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nadezda [96]
3 years ago
10

A 0.49 m copper rod with a mass of 0.15 kg carries a current of 13 A in the positive y direction. Let upward be the positive dir

ection. What is the magnitude of the minimum magnetic field needed to levitate the rod
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

Magnetic field, B = 0.23 T          

Explanation:

Given that,

Length of the copper rod, L = 0.49 m

Mass of the copper rod, m = 0.15 kg

Current in rod, I = 13 A (in +ve y direction)

When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

BIL=mg\\\\B=\dfrac{mg}{IL}\\\\B=\dfrac{0.15\times 9.8}{13\times 0.49}\\\\B=0.23\ T

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.

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Match the lithification processes.
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Answer:

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Precipitation of bonding agents between grains.

Cementation is the process in which the minerals in the form of fluid fills in spaces between the grains and binds them together upon crystallizing.

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A force F with arrow= (5xî+ 4yĵ), where F with arrow is in newtons and x and y are in meters, acts on an object as the object mo
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Answer:

w= 62.75 J

Explanation:

Given that

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The correct answer is A. He created a way of naming organisms based on genus and species

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