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amm1812
3 years ago
9

What adaptation does a wood frog have that enable it to survive in deciduous forest ?

Physics
2 answers:
KATRIN_1 [288]3 years ago
6 0

Answer:

The temperature in deciduous forest can range from -30° C to 30°C with 10°C being an average temperature.

Wood frog has adapted to survive in freezing temperatures during winters. It does so by practicing thawing and freezing in a cyclic or systematic manner.

It helps the frog to accumulate glucose and urea in the body. These substances then act as cryoprotectant that is, they prevent the freezing of the cells or tissues (by lowering the freezing point) of the organism and thus prevent the loss of water from the cells.

On arrival of spring, the warm temperature thaws the frog after which the leftover glucose is converted back to glycogen and urea is excreted out of the body.

sweet [91]3 years ago
5 0

Wood frogs have this adaptation where they accumulate urea in their bodies and convert their liver glycogen to glucose to act as cryoprotectants. This prevents the formation of ice crystals in their bodies that could cause damage cells during freezing in winter.  

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b) 2ft/s

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3 years ago
A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
2 years ago
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Mandarinka [93]

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