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3 years ago

6

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

held by the musculature in its socket.If the mass of the eyeball is 7.7 g, what is the effective spring constant of the musculature attached to the eyeball?

1 answer:

Sidana [21]3 years ago

6 0

Answer:

255.4 N/m

Explanation:

We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where in this case, we know:

f = 29 Hz is the frequency of oscillation

k is the spring constant, which is unknown

m = 7.7 g = 0.0077 kg is the mass of the eyeball

Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

$k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m$

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Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

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Answer:

Explanation:

Calculating the exit temperature for K = 1.4

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we can rewrite the above equation as :

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1 year ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

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Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

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The radius of the cord around the pulley is $r = 1.5 m$

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=> $T = mg -ma$

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$\tau = I \alpha$

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$\tau = T r$

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$Tr = I \alpha$

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$a = r\alpha$

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$v = \frac{d}{t}$

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$a = \frac{d}{t^2}$

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substituting value

$t = \sqrt{\frac{8}{8.78}}$

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Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

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$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

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60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

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Answer:

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Explanation:

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