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worty [1.4K]
3 years ago
11

Longitudinal waves transfer energy ___________ to the direction of the wave motion.

Physics
2 answers:
andriy [413]3 years ago
6 0

The correct answer to the question is parallel.

EXPLANATION:

Before answering the question, first we have to understand a longitudinal wave.

A longitudinal wave is a type of mechanical wave in which the particle vibration is parallel to the direction of wave propagation.

In this type of wave, one will find compression and rarefaction . Compression is the high pressure region where particles are closely aggregated to each other while rarefaction is the low pressure region where particles are far away from each other. Hence, a longitudinal wave is a series of compression and rarefaction.

The transfer of energy occurs from particle to particle. On the other hand, we may say that transfer of energy takes place from compression to rarefaction, and this process is repeated.

Hence, the longitudinal waves transfer energy parallel to the direction of the wave motion.


viva [34]3 years ago
4 0
Longitudinal waves transfer energy parallel to the direction of the wave motion
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3 years ago
Definition These are the components of the reservoirs of oxygen that are exchanged in our environment.
andrew11 [14]

Answer:

Oxygen cycle

Explanation:

The components of the reservoirs of oxygen that are exchange in our environment is the oxygen cycle

It suggests the movement of oxygen between the living and non-living parts.

  • The cycle does not account for oxygen that is trapped and cannot be exchanged in nature.
  • Oxygen is important component of the atmosphere.
  • Gaseous exchange between living organisms and atmosphere involves oxygen to a very large extent.
4 0
3 years ago
The primary source of evidence proposed by many scientist to support the theory of an ancient earth is ____ dating.
vova2212 [387]
It depends on your definition of “ancient.” Radiometric dating using Carbon-14 can reliably date back to about 50,000 years, uranium-lead or lead-lead dating can date back multiple millions, potassium-argon dating can reach 1.5 billion, and rubidium-strontium can reach 50 billion (nearly 4x the age of the universe). It depends on the context in which this question is being asked.
7 0
3 years ago
S A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected
Flura [38]

The power of is series combination is Vn^2 times that of a parallel combination.

For series combination :

Req = R + R + R + ............... n times = nR

I = Δv/nr

Power = (Δv/nr)^2 × nr = Δv^2/nr

For parallel combination

1/req = 1/R + 1/R + 1/R +................(n times) = n/R

Req = R/n

Power = Δv/(R/n) = nΔv^2/R

Ratio = Δv^2/nr/n·Δv^2/R = 1/n^2

Hence, power of is series combination is Vn^2 times that of a parallel.

Learn more about parallel combination here:

brainly.com/question/12400458

#SPJ4

3 0
2 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
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