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mihalych1998 [28]
3 years ago
6

PLEASE HELP I'LL GIVE BRAINLIEST: Consider a sound wave produced by a tuning fork. Through which materials would the sound trave

l the fastest? Order the materials shown according to the speed at which a sound wave would travel through them, from slowest to fastest.

Physics
2 answers:
ra1l [238]3 years ago
8 0

Answer:

  1. Cold air
  2. Hot air
  3. Ocean
  4. Glass
irina1246 [14]3 years ago
7 0

Answer:

cold air

hot air

the ocean

glass

Sound is a moving density wave. it can travel through basically any normal form of matter. What does this mean density wave? well what this means is that you have a domino effect bascally, one air molecule is moving, it bumps its neighbor then its naghborr bumps its other neighbor and it keeps going until the sound energy is dispersed or turned into heat.

The wave speed of sound depends upon how close the molecules are to each other. It is like plucking a string. if you pull it tight the string vibrates much more frequently than if you hold it very loosely. In air, the molicules are quite far apart from each other, so the sound travels a lot slower but in a metallic material, the atoms are shoulder to shoulder and the speed of sound is about 1300 mph were as in the air it's about 300 mph.

Hope it helped

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HALP me!! This is a physics question.
emmasim [6.3K]
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

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4 0
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Read 2 more answers
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

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