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Andrew [12]
3 years ago
12

Climate depends on such factors as an areas

Geography
1 answer:
ira [324]3 years ago
4 0
I think it would depend on how much direct sunlight is on the area, altitude, proximity to bodies of water and land shape.
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Why people use so much water than they used to?<br>PLEASE HELP ME ​
Ne4ueva [31]

Answer:

1: Drinking

2: tub

3: Washing plates

4: Flushing toilet

5: Dinner

6: Boiling foods

7: Cleaning

8: Hand Washing

9: Electricity

10: Much more to go

Explanation:

Every day we use huge amounts of water for different purposes. We using tap water, plates, a tub, flushing the toilet, preparing dinner, and much more. Nevertheless, the water is not only used to domestic needs, but in manufacturing and agriculture citizens still use it.

<em><u>Hope this helps.</u></em>

8 0
4 years ago
How do soil regions influence human activities?<br><br>Help pleaseeeee
Juliette [100K]
Different soil regions influenced human activities because of the fertility of the soil. The more fertile the soil the more agriculture and more agricultural means more food.
4 0
3 years ago
A vocabulary in context exercise in which students match words to definitions describing elliptical planetary orbits, applying i
Natali5045456 [20]

Answer:

1. Focus

2. Aphelion

3. Ellipse

4. Semimajor axis

5. Perihelion

6. Eccentricity

Explanation:

In Kepler's first law of planetary motion, he says that planets orbit around the sun, not in perfect circles, but in ellipses (3.) the Earth also orbits the sun in this shape. The shape of the ellipse is determined by two focii. The closer these two points are to each other, the more the ellipse will resemble a circle.

The moon for example, orbits around the Earth. The moon is not always at the same distance from the Earth in all points of the orbit. So the shape is an ellipse, which is made with Earth as one of its focus(1.).

Eccentricity(6.) describes the flatness the orbit is, or how elliptical the orbit is. Remember Kepler's first law, he says that planets move in elliptical orbits and not perfect circles. So it is simply how different from a perfect circle an orbit is. Pluto is known to have a large eccentricity.

Because of the shape of an ellipse, there are going to be points where a planet is closest to the sun and where a planet is farthest from the sun. This is where Kepler's 2nd law of planetary motion comes in. When the planet is closest to the sun, this is called the perihelion (5.) where a planet would be traveling the fastest at this point of the orbit. When the planet is at its farthest point from the sun, this is called the aphelion (2.) where a planet would be traveling the slowest.

The third law of Kepler describes the relationship between the period and the average distance of the sun. The semimajor axis(4.) is the average distance of the planet object from the sun and is measured in Astonomical units (AU). The period is measured in years.  

6 0
4 years ago
a rural kansas watershed that is ungauged has an area of 475 acres and a main channel length of 6870 feet with an average slope
dlinn [17]

Answer:

\mathbf{Q_p  =682 \ \  ft^3/s}

Explanation:

Given that:

Area = 475 acres

The length of the channel (L) = 6870 feet

The average water shield slope (S) = 100 feet/mile

Since; 1 mile = 5280 feet

Burst duration D = 15 min

∴

100 feet/mile = 100/5280

The average water shield slope (S) = 5/264

Using hydrograph method:

The time of concentration t_c = 0.0078L^{0.77} S^{-0.385}

where;

L = 6870

S = 5/264

t_c = 0.0078(6870)^{0.77} (\dfrac{5}{264})^{-0.385}

t_c =32.34 min

Since 60 min = 1 hour

32.34 min will be (32.34*1)/60

= 0.539 hour

Lag time T_l = 0.67\times t_c

T_l = 0.67\times 32.34

T_l = 21.6678\ min

The time to peak i.e

T_p = \dfrac{D}{2}+ T_L \\ \\  T_p = \dfrac{15}{2}+ 21.6678 \\ \\  T_p = 29.168 \ min

T_r = \dfrac{T_p}{5.5} \\ \\  T_r = \dfrac{29.1678}{5.5} \\ \\ T_r = 5.30 \ min

Since D = 15 min is not equal to T_r, then we hydrograph apart from T_r duration lag time.

Then;

T_p \ ' = T_p + \dfrac{D-t_r}{4} \\ \\ T_p \ ' = 29.168 + \dfrac{15-5.30}{4} \\ \\ T_p \ ' = 31.593

Now, we need to determine the peak discharge Q_p by using the formula:

Q_p  = \dfrac{484 \times A}{T_p \ '}

where

484 = peak factor

Recall that A = 475 acres, to miles, we have:

A = 0.7422 mile²

T_p \ ' = 31.593/60

∴

Q_p  = \dfrac{484 \times 0.7422}{\dfrac{31.593}{60}}

\mathbf{Q_p  =682 \ \  ft^3/s}

7 0
3 years ago
In the mid-thirteenth century, Baghdad was destroyed by which group?
vodomira [7]
The correct answer is B, the mongols. They violently sacked Baghdad in 1258. 
3 0
4 years ago
Read 2 more answers
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