All categories

3 years ago

The common edible frog of europe is a hybrid between two species

Biology

1 answer:

Misha Larkins [42]3 years ago

6 0

<h2>Frog of Europe </h2>

Explanation:

The common edible frog of Europe is a hybrid between two species: Rana lessonae and Rana ridibunda

This hybrid was first described in 1758 and have a wide distribution, from France across central Europe to Russia
Both male and female hybrids exist, but when they mate among themselves, they are rarely successful in producing offspring
Postzygotic isolation exists between the two frog species which means a hybrid cannot develop into a viable and fertile adult

You might be interested in

Some plant seeds live through cold winters but do not sprout until spring. in winter these seeds are _____.

Anna71 [15]

The seeds become dormant to preserve the energy that they would not get during the winter

3 0

3 years ago

Read 2 more answers

compare a human body's daring exercise to change the occur of a dogs body when exposed to a hot environment- since class it says

Romashka-Z-Leto [24]

The human body attempts to cool down during exercise by producing sweat, while a dog's body attempts to cool down by panting

8 0

3 years ago

Carbon is used to make *

Bezzdna [24]

Answer: Lysozymes

Explanation:

6 0

3 years ago

Read 2 more answers

Please Help (will give brainly)

Vaselesa [24]

C is the answer.We are using to classify organisms beyond molecular setting not by the physical structure.note-applicable for this question only

4 0

3 years ago

Read 2 more answers

Mars had an orbital period of 1.88 years. In two or more complete sentences,explain how to calculate the average distance from m

Vanyuwa [196]

This is an interesting problem involving astronomy, in fact, simple physics.
Let r=distance of sun to mars, in metres

Mars had an orbital period of 1.88 years.
=>
tangential velocity, v, of the planet, in m/s is
$v=\frac{2\pi{r}}{T}$
 $=\frac{2\pi{r}}{1.88*365.25*86400}$ m/s, accounting for leap years
$=3.371*10^{-8}\pi{r}$ m/s

The centripetal force, Fc, generated is
$Fc=\frac{mv^2}{r}$
where m=mass of mars = 6.39*10^(24) kg
$=\frac{mv^2}{r}$
$=\frac{6.39*10^{24}v^2}{r}$
$=7.26168*10^9\pi^2r$

The gravitation pull from the sun, Fg, is given by
$Fg=\frac{GMm}{r^2}$
where G=grav. const., =6.67408*10^(-11) m^3 kg^(-1) s^(-2)
M=mass of sun=1.989*10^(30) kg
$=\frac{6.67408*10^{-11}1.989*10^{30}6.39*10^{24}}{r^2}$
$=\frac{8.4826*10^44}{r^2}$

Since the radial distance is in equilibrium, the average distance, r can be found by equation Fc=Fg and solving for r:
Fc=Fg
=>
$7.26168*10^9\pi^2r=\frac{8.4826*10^44}{r^2}$
Solving for the real root:
$r^3=\frac{8.48256*10^44}{7.26168*10^9*%pi^2}$
$=\frac{1.1681263*10^{35}}{\pi^2}$
=2.279*10^11 m

5 0

3 years ago