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3 years ago

13

Describe what happens to water waves when they strike a flat surface?

1 answer:

shutvik [7]3 years ago

6 0

Answer:

In this case reflection occurs.

Explanation:

This happens when they bounce on objects that they can not pass through.

Hope it helps

okay.

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Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq

Anna35 [415]

Water boilingis the answer

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3 years ago

Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert this

Paladinen [302]

1 m = 1 000 000 ym

converted other way we can say that:

1 ym = $10^{-6}$ m

Now, since we have ym^2 which is ym*ym which means:
1 ym^2 = $(10^{-6}) ^2 = 10^{-12}$ m

we have 1,5 ym^2 which means that answer is:
$1.5* 10^{-12} m$

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4 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flows through the bulb. What is t

Naddik [55]

Answer: 240ohms

Explanation: P=IV

P=60watts

I=0.5A

V=120volts

From ohms law

V=IR

R=V/I

R=120/0.5

R= 240ohms

7 0

3 years ago