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3 years ago

One of the leading causes of permanent damage leading to hearing impairment is _____.

Physics

2 answers:

Dafna1 [17]3 years ago

8 0

<h3><u>Answer;</u></h3>

loud noise that damages hair cells

One of the leading causes of permanent damage leading to hearing impairment is<em><u> loud noise that damages hair cells</u></em>.

<h3><u>Explanation</u>;</h3>

Loud noise is the among the leading cause of permanent damage of the ear which impairs hearing.
<em><u>Loud noise may cause damage to the cells and the membranes in the cochlea of the ear. Listening to loud noise for a very long time overworks the hair cells that may make them to die.</u></em>
The damage to the cochlea of the ear causes permanent hearing loss since all or part of the inner ear has been damaged.

Maslowich3 years ago

6 0

One of the leading causes of hearing impairment that's due to damage
of the ear machinery is prolonged exposure to loud noise or loud music,
which damage the tiny hair cells that stimulate nerve endings in the inner ear.
Trust me.

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A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4

denpristay [2]

Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u

Mb=Ma
48=6u
u=8m/s

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2 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

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Answer:

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