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solong [7]
3 years ago
14

A radio wave has a frequency of 8.6 × 108 hz. what is the energy of one photon of this radiation (h = 6.63 × 10–34 j • s)?

Physics
1 answer:
My name is Ann [436]3 years ago
5 0

Answer: 57.018\times10^{-26}J

Energy is directly proportional to the frequency.

The energy of a photon is given by:

E=h\nu

Where, E is the energy and \nu is the frequency.

Frequency of the radio wave is given:

\nu=8.6\times10^8 Hz

h=6.63\times10^{-34}J.s

Multiply the above two:

Energy,

E=h\nu=6.63\times10^{-34}J.s\times8.6\times10^8 Hz= 57.018\times10^{-26}J

Hence, the energy of each photon of radio wave having frequency  \nu=8.6\times10^8 Hz is 57.018\times10^{-26}J

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The study and analysis of light according to its component wavelengths is called
BARSIC [14]

ANSWER:

The study and analysis of light according to its component wavelengths is called spectroscopy.

EXPLANATION:

Spectroscopy is the branch of science  that is concerned with the investigation and measurement of spectrum produced when matter interacts with or emits electromagnetic radiation.It helps us to identify atoms and molecules in the object.Spectroscopy is used to find out Dopplers effect (the red shift and blue shift),which tells how fast the object is comming towards earth or moving away from the earth.

4 0
3 years ago
A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo
Sever21 [200]
The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle. 
                        x-component of speed = (31 m/s)(cos 35°) 
                                                               = 25.39 m/s
Thus, the horizontal velocity component of the ball is 25.39 m/s. 
8 0
3 years ago
Read 2 more answers
You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m
ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

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creativ13 [48]

Your experiment should keep one thing constant and measure the other. So vary the temp and measure the pressure. You will get a set of data that relates pressure with temp.
<span>PV = nRT
So
P and T are directly proportional.
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8 0
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Which objects will likely have the smallest gravitational force between them?
lora16 [44]
I’d say two soccer balls that are touching each other, I hope that helps!
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