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wariber [46]
4 years ago
7

A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=

1s−1. calculate the broadcast wavelength of the radio station 88.70 fm.

Physics
2 answers:
Arte-miy333 [17]4 years ago
7 0

The broadcast wavelength of radio waves is \fbox{3.38\,{\text{m}}} or \fbox{338\,{\text{cm}}}.

Further Explanation:

Radio waves are one of the types of the electromagnetic radiation. It travels with the speed of light in vacuum. The range of frequency of radio waves is very large ranging from 30\,{\text{Hz}} - 300\,{\text{GHz}}.

Radio waves are used to transmitting the signal over a large distance on the Earth. Radio waves can go beyond the line of sight through diffraction and reflection. Radio waves are non ionizing radiation. They have not sufficient energy to separate electrons from atoms or molecules.

Transmitters are used to generating radio waves artificially and antennas are used to receiving the waves.

Given:

The frequency of radio waves is 88.7\,{\text{MHz}}.  

Concept:

The relationship between frequency and wavelength given by:

\fbox{\begin\\\lambda=\dfrac{c}{f}\end{minispace}}

Here, c is the speed of light, f is the frequency of radio waves, and \lambda is the wavelength of radio waves.

Substitute 88.7 \times {10^6}\,{\text{Hz}} for f and 3 \times {10^8}\text{ m}/\text{s} for c in above equation.

\begin{aligned}\lambda&=\frac{{3\times{{10}^8}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}}{{88.7\times{{10}^6}\,{\text{Hz}}}}\\&=3.38\,{\text{m}}\\&= 338}}\,{\text{cm}}\\\end{aligned}

Thus, the broadcast wavelength of the radio station is \fbox{3.38\,{\text{m}}}.

Learn more:

1.  Threshold frequency of cesium atom brainly.com/question/6953278

2.  Stress developed in a string brainly.com/question/12985068

3. Components of a lever brainly.com/question/1073452

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetic waves

Keywords:

Frequency, broadcast wavelength, radio waves, radio station, 88.7fm, speed of light, broadcast, electromagnetic waves, channel. transmitters, 3.38m, 338 cm

AURORKA [14]4 years ago
6 0
The frequency of the radio station is
f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength \lambda is
\lambda= \frac{c}{f}
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m
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andriy [413]

Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

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  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

7 0
3 years ago
A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo
DIA [1.3K]

Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

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olga2289 [7]

Answer:

NO_{1.499}

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n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }

n_{N} = 2.632\,kmol

n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }

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n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
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a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
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Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

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X = horizontal distance traveled = 100 m

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v_{o} = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

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