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alex41 [277]
3 years ago
6

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

f 0.640 s. What is the spring constant of the spring
Physics
1 answer:
inysia [295]3 years ago
6 0
The answer is B. 2.45 N/m.
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Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_f is the final angular speed

\omega_i is the initial angular speed

\Delta t is the time interval

For the fan in this problem,

\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s

Substituting,

\alpha = \frac{180-0}{4}=45 rpm/s

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

\omega' = \omega_i + \alpha t

where

\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s

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\omega' = 0 + (45)(5)=225 rpm

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