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galben [10]
3 years ago
12

Lance is a 60-kg runner who, on a good day, runs at a speed of 5 m/s. What is his kinetic energy when he runs at this speed? *

Chemistry
1 answer:
Contact [7]3 years ago
7 0

Answer:

His kinetic energy when he runs at that 5 m/s is 750 joules

Explanation:

The given information are;

The mass of, Lance, the runner = 60-kg

The speed by which Lance is running = 5 m/s

The kinetic energy, KE, is given by the relation;

KE = 1/2×m×v²

Where;

m = The mass = 60-kg

v = The instantaneous speed of motion = 5 m/s

The kinetic energy can be found by substituting the values of the variables in the kinetic energy equation as follows;

KE = 1/2 × 60 kg × (5 m/s)² = 750 joules

His kinetic energy when he runs at that 5 m/s is 750 joules.

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What does the graph show about the rate of temperature change over time?
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An unsaturated solution is formed when 80 grams of a salt is dissolved in 100 grams of water at 40. This salt could be ?
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Read 2 more answers
g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of
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Answer:

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Explanation:

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p*∆V

Where:

  • W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
  • p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
  • ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)

In this case:

  • p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
  • ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)

Replacing:

W system= -1.013*10⁶ Pa* (-0.018 m³)

Solving:

W system= 18234 J

<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>

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