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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

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An electrical power plant manages to send 88% of the heat produced in the burning of fossil fuel into the water-to-steam convers
Naya [18.7K]

Answer:

A

Explanation:

Let the x represent the amount of heat generated from the fossil fuel.

88% of x = 0.88 x

0.88 x was used to convert water to steam.

heat carried by steam = 40% × 0.88 x = 0.352 x

efficiency of the heat -to- work conversion = work output / work input = 0.352 x / x = 0.352 × 100 = 35.2 % which is less than 40 %

4 0
3 years ago
May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
A car travels a distance of 98 meters in 10 seconds. What is the average speed of the car?
stira [4]

Answer:

C

Explanation:

4 0
3 years ago
Read 2 more answers
.
lesya692 [45]

Answer:

John Dalton

Explanation:

Dalton's atomic theory was the foundation for a new understanding of chemical structures. He proposed that matter was constituted by indivisible and indestructible particles "atoms." He theorized that all atoms of a particular substance were equal, and the atoms of different substances had atoms of different sizes and masses.

He also proposed that all compounds of elements were combinations of elements but in a very precise ratio.

7 0
3 years ago
Read 2 more answers
| A 1.0 kg stone is dropped from a bridge 100 m above a canyon. What will be the kinetic energy of the stone after it
Mnenie [13.5K]

Answer:

Option D

490 J

Explanation:

When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.

Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then

PE initial = 1*9.81*100= 981 J

At 50 m, PE will be 1*9.81*50=490.5 J

Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence

981-490.5= 490.5 J

Approximately, 490 J

8 0
3 years ago
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