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Len [333]
3 years ago
9

Which statement describes the energy that a transverse wave carries as its amplitude increases?

Physics
2 answers:
stealth61 [152]3 years ago
3 0

Answer:

It increases and is perpendicular to the motion of the wave.

Explanation:

- A transverse wave is a wave in which the oscillation occurs in a direction perpendicular to the motion of the wave (example of transverse waves are electromagnetic waves)

- A longitudinal wave is a wave in which the oscillation occurs in a direction parallel to the motion of the wave (example of longitudinal waves are sound waves)

- The amplitude of a wave is defined as the maximum displacement of the wave relative to the equilibrium position, and the energy carried by the wave is proportional to the square of the amplitude:

E\propto A^2

Therefore, as the amplitude of the wave increases, the energy increases as well.

barxatty [35]3 years ago
3 0

Answer:

It increases and is perpendicular to the motion of the wave.

Explanation: I TOOK THE TEST

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A small ball is attached to one end of a spring that has an un- strained length of 0.200 m. The spring is held by the other end,
masya89 [10]

Answer:

\Delta x=0.002287\ m=2.287\ mm

Explanation:

Given:

  • un-stretched length of the spring, l=0.2\ m
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  • length stretch during the motion, \Delta l=0.01\ m

<u>Now the radius of revolution of the ball:</u>

r=l+\Delta l

r=0.2+0.01

r=0.21\ m

<u>Now in this case the centrifugal force is equal to the spring force:</u>

F_c=F_s

m.\frac{v^2}{r} =k.\Delta l

where:

m = mass of the ball

k = spring constant

m\times \frac{3^2}{0.21} =k\times 0.01

k=(4285.714\times m)\ N.m^{-1}

<u>Now the extension in the spring upon hanging the ball motionless:</u>

m.g=k.\Delta x

9.8\times m=(4285.714\times m)\times \Delta x

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6 0
3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

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h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

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