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vlada-n [284]
4 years ago
6

A 90dB sound wave strikes an eardrum whose area is 5.0×10^-5m^2. How much energy is absorbed by the eardrum per second

Physics
1 answer:
MakcuM [25]4 years ago
6 0

-- The reference level for sound wave intensity is the threshold of human hearing, defined as  10⁻¹² W/m² .

-- This question is talking about a 90dB sound.  So its intensity is 10⁻³ W/m² .

-- The question tells us the size of the area that this sound is hitting.  So the power absorbed by that area is ...

Power = (intensity) x (area)

Power = (10⁻³ W/m²) x (5 x 10⁻⁵ m²)

Power = (10⁻³ x 5 x 10⁻⁵) W

<em>Power = 5 x 10⁻⁸ Watts</em>

(That's 5 x 10⁻⁸ Joules of energy per second.)

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Work done in lifting to a height of 1.5 m is 42.63 J.

Work done in carrying it to 50 m is 0 J

Explanation:

Given:

Mass of the pumpkin (m) = 2.9 kg

Vertical displacement of the pumpkin (y) = 1.5 m

Horizontal displacement of the pumpkin (x) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done by a force is given by the formula:

W=FS\cos\theta

Where,

F\to force\ applied\\\\S\to displacement\ caused\\\\\theta\to angle\ between\ F\ and\ S

  • Now, the work done is maximum when both force and displacement is in same direction. (\theta=0^\circ)
  • Work done is minimum when both force and displacement are in opposite direction. (\theta=180^\circ)
  • Work done is zero when force and displacement are perpendicular to each other. (\theta=90^\circ)

Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.

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Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.

So, Force = mg = 2.9\times 9.8 = 28.42\ N

So, work, W=Fy=28.42\times 1.5=42.63\ J

So, work done in lifting the pumpkin is 42.63 J.

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So, the force and displacement are perpendicular to each other. Thus, the work done is zero.

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