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lozanna [386]
3 years ago
13

Which of the following suggestions would best help alleviate the Gulf of Mexico dead zone?

Engineering
1 answer:
JulijaS [17]3 years ago
3 0

The answer has to be cut down on gas emissions. This can cut down emissions to the dead zone.

You might be interested in
5. Identify the pros and cons of<br> manufactured siding.
lapo4ka [179]
Aluminum
Aluminum siding is out of date and really has no advantages in today’s market. It dents, it’s difficult to keep clean, and it’s hard to update it once it’s installed. Aluminum siding is not recommended, and most professional contractors won’t even install it.

Vinyl
Today’s mainstream siding material is vinyl. It is the most economical type of siding, coming in at $4 to $5 per square foot.

Available in a wide variety of colors, vinyl siding not only protects your home, it can make a statement. You can choose from whites, ivories, grays, browns, reds, greens and even blues — there’s a color for everyone.

In addition, vinyl is a complete self-flushing water-exclusion system. It is made to be water repellent and durable in any kind of weather. Its lightweight footprint makes it easy to install, which minimizes expenses.

Vinyl is low-maintenance and requires only periodic cleaning with a power washer every few years – more if you want it to really shine.

Premium vinyl
A step up from regular vinyl, premium vinyl gives you a more wood-like appearance. It’s usually a thicker vinyl that looks more rigid because it has no oil canning, as lighter vinyl does.

The upgrade to premium gives you more strength and longer lifespan. Many premium vinyl sidings also have integrated insulation or r-values (insulation grade) of 5 to 6.

Premium vinyl cost is generally higher than vinyl, at around $7 to 8 per square foot. Premium vinyl also comes in a rainbow of colors for customization and personalization

Natural wood
Wood siding provides a more rustic look for your home, with a more customized result.

The siding can be in long horizontal pieces or it can be shaped to suit your individual design needs. That’s the beauty of wood. You may want the look of clapboard – overlapping pieces of wood installed horizontally – or shingles – smaller pieces of wood overlapping like a shingled roof. Wood siding is a good choice for this façade.

Fiber cement siding
Finally, you may hear about fiber cement siding. Today’s version is a combination of wood pulp, cement, clay and sand. It can be made to look like wood siding, and it is installed and wears much like wood siding.

This is different from the fiber cement siding that was used in homes built prior to the late 1980s. Those older sidings contain asbestos, and any work with them should be done by a specialized asbestos-removal professional. Today’s fiber cement siding is safer and longer-lasting, if you’re willing to pay the price.

Fiber cement siding can be ordered pre-colored, eliminating the need to paint the siding. If you prefer to paint it, that is still an option, because the material does accept paint rather easily. Fiber cement siding requires a periodic recaulk of butt joints at trim terminations, which is a little more maintenance than a vinyl siding.

When to replace siding
When should you replace your siding? It should be replaced if it has any kind of water damage, mold or rotting. If it hasn’t been properly maintained, it might be time for new siding. And, of course, if you simply want a new look, new siding can be a side show for the neighborhood.
5 0
3 years ago
Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
atroni [7]

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1:  ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2:  ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3:  ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4:  ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5:  ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

6 0
3 years ago
A driver complains that his front tires are wearing
Margarita [4]

Answer:

The correct option is;

Neither A nor B

Explanation:

The location of the where the thread wears in tire that has too high inflation is at the thread pattern center due to the reduced size of the contact patch with the load of the car resting on the central portion of the tire's contact surface

When the wear occurs at the outer edges of the tire, the load of the car rests on the outer edges as the contact patch increases due to the tire being under-inflated

Camber is the slope provided in road pavement to drain off water from the road

Roads with camber has a raised middle portion and wear due to camber includes outer-edge tread wear, inner-edge tread wear and tire feathering

8 0
3 years ago
A heat recovery system​ (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop
blagie [28]

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

7 0
3 years ago
A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the
padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

A=\pi*D^2/4

Replacing the diameter the area results:

A= 17.76 in^2

Therefore the the stress results:

σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi

5 0
3 years ago
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