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Artemon [7]
2 years ago
8

Jonathan is riding a bicycle and encounters a hill of height 7.30 m. At the base of the hill, he is traveling at 6.00 m/s . When

he reaches the top of the hill, he is traveling at 1.00m/s. Jonathan and his bicycle together have a mass of 85.0 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road.(c) How much work does Jonathan do on the bicycle pedals within the Jonathan-bicycle-Earth system during this process?
Physics
1 answer:
Burka [1]2 years ago
6 0

W= mgh + 0.5m*vf^2- 0.5m*vi^2.

<h3>What is the Earth system?</h3>
  • The interactions between Earth's five systems—the geosphere, biosphere, cryosphere, hydrosphere, and atmosphere—create the conditions we are accustomed to.
  • The rock cycle, the food chain, the carbon cycle, the nitrogen cycle, the water cycle, and the energy cycle are a few examples found in the Earth system.
  • The major Earth circulation processes that move water, carbon, and nutrients renew essential elements for life and contribute to climate control. The continents, atmosphere, oceans, ice, and life on Earth are dynamic elements that constantly change and interact in a wide variety of ways.
  • A system is made up of four independent spheres. A change in one sphere may result in another change because the spheres interact. Fields are plowed by humans (biosphere) using farm equipment made of geosphere materials, while plants are watered by precipitation (hydrosphere) brought by the atmosphere.

How much work does Jonathan do on the bicycle pedals within the Jonathan-bicycle-Earth system during this process?W= mgh + 0.5m*vf^2- 0.5m*vi^2

To learn more about Earth's system, refer to:

brainly.com/question/1265737

#SPJ4

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heterogeneous mixture has components that are not evenly distributed. This means that you can easily distinguish between the different components.

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What is a charge-coupled device (CCD), and how is it used in astronomy?
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Yes

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7 0
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Elenna [48]

Answer:

\frac{g_{2}}{g_{1}} = \frac{1}{4}

Explanation:

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g - Gravity constant, in \frac{m}{s^{2}}.

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The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

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\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}

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