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Vikentia [17]
3 years ago
15

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

e adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven-the driving force is transferred to the object, which oscillates instead of the entire building X 50%
Part (a) What effective force constant, in N/m, should the springs have to make them oscillate with a period of 1.2 s? k = 9.5 * 106 9500000 X Attempts Remain 50%
Part (b) What energy, in joules, is stored in the springs for a 1.6 m displacement from equilibrium?
Physics
1 answer:
Vikki [24]3 years ago
7 0

Answer:

The force constant is  k =1.316 *10^{7} \  N/m

The energy stored in the spring is  E =  1.68 *10^{7} \ J

Explanation:

From the question we are told that

   The mass of the object is  M  = 4.8*10^{5} \ kg

    The period is T  = 1.2 \ s

The period of the spring oscillation is  mathematically represented as

         T  =2 \pi \sqrt{ \frac{M}{k}}

where  k is the force constant

   So making k the subject

       k = \frac{4 \pi ^2 M }{T^2}

substituting values

       k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}

      k =1.316 *10^{7} \  N/m

The energy stored in the spring is mathematically represented  as

       E =  \frac{1}{2} k x^2

Where x is the spring displacement which is given as

        x =  1.6 \ m

substituting values

      E =  \frac{1}{2} (1.316 *10^{7}) (1.6)^2

       E =  1.68 *10^{7} \ J

   

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The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

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K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

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v = 50 m/s

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Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

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Answer:

The ratio of the diameter of iron to Cu is;

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Explanation:

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from the question the two materials have the same resistance per unit length.

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\frac{R}{L}   for iron = \frac{R}{L}  for copper

This means we can equate ρ/A for both materials.

\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }

re-arranging the equation we have,

\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }

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\frac{d^{2}{Fe}   }{ d^{2}{Cu}   } =\frac{p_{Fe} }{ p_{Cu} }

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