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Goryan [66]
3 years ago
15

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

hin rim with radius 0.54 cm , connected to the balance staff by thin spokes of negligible mass. The total mass of the balance wheel is 0.95 g .Part AWhat is the moment of inertia of the balance wheel about its shaft?Express your answer using two significant figures.Part BWhat is the torsion constant of the hairspring?Express your answer using two significant figures.
Physics
1 answer:
Semenov [28]3 years ago
3 0

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

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Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

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E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

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An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. how high does it rise (v = 0 cm
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Determine the final temperature of a gold nugget (mass = 376 g) that starts at 398 K and loses 4.85 kJ of heat to a snowbank whe
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The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

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