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vagabundo [1.1K]
3 years ago
6

Which of these is part of the middle ear? A. External auditory meatus B. Tympanic membrane C. organ of corti D. The ossicles

Physics
2 answers:
Alenkasestr [34]3 years ago
3 0
D. The ossicles.......
Vlad1618 [11]3 years ago
3 0
The best answer is D which is The ossicles
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A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0
3 years ago
A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0
3 years ago
A plane flying at a steady speed of 100 m/s accelerates to 150 m/s in 10 seconds. What is the plane’s acceleration?
Rashid [163]

A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

<h3>What is acceleration?</h3>

Acceleration is the change in velocity over time.

A plane is flying initially at 100 m/s (u) and it accelerates to 150 m/s (v) in 10 s (t). We can calculate its acceleration using the following expression.

a = v - u / t = (150 m/s - 100 m/s) / 10 s = 5 m/s²

A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

Learn more about acceleration here: brainly.com/question/14344386

#SPJ1

5 0
1 year ago
During combustion, a fuel’s electromagnetic energy is converted into thermal energy.
swat32
True I believe..................
6 0
3 years ago
If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
20/2= 10mph
5 0
3 years ago
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