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mylen [45]
3 years ago
11

The annoying sound from a mosquito is produced when it beats its wings at the average rate of 600 wingbeats per second. What is

the frequency of the soundwaves?
Physics
1 answer:
podryga [215]3 years ago
6 0
Frequency is given by the formula cycles÷seconds

So if the mosquito can complete 600 cycles in one second, our formula will be

f=600÷1
f=600!

The frequency is 600 Hz
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NeX [460]

Answer:

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Explanation:

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4 0
3 years ago
When two objects of different masses, different temperatures, and different sizes are placed in thermal contact, energy will alw
denpristay [2]

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a) from the hotter object to the cooler object

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6 0
3 years ago
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which o
Sergeeva-Olga [200]

Answer:

  • on the moon, they will fall at the time
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Explanation:

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4 0
3 years ago
What is air pressure
vesna_86 [32]
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6 0
3 years ago
Read 2 more answers
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
4 years ago
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