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3 years ago

Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a

Physics

1 answer:

yKpoI14uk [10]3 years ago

3 0

Answer:

Wavelength, $\lambda=2.94\ m$

Explanation:

It is given that,

Speed of radio waves is $v=3\times 10^8\ m/s$

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

$v=f\lambda$

$\lambda$ is wavelength

$\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m$

So, the wavelength of WFNX’s radio waves is 2.94 m.

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A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

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Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}$

$d=\frac{1474*8.8}{2}$

d= 6485.6m

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2 years ago

What are challenges of securing a crime scene

kotegsom [21]

At any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

<h3>What is the meaning of physical evidence?</h3>

In evidence law, physical evidence (also called real evidence or material evidence) is any material object that plays some role in the matter that gave rise to the litigation, introduced as evidence in a judicial proceeding (such as a trial) to prove a fact in issue based on the object's physical characteristics.

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Hence, at any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

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1 year ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

2 years ago