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Nataliya [291]
2 years ago
7

The table shows information about four students who are running around a track. Which statement is supported by the information

in the chart?
Autumn has more kinetic energy than Chiang.
Mohammed has less kinetic energy than Autumn.
Lexy has more kinetic energy than Mohammed.
Chiang has less kinetic energy than Lexy.

Physics
2 answers:
Eddi Din [679]2 years ago
7 0

Answer:

its b

Explanation:

got in right on edge

Vikentia [17]2 years ago
3 0

Answer:

<em>Correct option: Mohammed has less kinetic energy than Autumn.</em>

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its motion. If the object has a mass m and travels at a speed v, then the kinetic energy K is:

\displaystyle K=\frac{1}{2}mv^2

The information about four students includes their mass and velocity as follows:

Autumn has a mass of m1=50 kg and a velocity (magnitude) of v1=4 m/s, thus their kinetic energy is:

\displaystyle K_1=\frac{1}{2}50\cdot 4^2

K_1=400\ J

Mohammed has a mass of m2=57 kg and a velocity (magnitude) of v2=3 m/s, thus their kinetic energy is:

\displaystyle K_2=\frac{1}{2}57\cdot 3^2

K_2=256.5\ J

Lexy has a mass of m3=53 kg and a velocity (magnitude) of v3=2 m/s, thus their kinetic energy is:

\displaystyle K_3=\frac{1}{2}53\cdot 2^2

K_3=106\ J

Chiang has a mass of m4=64 kg and a velocity (magnitude) of v4=5 m/s, thus their kinetic energy is:

\displaystyle K_4=\frac{1}{2}64\cdot 5^2

K_4=800\ J

Sorted from lower kinetic energy to higher:

Lexy, Mohammed, Autumn, Chiang. Thus:

Autumn has more kinetic energy than Chiang. False

Mohammed has less kinetic energy than Autumn. True

Lexy has more kinetic energy than Mohammed. False

Chiang has less kinetic energy than Lexy. False

Correct option: Mohammed has less kinetic energy than Autumn.

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25.82 m/s

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Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
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Answer:

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Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

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