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Tema [17]
3 years ago
5

A changing climate has strong implications for biodiversity. Studies of fossil and pollen distribution show that species are ver

y sensitive to climate changes. When the climate changes, species often die out in their present areas and colonize new areas. Therefore, as the climate changes in the future, there will be disruption of natural communities and extinction of populations and species.
Based on the information above, which of these would most likely be affected by a global warming?
A) Rattlesnakes in the desert.
B) Polar bears living on the sea ice.
C) Earthworms burrowing in the ground.
D) Spider monkeys in the tropical rain forests.
Physics
2 answers:
Eva8 [605]3 years ago
6 0

Answer: B) Polar bears living on the sea ice.  

Global warming is a natural phenomena which involves the increase in the global atmospheric temperatures because of the human activities such as deforestation, burning of the fossil fuels and others. Terrestrial as well as aquatic animals and other life forms are directly or indirectly are affected by global warming. Global warming is related with the melting of the global glaciers.

The polar bear is a species of bear which is a native of Arctic glacier region. These animals are adapted to survive in the harsh winter conditions of the Arctic region. The change in the climatic conditions in the Arctic region due to global warming will directly influence the population of these animals in a negative sense. These organisms have thick coat and consists of heavy body fat, they cannot survive in hot summer conditions.

ivolga24 [154]3 years ago
4 0
The answer is B i believe.
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adell [148]

Answer:

Part a)

v_f = 7.99 m/s

Part b)

y = 3.25 m

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

v_f^2 - v_i^2 = 2 a y

here we have

v_i = 2.22 m/s

y = -3 m

v_f^2 - (2.22)^2 = 2(-9.81)(-3)

v_f = 7.99 m/s

Part b)

at highest point of his motion the final speed will be zero

so we will have

v_f^2 - v_i^2 = 2 a (\Delta y)

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7 0
3 years ago
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11
Arada [10]

Answer:

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

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