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o-na [289]
3 years ago
3

Please help me!!

Physics
1 answer:
aliya0001 [1]3 years ago
5 0

Answer:A sound wave is a pressure disturbance that travels through a medium by means of particle-to-particle interaction.As one particle becomes disturbed,it exerts a force on the next adjacent particle, disturbing that particle from rest and transporting the energy through the medium.Like any wave,the speed of a sound wave refers to how fast the disturbance is passed from particle to particle.While frequency refers to the number of vibrations that an individual particle makes per unit of time, speed refers to the distance that the disturbance travels per unit of time.

Since the speed of a wave is defined as the distance that a point on a wave (such as a compression or a rarefaction)travels per unit of time,it is often expressed in units of meters/second (abbreviated m/s). In equation form, this is speed = distance/time

The faster a sound wave travels,the more distance it will cover in the same period of time.If a sound wave were observed to travel a distance of 700 meters in 2 seconds,then the speed of the wave would be 350 m/s. A slower wave would cover less distance - perhaps 660 meters - in the same time period of 2 seconds and thus have a speed of 330 m/s.Faster waves cover more distance in the same period of time.

The speed of any wave depends upon the properties of the medium through which the wave is traveling.Typically there are two essential types of properties that affect wave speed - inertial properties and elastic properties.Elastic properties are those properties related to the tendency of a material to maintain its shape and not deform whenever a force or stress is applied to it.A material such as steel will experience a very small deformation of shape and dimension when a stress is applied to it.Steel is a rigid material with a high elasticity.On the other hand,a material such as a rubber band is highly flexible;when a force is applied to stretch the rubber band,it deforms or changes its shape readily.A small stress on the rubber band causes a large deformation.Steel is considered to be a stiff or rigid material,whereas a rubber band is considered a flexible material. At the particle level,a stiff or rigid material is characterized by atoms and/or molecules with strong attractions for each other.The phase of matter has a tremendous impact upon the elastic properties of the medium.For this reason,longitudinal sound waves travel faster in solids than they do in liquids than they do in gases.Even though the inertial factor may favor gases,the elastic factor has a greater influence on the speed (v) of a wave,thus yielding this general pattern:solids > liquids > gases The density of a medium is an example of an inertial property.The greater the inertia (i.e.mass density) of individual particles of the medium,the less responsive they will be to the interactions between neighboring particles and the slower that the wave will be. However,within a single phase of matter,the inertial property of density tends to be the property that has a greatest impact upon the speed of sound.Like any liquid,water has a tendency to evaporate.As it does,particles of gaseous water become mixed in the air.The temperature will affect the strength of the particle interactions an elastic property.At normal atmospheric pressure,the temperature dependence of the speed of a sound wave through dry air is approximated by the following equation:

v = 331 m/s + (0.6 m/s/C)•T

where T is the temperature of the air in degrees Celsius.Using this equation to determine the speed of a sound wave in air at a temperature of 20 degrees Celsius yields the following solution.

v = 343 m/s

(The equation itself does not have any theoretical basis;it is simply the result of inspecting temperature-speed data for this temperature range.Other equations do exist that are based upon theoretical reasoning and provide accurate data for all temperatures.The speed of light as it travels through air and space is much faster than that of sound;it travels at 300 million meters per second or 273,400 miles per hour.Visible light can also travel through other things besides through air and through space.The speed of light in water is approximately 2.26×108 2.26 × 10 8 meters per second.

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Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
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