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VARVARA [1.3K]
3 years ago
7

Bats use a process called echolocation to find their food. This involves giving out sound waves that hit possible prey or food.

When the waves hit an object, they bounce back to the bat, allowing the bat to locate its prey. What type of wave interaction is echolocation?
Physics
2 answers:
cupoosta [38]3 years ago
8 0
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
Neko [114]3 years ago
4 0

reflection is the correct answer

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A 2000 kg car is moving at 15 m/sec when it collides with a 1200 kg car sitting still. Briefly compare the impulse imparted on t
Roman55 [17]

Answer:

The 1200 kg car.

Explanation:

The 1200 kg car had bo momentum at the initial stage of the journey because it wasn't on motion. But after collision with the 2000 kg car it gained alot of momentum.

4 0
3 years ago
Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.
horsena [70]

Answer:

a) \Delta x=56.25 m

b) imagen adjunta

Explanation:

a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.

Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:

v_{f}^{2}=v_{0}^{2}+2a \Delta x

Queremos encontrar la posición hasta detenerse, osea vf = 0.

\Delta x=\frac{-v_{0}^{2}}{2a}

\Delta x=\frac{-22.5^{2}}{-2*4.5}

\Delta x=56.25 m

b) Para este caso el gráfico se encuentra adjunto.                                      

Espero que te sirva de ayuda!                                                                                                                                                                          

Download pdf
8 0
3 years ago
A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w
katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85  = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x  5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788  x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

4 0
3 years ago
A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate
babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

\phi = NBA

here we have

A = \pi R_s^2

B = \mu_o N i/L

now we have

\phi = N(\mu_o N i/L)(\pi R_s^2)

now the induced EMF is rate of change in magnetic flux

EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

now for induced electric field in the coil is linked with the EMF as

\int E. dL = EMF

E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L

E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}

E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}

E = 58.7 V/m

3 0
3 years ago
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. Th
Dima020 [189]

Answer:

Explanation:

Let the thickness of the film is t and the refractive index of the material of film is n.

When light travels through a sheet of thickness t, the optical path traveled is nt.

When the path of one of slit is covered by a sheet of thickness t, the optical path becomes

x = ( n - 1) t

As the one fringe is shift, so the optical path changed by one wavelength.

i.e., x = λ

So, λ = ( n - 1) t

t=\frac{\lambda }{n-1}

7 0
3 years ago
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