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Gelneren [198K]
1 year 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]1 year 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

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A 4.6 kilogram block of ice would absorb how much energy
nataly862011 [7]

Answer: 45 joules of energy

Explanation:

6 0
3 years ago
A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

4 0
3 years ago
(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12
Elodia [21]

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

4 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
Who was the scientist who gave us the laws of motion?.
timurjin [86]

Answer:

Isaac Newton

Explanation:

Newton's laws of motion, three statements describing the relations between the forces acting on a body and the motion of the body.

3 0
2 years ago
Read 2 more answers
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