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juin [17]
3 years ago
13

6. Directions: Drag each item to the correct location on the diagram. Many of the astronomical events that can be witnessed from

the Earth's surface can be explained by the Earth's revolution and its rotation.
Which of the following phenomena are best explained by the Earth's revolution around the Sun, and which are best explained by the Earth's rotation along its axis?

Which of these go under- Earths Rotation and which go under- Earths Revolution

the Moon's movement
during one night

changing seasons

a sunrise

the Sun's changing
midday height over
several days

passing years

a sunset
Physics
2 answers:
algol133 years ago
4 0
Rotation- sunset,sunrise,moons movement

all the others are revolution

leave a thanks if i answered your question 
slega [8]3 years ago
3 0
Earth's daily rotation:
--  <span>the Moon's movement during one night
--  </span>a sunrise
--  a sunset

Earth's yearly revolution:
<span>--  changing seasons
--  the Sun's changing midday height over several days
--  passing years</span>




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Answer:

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<u>Composition of the Atmosphere</u>

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One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F
Alja [10]

Answer:

\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}

Explanation:

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A harmonic is an integral multiple of the fundamental frequency.

\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}

f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}

b) Wave speed

(i) Calculate the wavelength

In a  fundamental vibration, the length of the string is half the wavelength.

\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}

(b) Calculate the speed s

\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}

\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}

\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}

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hope this helps
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