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DENIUS [597]
3 years ago
8

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

oscillate at 1.00 Hz?A) 39.5 kgB) 2.00 kgC) 1.00 kgD) 1.56 kgE) 6.29 kg
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
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159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

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b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

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c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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