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3 years ago

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

1 answer:

7 0

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

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

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

$k = 1.38N/m$

Therefore the spring constant is 1.38N/m

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3 years ago

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Mars2501 [29]

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3 years ago

Which of the following is NOT accelerating?

ivolga24 [154]

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3 years ago

Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

$A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2$

here $\omega _{o}^{2}= k/m$ = (6.30)/(0.135) = 46.67 N/m kg

$F / mA$ = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

$\omega ^{2}= \omega _{o}^{2}\pm F/m$

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

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3 years ago

A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,

AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

$Q_{water}=Q_{methanol}$