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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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Acid compounds contain a.Oxygen B. Ion c. Hydrogen
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Answer:

c hydrogen

Explanation:

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Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second
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To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

a sin\theta = m\lambda

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

\theta = Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

sin\theta = \frac{y}{d}

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

\theta = sin^{-1}(\frac{y}{d})

\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})

\theta = 0.2673\°

PART B) Equation both equations we have

a sin\theta = m\lambda

a \frac{y}{d} = m\lambda

Re-arrange to find a,

a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}

a = 1.65*10^{-4}m

8 0
3 years ago
What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
Andru [333]
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
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3 years ago
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A car driving at a constant speed of 64 mi/h travels 68 miles. How many hours did this take?
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64 miles/hour
Therefore 1/64 hours/mile
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A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m
Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A  where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

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V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

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