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pav-90 [236]
3 years ago
5

Over 95% of incoming UV-B radiation is absorbed in the stratosphere true or false

Geography
2 answers:
Sauron [17]3 years ago
7 0
This sentence is true.<span />
Genrish500 [490]3 years ago
5 0

The correct answer is - True.

The stratosphere manages to absorb around 95% of the UV radiation that comes from the Sun. The stratosphere acts like a protective layer in our atmosphere for the living organisms on our planet. If the stratosphere was not able to absorb the UV radiation in that big of an amount and was letting through much more of it to reach the Earth, than life would be almost impossible on our planet, and most of the organisms would die out because the UV radiation will cause extremely big damage and lead to death.

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Does an unusually cold day in summer indicate that a region's climate has changed? Why?
Gwar [14]

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3 years ago
Read 2 more answers
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
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