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den301095 [7]
2 years ago
8

An overpass is being built over the PVI of an existing equal-tangent sag curve. The sag curve has a 70 mi/h design speed and G1=

-5%, G2=+3%. Determine the minimum necessary clearance height of the overpass and the resultant elevation of the bottom of the overpass over the PVI. Ignore the cross-sectional width of the overpass.

Engineering
1 answer:
vivado [14]2 years ago
8 0

Answer: Resultant elevation will be 30.98 ft

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2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
gladu [14]

Answer:

we use

sigma=force /area\\A=\pi r^2\\D=2*r

5 0
2 years ago
1. A spur gear made of bronze drives a mid steel pinion with angular velocity ratio of 13 /2 : 1. Thepressure angle is 14 1/2° .
Alik [6]

Answer:

Given data

w1/w2=6.5/1

Power=5 KW

wp=1800 rpm

angle=14 degrees

Based on above values,the minimum diameter=30 mm

5 0
2 years ago
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting
Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = \frac{5L}{3}T, CD = \frac{4L}{3} C

B) AB = AD = BC = BD = 0

   AC = \frac{5L}{3} T, CD = \frac{4L}{3} C

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = \frac{5L}{3} T,  CD = \frac{4L}{3}C

7 0
2 years ago
An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)    

v=total volume i.e. volume of container"80L"   80L=0.08 cubic meter

vf=vol of liquid phase  vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg          hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0
2 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose
Delicious77 [7]

Answer: 0.93 mA

Explanation:

In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.

We can express the resistance as follows:

R = ρ.L/A

In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2

Replacing by the values, we get R as follows:

R = 1.4 1010 Ω

Applying Ohm’s Law, and solving for the current I:

I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA

7 0
2 years ago
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