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tiny-mole [99]
3 years ago
15

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amp

litude of 7.2 x10-7 m and a maximum speed of 3.6 x10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibrations
Physics
1 answer:
Paraphin [41]3 years ago
6 0

Answer:

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf)............... Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA................ Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

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First of all, let's just talk about the speed, and not get wound up
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If a fly is sitting on the rim of the wheel and the wheel is rotating, then for
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So if the wheel is going, let's say 71 revs per minute (RPM), a point
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Another way to say it:

Speed of a point on the circle = (2 π) x (rotation frequency) x (radius).

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7 0
3 years ago
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An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
List five situations in which a lesser amount of friction would be beneficial. Overall, would a high or low coefficient of frict
Natalka [10]
Low coefficient of friction

1. flying a plane (friction between air and plane)
2. ice skating (friction between ice and skate blade)
3. swimming (water & skin)
4. rowing a boat (water and boat)

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3 years ago
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Question 1 of 20
ad-work [718]

Answer

D. move a small magnet back and forth within a section of the coiled wire.

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A pitcher throws a curveball that reaches the catcher in 0.61 s. The ball curves because it is spinning at an average angular ve
sashaice [31]

Answer:

22.36 rad

Explanation:

Applying,

ω = θ/t.............. Equation 1

Where ω  = angular velocity, θ = angular displacement of the baseball, t = time

make θ the subject of the equation

θ = ωt............... Equation 2

From the question,

Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s

Substitute these values into equation 1

θ = 0.61(36.652)

θ = 22.36 rad

Hence the angular displacement of the baseball is  22.36 rad

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3 years ago
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