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mamaluj [8]
3 years ago
15

Some bats have specially shaped noses that focus ultrasound echolocation pulses in the forward direction. Why is this useful?

Physics
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

The evolutionary success of bats is accredited to their ability, as the only mammals, to fly and navigate in darkness by echolocation, thus filling a niche exploited by few other predators. Over 90% of all bat species use echolocation to localize obstacles in their environment by comparing their own high frequency sound pulses with returning echoes. The ability to localize and identify objects without the use of vision allows bats to forage for airborne nocturnal insects, but also for a diverse range of other food types including motionless perched prey or non-animal food items.

The agility and precision with which bats navigate and forage in total darkness, is in large part due to the accuracy and flexibility of their echolocation system. The echolocation clicks of the few echolocating Pteropodidae (Rousettus) are fundamentally different from the echolocation sounds produced in the larynx that we focus on here, and thus not part of this review. Many studies have shown that bats adapt their echolocation calls to a variety of conditions, changing duration and bandwidth of each call and the rate at which calls are emitted in response to changing perceptual demands . In recent years the intensity and directionality of echolocation signals has received increasing research attention and it is becoming evident that these parameters also play a major role in how bats successfully navigate and forage. To perceive an object in its surroundings, a bat must ensonify the object with enough energy to return an audible echo. Hence, the intensity and duration of the emitted signal act together to determine how far away a bat can echolocate an object. Equally important is signal directionality. Bat echolocation calls are directional, i.e., more call energy is focused in the forward direction than to the sides (Simmons, 1969; Shimozawa et al., 1974; Mogensen and Møhl, 1979; Hartley and Suthers, 1987, 1989; Henze and O'Neill, 1991). An object detectable at 2 m directly in front of the bat may not be detected if it is located at the same distance but off to the side. Consequently, at any given echolocation frequency and duration, it is the combination of signal intensity and signal directionality that defines the search volume, i.e., the volume in space where the bat can detect an object.

The aim of this review is to summarize current knowledge about intensity and directionality of bat echolocation calls, and show how both are adapted to habitat and behavioral context. Finally, we discuss the importance of active motor-control to dynamically adjust both signal intensity and directionality to solve the different tasks faced by echolocating bats.

Explanation:

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Feliz [49]

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

v=u+at

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

v=16 + (3)(10)=46 m/s

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

s=ut+\frac{1}{2}at^2

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

t = 10 s is the time

Substituting into the equation,

s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m

4 0
3 years ago
The four principle surveying methods to spatially determine the position of features are: Select one: ut of a. triangulation, tr
Natalija [7]

A: perpendicular offset

B: radiation

C: transliteration

D: theodolite

7 0
3 years ago
24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect
Tatiana [17]

Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

where:

E is the magnitude of the electric field

A is the area of the surface

\theta is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

L = 2.0 m is the side of the sheet, so the area is

A=L^2=(2.0)^2=4.0 m^2

\theta=0^{\circ}, since the electric field is perpendicular to the surface

Therefore, the electric flux is

\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C

4 0
2 years ago
A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by v_{max}=A\omega where A is amplitude and \omega is angular velocity

Angular velocity is given by \omega=\sqrt{\frac{k}{m}}  where k is spring constant and m is mass

So v_{max}=A\sqrt{\frac{k}{m}}

A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m

3 0
3 years ago
Read 2 more answers
If a short-wave radio station broadcasts on a frequency of 9.065 megahertz (MHz), what is the wavelength of
Yakvenalex [24]

Answer:

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one ... Typical radio wave frequencies are about 88~108 MHz .

Explanation:

To calculate the wavelength of a radio wave, you will be using the equation: Speed of a wave = wavelength X frequency.

Since radio waves are electromagnetic waves and travel at 2.997 X

10

8

meters/second, then you will need to know the frequency of the radio wave.

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one million hertz. If the radio wave is from an AM radio station, these are in kilohertz (there are one thousand hertz in a kilohertz). Hertz are waves/second. Hertz is usually the label for the frequency of electromagnetic waves.

To conclude, to determine the wavelength of a radio wave, you take the speed and divide it by the frequency.

Typical radio wave frequencies are about

88

~

108

MHz

. The wavelength is thus typically about

3.41

×

10

9

~

2.78

×

10

9

nm

.

7 0
2 years ago
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