All categories

3 years ago

Which sound wave-object interaction is used by animals in echolocation?

Physics

2 answers:

iris [78.8K]3 years ago

8 0

Answer:

Explanation:

Animals who use echolocation send a special kind of sound, when this sound wave collides with the object it reflect back and animal detect the location, shape and size of object.Example of such animal is bats, whales, dolphins and oilbird. Bats uses echolocation to reach their pray and for their movement at night.

tekilochka [14]3 years ago

5 0

The answer is reflection

You might be interested in

The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure

Anuta_ua [19.1K]

Crazy Wally Ok Ok ok hhahahaha

4 0

2 years ago

List 3 indicators that a chemical reaction has occurred.

mina [271]

Hello! There are many indicators to show a sign that a chemical reaction has occured. These are 3 as followed.

1. An odor is produced.

2. A change in temperature.

3. A color change.

Others can be a formation of a solid or a gas.

8 0

3 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

7 0

3 years ago

Read 2 more answers

The lightbulb is an example of

AlexFokin [52]

A object that has been reinvented so it is more energy efficient

4 0

3 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

3 years ago