1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gelneren [198K]
2 years ago
10

Which trait shared by dolphins and bats possibly lead to the evolution of echolocation in these two animal groups? the need to m

ove quickly through dark environments the need to be able to send sound waves through the air the need to be able to send sound waves through the water the need to communicate with other species through echolocation.
Physics
1 answer:
Lisa [10]2 years ago
3 0

A trait shared by dolphins and bats that possibly led to the evolution of echolocation in these two animal groups will be the need to move quickly through dark environments.

<h3>What is the evolution of echolocation?</h3>

Our understanding of the evolution of echolocation in bats has shifted as a result of recent molecular phylogenies. These phylogenies imply that bats with advanced echolocation

According to one interpretation of these trees, laryngeal echolocation originated in the ancestor of all living bats. Echolocation may have been lost in Old World fruit bats

The vast adaptive radiation in echolocation call design is substantially controlled by ecology, demonstrating how environmental perceptual problems influence call design.

A trait shared by dolphins and bats that possibly led to the evolution of echolocation in these two animal groups will be the need to move quickly through dark environments.

Hence option A is correct.

To learn more about the evolution of echolocation refer to the link;

brainly.com/question/20789287

You might be interested in
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

6 0
3 years ago
A car accelerates at 3 m/s*2. Assuming the car starts from rest, how much time does it need to
uranmaximum [27]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the time take by a car.

<h3><u>Solution:-</u><u> </u></h3>

According to the first equation of motion:

v = u + at

★ Substituting the values in the above formula,we get:

⇒ 12 = 0 + 3 × t

⇒ 12 = 3t

⇒ 3t = 12

⇒ t = 12/3

⇒ t = 4 sec

5 0
3 years ago
Read 2 more answers
A _____ is NOT a compound machine.
UNO [17]

A bolt, all these besides bolt are a combo of simple machines

6 0
3 years ago
Read 2 more answers
two projectile are thrown at the same intial velocity the angle of the one is theta ,the angle of the oher is 90-theta /can both
Sauron [17]

Answer:

Yes, if θ = 45°

Explanation:

3 0
3 years ago
You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in
Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

      Volume = \frac{\text{mass}}{\text{density}}

So,       V = \frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}

               = 5.181 \times 10^{-8} m^{3}

Now, we will calculate the area as follows.

      Area = \frac{\text{volume}}{\text{length}}

               = \frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}

               = 2.15 \times 10^{-11} m^{2}

Formula to calculate the resistance is as follows.

         R = \rho \frac{l}{A}

             = \frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}

             = 2.71 \times 10^{6} ohm

Thus, we can conclude that the resistance of given wire is 2.71 \times 10^{6} ohm.

4 0
3 years ago
Other questions:
  • A student will study german at last 3 years
    8·1 answer
  • Do you think the universe has a center
    14·1 answer
  • How many grams of CO-60 result in 1 Millicuire of activity? How many years until the activity decays to 1 microcuire tl/2 =5.3 Y
    12·1 answer
  • If the maximum time between the emitted and received pulse is 1/ 10 second, what is the farthest distance you could measure with
    12·1 answer
  • Which sections of the heating curve illustrate this process?
    14·1 answer
  • A free positive charge released in an electric field will _____
    15·1 answer
  • A 260-m length of wire stretches between two towers and carries a 115-a current. determine the magnitude of the force on the wir
    5·1 answer
  • The amount of gravity between 1kg of lead and Earth is ___ the amount of gravity between 1kg of marshmallows and Earth.
    8·1 answer
  • Why do we use the two-body problem to solve interplanetary trajectories, instead of including all of the appropriate gravitation
    15·1 answer
  • Question 2 of 25
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!