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Citrus2011 [14]
3 years ago
12

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

aced from its equilibrium position and undergoes simple harmonic oscillations. What is the period of the oscillations
Physics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

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Answer:

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Explanation:

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3 0
2 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
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Answer:

1.195 m

2.8375 s

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Explanation:

d = Distance = 2.39 m

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t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

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The angular speed of the motion is 2.21433 rad/s

4 0
3 years ago
A typical meteor that hits the earth's upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at
attashe74 [19]

Answer:

Answer:u=66.67 m/s

Explanation:

Given

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Kinetic Energy of Meteor

K.E.=\frac{mv^2}{2}

K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}

K.E.=2\times 10^6 J

Kinetic Energy of Car

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=\frac{1}{2}\times 900\times u^2

\frac{1}{2}\times 900\times u^2=2\times 10^6  

900\times u^2=4\times 10^6

u^2=\frac{4}{9}\times 10^4

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In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react
wlad13 [49]

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

hope it helped u buddy

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