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sasho [114]
2 years ago
5

Daniel Kipper is a produce farmer in northern California. His major customers are grocery stores in the Midwest. Daniel's produc

t is a perishable item and will only last for about 2 weeks after it has been picked, so Daniel is concerned with getting his product to his customers quickly. He ships almost daily when his produce is in season. However, he also needs to be aware of the cost of shipping. Which form of shipping is Dave most likely to use?A. TrucksB. ShipsC. AirD. Rail
Business
1 answer:
Scrat [10]2 years ago
5 0

Answer:

The correct answer is A. Trucks.

Explanation:

To think about transport is to think of trucks. Ground transportation is omnipresent in our lives and roads and trucks are its backbone. But ground transportation has many different forms, which translates into very different types of vehicles. Depending on the type of merchandise to be transported, the dimensions or structure of the vehicle, we must choose the one that best suits our needs.

In land transport, the shape and equipment of trucks make them more appropriate for some types of shipments or materials.

Canvas truck (or tauliner) : This type of trailer is the most common and its semi-trailer is covered by the sides and above with tarps, which can be removed. This allows you to be very comfortable for loading and unloading and can be adapted to a large number of materials, as well as being suitable for transporting products that are difficult to load.

Open platform : In this case the platform where the load goes is open (although it may be partially covered by the sides depending on the merchandise and the possibility that it moves). Although it also offers many options, it is usually used for heavy goods, construction, etc.

Refrigerators : As the name implies, refrigerators are the trucks in charge of transporting the merchandise to be transported refrigerated by land. Depending on whether they have cold generation systems or only insulation with the outside they can be refrigerators, refrigerated or isothermal. Food is the most common customer of this type of vehicles.

Although they are less frequent, there are also calorific trucks to keep the temperature above a certain number of degrees.

Tank : Very common ADR merchandise, tanks are used to transport liquid, gaseous and chemical products, which have particular safety requirements.

Closed : closed trucks, as opposed to tarpaulins, have a rigid structure in the cargo compartment. This means that they can only be loaded from the rear, while the canvas ones can also be loaded from the sides. They are vehicles commonly used for urban delivery and parcel delivery, although there are also closed box trailers.

Car holder : These trucks are specifically designed for land car transport. There are two models of car carriers: the open, which is the most common, and the closed. The latter does not allow vehicles to be seen from outside and is sometimes used to transport the most expensive cars.

Cage Truck : Cage trucks are used to transport live animals. To do this, they have part of the sides or roof open so that air reaches the animals and there is sufficient ventilation.

Containers : The containers are watertight cargo structures, which allows them to protect the merchandise from the weather. They are used for multimodal transport and to facilitate loading and unloading of container ships.

The infinity of materials to transport makes the variety of trucks continue and is very long: hopper (often used for earthmoving), stakes (log transport, among others), concrete mixer, armored (money movements), vehicles special for heavy machinery and a long etcetera.

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A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is kno
raketka [301]

Here is the correct question.

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to \dfrac{1}{0.85} due to the stress that causes the failure (strength)  is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load  must be reduced to \dfrac{1}{1.1} because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to \dfrac{1}{0.95}  because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

\sigma' = \dfrac{P'}{A'}

where,

\sigma = stress

P = load

A = cross-sectional area of the cylinder

∴

\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}

replacing P' with \dfrac{1}{1.1}P   and d' with \dfrac{1}{0.95}d, we have:

\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }

\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }

\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }

\sigma' =\sigma \times 0.82045

\dfrac{\sigma' }{\sigma } =0.82045

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

n_d = \dfrac{loss  \ of  \ function \  parameter }{maximum \  allowable \ parameter}

where;

the design factor = n_d

n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}

the design factor  n_d = 1.434.

Thus,  the minimum design factor that will guarantee that the part will not fail. = 1.434

7 0
2 years ago
Hedge Fun is a landscaping firm that specializes in topiary. It contracts with the owners of 125 local homes and provides its se
tekilochka [14]

Answer:

Break-even level of output = 56

Explanation:

Given:

Annual Revenue = $1,300

Total Fixed cost = $28,000

Variable cost = $800

Computation of contribution:  

Contribution = Sales - Variable cost

Contribution = Revenue - Variable cost

Contribution = $1,300 - $800

Contribution = $500

Computation of Break-even level of output:

Break-even level of output = Total Fixed cost / Contribution

Break-even level of output = $28,000 / $500

Break-even level of output = 56

3 0
2 years ago
Fairchild Garden Supply expects $700 million of sales this year, and it forecasts a 15% increase for next year. The CFO uses thi
vazorg [7]

Answer:

D) 3.48

Explanation:

Current Year Sales = $700

Growth rate = 15%

Projected Sales=$700*15% +$700

Which is $805

Required inventory = $30.2 + 0.25*projected sales

Req.Inv = $30.2 + 0.25($805)

Req.Inv = $231.45

Inventory turn over = projected sales/Req.inv

$805/$231.45

Inventory turn over = 3.48 times

8 0
1 year ago
"Swiss Clothing Store had a balance in the Accounts Receivable account of $920,000 at the beginning of the year and a balance of
Fudgin [204]

Answer:

Receivable days are 52 days.

Explanation:

Receivable days can be found from the following formula:

Receivables days = Receivables / Credit Sales * 365

The credit sales here is $6,650,000 during the year and the average receivables days is $950,000 [(950,000 + 980,000)/2] during the year. By putting the values we have:

Receivables days = $950,000 / $6,650,000  * 365 = 52 days

So the average receivable collection days were 52 days during the year.

6 0
2 years ago
Nezzie invests in 300 shares of stock in the fund shown below. name of fund nav offer price lkit mid-cap $16.58 $16.99 nezzie pl
pochemuha

The correct answer is D) 33.66

5 0
1 year ago
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