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Varvara68 [4.7K]
3 years ago
14

With an elevation of 5,334 m above sea level, the village of Aucanquilca, Chile is the highest inhabited town in the world. What

would be the gravitational potential energy associated with a 64 kg person in Aucanquilca?​
Physics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

The gravitational potential energy, P.E = 3345484.8 J

Explanation:

Given,

The mass of the person, m = 64 kg

The elevation of the village, h = 5334 m

The gravitational potential energy is the energy possessed by the body due to its height from the surface of the Earth at sea level.

Therefore, the gravitational P.E at height h above sea level is given by the relation

                             <em>P.E = mgh joules</em>

Substituting the given values,

                              P.E = 64 kg x 9.8 m/s² x 5334 m

                                     = 3345484.8 J

Hence, the gravitational potential energy associated with the person, P.E =  3345484.8 J

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F \  \alpha \  \frac{1}{D^{2}} for a distance D

If we move them so that D is doubled:

\frac{1}{2^{2}.D^{2}  }= \frac{1}{4} \eq  \frac{1}{.D^{2}  } \eq

Then the force they experience is one fourth of the original.

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How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
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Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

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Given;

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radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

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current in the solenoid, I = 0.8 A

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The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

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E = ¹/₂LI²

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Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

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Determine the value of the resultant and its location from O.<br>see attach image.​
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Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L  

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L  

the magnitude of the resultant will be

F = w₀L  √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

     (√⅛)[d] = ½         [⅔L] + ¼      [⅔L/2] - ¼      [L - ⅓L/2]

     (√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

     (√⅛)[d] =      ⅓L  +    ⅟₁₂L     -  ¼L + ⅟₂₄L  

     (√⅛)[d] = 5L/24

               d = 5L/24 / (√⅛)

               d = 5√⅛L/3

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