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babunello [35]
3 years ago
9

How could you identify the type pf mineral

Geography
1 answer:
Alja [10]3 years ago
3 0
Color hardness Streack clevage or fracture
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In recent years , the number of refugees has been increasing. However most refugeses come from Just a few countries.
alexandr1967 [171]

Explanation:

In 2018, the increase was driven particularly by internal displacement in Ethiopia and asylum-seekers fleeing Venezuela. The proportion of the world's population who were displaced also continued to rise, as the world's forcibly displaced population grew faster than the global population.

5 0
3 years ago
What is the area where an oceanic plate descends into the upper mantle called?
GarryVolchara [31]
The are where an oceanic plate descends into the upper mantle is called a subduction zone.
6 0
3 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
How much does the air weigh within a column that is 1 square inch in area that extends from sea level all the way to the top of
iren2701 [21]

Answer:

option (c) 14.7 lbs

Explanation:

Data provided in the question:

Area of the column = 1 square inch

Now,

we know Air pressure at the ground surface level from the atmosphere

= 14.7 Psi

also, it can be represented as

14.7 Psi = 14.7 pounds per square inch

Therefore,

Weight of the air weigh within a column that is 1 square inch in area

= Air pressure × Area

= 14.7 pounds per square inch × 1 square inch

= 14.7 pounds i.e 14.7 lbs

Hence,

The answer is option (c) 14.7 lbs

5 0
3 years ago
Secondary sources or primary an original copy of the Declaration of Independence
igor_vitrenko [27]
Primary source because it is the original copy of it. Hope this helps and have an awesome rest of your day!
8 0
3 years ago
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