If new research showed that the standard heights for crest vertical curve design were H1=3.0 ft and H2=1.5 ft, respectively, by what percent would the minimum length of curvature increase or decrease for a crest curve with a stopping sight distance of 500 ft and grades of +3% and -2%, respectively.

Answers

Answer 1
Answer:

Answer:

A = 5

S<L, L = 714.89ft

S>L, L = 650.29ft

L = 115.85ft

Percentage min. Length of curvature = 6.2 %

Explanation: see explanation at the attached file


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Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Answers

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa    ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = ((1,0,0) \cdot (1,1,0))/(1 * \sqrt2)$

       $=(1)/(\sqrt2 )$

$\cos \lambda = ((1,0,0) \cdot (1,-1,1))/(1 * \sqrt3)$

       $=(1)/(\sqrt3 )$

  τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

ii. Slip plane  --- (1 1 0)

   Slip direction -- [1 1 1]

  $\cos \phi = ((1, 0, 0) \cdot (1, -1, 0))/(1 * \sqrt2) =(1)/(\sqrt2)$

   $\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

 τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

iii. Slip plane  --- (1 0 1)

    Slip direction --- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, 1))/(1 * \sqrt2) =(1)/(\sqrt2)$

   $\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

    Slip direction  ---- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, -1))/(1 * \sqrt2)=(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, -1, 1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?

Answers

Find the given attachment

Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold climates because it is light, fashionable, and a very effective thermal insulator. So, it is no surprise that such clothing has largely replaced thick and heavy old-fashioned coats. Consider a jacket made up of six layers of 0.1 mm thick synthetic fabric (k = 0.026W/m.K) with 1.2 mm thick air space (k = 0.026 W/m.K) between the fabric layers. Assuming the inner surface temperature of the jacket to be 25˚C and the surface area to be 1.25 m2 , determine the heat loss through the jacket when the temperature of the outdoors is -5˚C and the heat transfer co-efficient of outer surface is 25 W/m2 .K. What would be the thickness of a wool fabric (k = 0.035W/m.K) if the person has to achieve the same level of thermal comfort wearing a thick wool coat instead of a jacket. (30 points)

Answers

Answer:

Q=127.66W

L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

To solve any problem related to heat transfer, the general equation is used

Q = delta / R

Where

Q = heat

Delta = the temperature difference

R = is the thermal resistance by conduction, convection and radiation

to solve this problem we propose the previous equation

Q = delta / R

later we find R

R=[tex]r=(6L1)/(AK1) +(5L2)/(AK2)+(1)/(Ah)

R=(6(0.0001))/((1.25)(0.026)) +(5(0.012))/((1.25)(0.026))+(1)/((25)(1.25)) =0.235 K/w

Q=(25-(-5))/0.235=127.66W

part b

we use the same ecuation with Q=127.66

Q = delta / R

ΔR=(L)/(KA) +(1)/(hA) \nR=(L)/((0.035)(1.25)) +(1)/((25)(1.25))\n R=22.85L+0.032\nQ=(T1-T2)/R\n\n127.66=(25-(-5))/(22.85L+0.032)\nsolving for L\nL=9.2mm

Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a vehicle with a mass of 1000 kg. a. For a low speed test (v = 2.5 m/s), compute the energy in the vehicle just prior to impact. If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm? b. At a higher speed impact of v = 25 m/s, considerable deformation occurs. To absorb the energy, the front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)? If it is desired to limit the deformation to 10 cm, what level of resistance force is required? What is the deacceleration of the vehicle in this condition?

Answers

Answer:

Explanation:

The concept of Hooke's law was applied as it relates to deformation.

The detailed steps and appropriate substitution is as shown in the attached file.

Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment with a row of mailboxes in the post office. These mailboxes were numbered 1 through 150, and beginning with mailbox 2, he opened the doors of all the even-numbered mailboxes, leaving the others closed. Next, beginning with mailbox 3, he went to every third mail box, opening its door if it were closed, and closing it if it were open. Then he repeated this procedure with every fourth mailbox, then every fifth mailbox, and so on. When he finished, he was surprised at the distribution of closed mailboxes. Write a program to determine which mailboxes these were.

Answers

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

A chemical process converts molten iron (III) oxide into molten iron and carbon dioxide by using a reducing agent of carbon monoxide. The process allows 10.08 kg of iron to be produced from every 16.00 kg of iron (III) oxide in an excess of carbon monoxide. Calculate the percentage yield of iron produced in this process.

Answers

Answer:

percentage yield = 63%

Explanation:

The yield efficiency or percentage yield measure the amount of products that are formed from a given amount of reactant. For a percentage yield of 100, all the reactants are completely converted to product. Mathematically, the percentage yield is given by:

percentage\ yield = (Actual\ yield)/(expected\ yield) * 100\nActual\ yield = 10.08kg\nExpected\ yield= 16.00kg\n\n\therefore percentage\ yield = (10.08)/(16.00) * 100 = 63 \%

Other Questions
Suppose there is a mobile application that can run in two modes: Lazy or Eager. In Lazy Mode, the execution time is 3.333 seconds. In Eager Mode, the app utilizes a faster timer resolution for its computations, so the execution time in Eager Mode is 2 seconds (i.e., Eager Mode execution time is 60% of Lazy Mode execution time).After finishing computation, the app sends some data to the cloud, regardless of the mode it’s in. The data size sent to the cloud is 600 MB. The bandwidth of communication is 15 MBps for WiFi and 5 MBps for 4G. Assume that the communication radio is idle during the computation time. Assume that the communication radio for WiFi has a power consumption of 75 mW when active and 15 mW when idle. Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization. Assume that the power consumption of the CPU is a linear function of its utilization. In other words: P = (Idle Power) + (Utilization)*(Power per unit Utilization). A configuration of the mobile app involves choosing a timer resolution (Lazy or Eager) and choosing a type of radio (WiFi or 4G). For example, faster timer resolution (Eager) and 4G network is a configuration, while slower resolution (Lazy) and WiFi is another. There are four possible configurations in all.Required:What is the average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G?