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OLEGan [10]
3 years ago
8

Sort the phrases based on whether they represent kinetic energy or potential energy.

Physics
2 answers:
alexandr1967 [171]3 years ago
8 0
Kinetic energy:
*the energy of a moving body*
rising water vapor.
the wings of flying hummingbird.
rolling marble.

Potential energy:
*the energy that is stored in a body so that any small change in position or state of the body, will result in this body movement by transforming all this potential energy into kinetic energy*
stone resting.
disconnected battery (the potential energy is what called voltage here)
stretched rubber band.

Hope this helps.
dangina [55]3 years ago
6 0

Answer:

Kinetic energy:

*the energy of a moving body*

rising water vapor.

the wings of flying hummingbird.

rolling marble.

Potential energy:

*the energy that is stored in a body so that any small change in position or state of the body, will result in this body movement by transforming all this potential energy into kinetic energy*

stone resting.

disconnected battery (the potential energy is what called voltage here)

stretched rubber band.

Explanation:

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Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the
Alex777 [14]
The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire. 

The magnetic field generated by wire 1 is:
B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}=  4 \cdot 10^{-5}T

The magnetic field generated by wire 2 is:
B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T

And so, the resultant magnetic field at the point midway between the two wires is
B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T
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Hi ChloeRenee12345,

Your Question:

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