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Sindrei [870]
3 years ago
5

Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the sp

eed of light.
A.) FM radio. Find the wavelength of an FM radio station signal broadcasting at a frequency of 102.5 MHz .
B.) X rays. X rays have a wavelength of about 0.200 nm . What is their frequency?
C.) The Big Bang. Microwaves with a wavelength of 1.20 mm , left over from soon after the Big Bang, have been detected. What is their frequency?
D.) Sunburn. Sunburn (and skin cancer) are caused by ultraviolet light waves having a frequency of around 1.04×1016 Hz . What is their wavelength?
E.) SETI. It has been suggested that extraterrestrial civilizations (if they exist) might try to communicate by using electromagnetic waves having the same frequency as that given off by the spin flip of the electron in hydrogen, which is 1.42 GHz . To what wavelength should we tune our telescopes in order to search for such signals?
F.) Microwave ovens. Microwave ovens cook food with electromagnetic waves of frequency around 2.45 GHz . What wavelength do these waves have?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Answer:

2.92682 m

1.5\times 10^{18}\ Hz

250000000000 Hz

2.88462\times 10^{-8}\ m

0.21126 m

0.12244 m

Explanation:

c = Speed of light = 3\times 10^8\ m/s

\lambda = Wavelength

f = Frequency

Wavelength is given by

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m

The wavelength is 2.92682 m

Frequency is given by

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz

The frequency is 1.5\times 10^{18}\ Hz

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{1.2\times 10^{-3}}\\\Rightarrow f=250000000000\ Hz

The frequency is 250000000000 Hz

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.04\times 10^{16}}\\\Rightarrow \lambda=2.88462\times 10^{-8}\ m

The wavelength is 2.88462\times 10^{-8}\ m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m

The wavelength is 0.21126 m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{2.45\times 10^9}\\\Rightarrow \lambda=0.12244\ m

The wavelength is 0.12244 m

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The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J

The amount of water that is needed to produce this energy can be found using the equation

E_{in}=mC\Delta T

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m is the amount of water

C=4186 J/kg^{\circ}C is the specific heat capacity of water

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And solving for m, we find:

m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg

Learn more about specific heat capacity:

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