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2 years ago

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

Physics

2 answers:

KATRIN_1 [288]2 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]2 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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Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$

7 0

3 years ago

Which of the following is a contact force?

erica [24]

Option C Magnetic force

5 0

1 year ago

Read 2 more answers

How do you calculate the magnitude of the resultant force of two

zlopas [31]

Answer:

Two forces that act in opposite directions produce a resultant force that is smaller than either individual force. To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force. The direction of the resultant force is in the same direction as the larger force.

3 0

2 years ago

Read 2 more answers

Throwing a football is an example of force because

siniylev [52]

Answer:

i found this provided by the San Francisco 49ers (found on Khan Academy)

Explanation:

I HOPE THIS HELPS!!!

So, what is a force? A force is a push or a pull exerted on one object from another. Forces make things move. You can make something start or stop when you push or pull an object.

There are many different types of forces in action in football. A player kicking a football is a force that makes the football fly through the air. A quarterback throwing a football is another example of a force that makes the football fly in a game.

When studying the concept of force, we can look to history to find mathematical principles that guide the laws of motion. Sir Isaac Newton was one of the most famous scientists of the 17th century to study the laws of forces and motion. Through careful study of how objects react to various forces, Newton developed the Three Laws of Motion. Below are explanations of each law and how these laws can be applied to football.

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2 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s

Mumz [18]

Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

$M = \dfrac{W}{g}$

$M = \dfrac{40}{9.8}$

M = 4.08 Kg

b) Mass on the surface of Lo

Mass of an object remain same.

Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

W' = m g'

W' =4.08 x 1.81

W' = 7.38 N

8 0

2 years ago