_r=0.35 m3/s[/tex]

Distance [tex]d=4km[/tex]

Elevation of the pumping station [tex]h_p= 140 m[/tex]

Elevation of the Exit point [tex]h_e= 150 m[/tex]

Generally the Steady Flow Energy Equation SFEE is mathematically given by

[tex]h_p=h_e+h[/tex]

With

[tex]P_1-P_2[/tex]

And

[tex]V_1=V-2[/tex]

Therefore

[tex]h=140-150[/tex]

[tex]h=10[/tex]

Generally h is give as

[tex]h=\frac{0.5LV^2}{2gD}[/tex]

[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

Therefore

[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]

[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]

[tex]D=0.41m[/tex]

Answer:

The answer of these questions is

Explanation:

b) NO

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]