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inna [77]
3 years ago
12

A pole-vaulter converts the kinetic energy of running to elastic potential energy in the pole, which is then converted to gravit

ational potential energy. If a pole-vaulter's center of gravity is 1.3 m above the ground while he sprints at 10.3 m/s, what is the maximum height of his center of gravity during the vault? For an extended object, the gravitational potential energy is U = mgh, where h is the height of the center of gravity
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Explanation:

bf means of cost of the day of the day of the day Happy Birthday to you and your family and friends and family members of a na please replay video call rate of your life is not a good night and good night and friends and you are right sir and friends for the day of the day of the day of my friends with you and your family and friends and family is fine for the day of the day of the year and and your family is a great friend of the day and you can see it and I have a good time to talk with you and your friends for the delay of your family is my friend of my friend of my friend and daughter of your mother boad and daughter and I have told you I

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To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

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r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

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