All categories

3 years ago

What can you do to prepare for a long distance trip

Engineering

2 answers:

IceJOKER [234]3 years ago

6 0

Answer:

Well, you ALWAYS want to plan ahead, how much gas money you'll need, to know the hotel you'll stay at (if so.) You also want to eat light and stay hydrated! I think the most important thing to prepare for a long distance trip is, to be comfortable! (:

Explanation:

JulsSmile [24]3 years ago

4 0

Answer:

plan where you are going and plan where you are going to stop for gas or either you are going to fly

Explanation:

You might be interested in

The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content

MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

$Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf$

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

$\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}$

Therefore, $V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}$

$V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy$

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since $1 yd^{3}= 27 ft^{3}$

The embankment requires water of 6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

$Unit_{weight}=\frac {17451720}{498594}=35.00186 lb$

1 gallon is approximately $8.35 yd^{3}$ hence

$\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}$

That's approximately 4.2 gallons

7 0

3 years ago

How can input from multiple individuals improve design solutions for problems that occur because of a natural disaster, such as

Alla [95]

Answer:

Map and avoid high-risk zones.

Build hazard-resistant structures and houses.

Protect and develop hazard buffers (forests, reefs, etc.)

Develop culture of prevention and resilience.

Improve early warning and response systems.

Build institutions, and development policies and plans.

Explanation:

5 0

3 years ago

The electricity generated by wind turbines annually in kilowatt-hours per year is given in a file. The amount of electricity is

Anika [276]

Answer:

Steps:

1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year

2. load the data file for example, in matlab, use ('fileame.txt') to load the file

3. create variables from each column of your data

for example, in matlab,

x=t{1}

y=t{2}

4. plot the wind velocity and electricity generated.

plot(x, y)

5. Label the individual axis and name the graph title.

title('Graph of wind velocity vs approximate electricity generated for the year')

xlabel('wind velocity')

ylabel('approximate electricity generated for the year')

5 0

3 years ago

Read 2 more answers

A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s

KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 ) ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity = (change in p - X sin∅ ) $D^{2}$ / 32 V

= ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2 / 32 * 0.90

= - 59.904 sin (-90) * 0.0035 / 28.8

= 0.1874 / 28.8

viscosity = 0.00650 Ib s /ft^2

8 0

4 years ago

Read 2 more answers

What are the three elementary parts of a vibrating system?

zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring - the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and zero elasticity.

7 0

3 years ago