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GenaCL600 [577]
3 years ago
7

Who first used the zodiac constellations? Will Mark Brainliest!!!

Physics
1 answer:
shepuryov [24]3 years ago
7 0
Astronomers aren't sure who first used the zodiac constellations, but it's believed to date back to ancient Greek or Roman times.
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The rotation of a planet around it's sun

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Page 40-44 earth science regents<br> just post the picture of the pages please
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WILL GIVE BRAINLIEST!!!
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Read 2 more answers
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
Drupady [299]

Answer:

4.5\times 10^{-5} T

Explanation:

We are given that

Current in wire=40 A

Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

We have R=29 cm=\frac{29}{100}=0.29 m

1 m=100 cm

Substitute the values in the given formula

B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

B=4.5\times 10^{-5} T

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

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3 years ago
A hot-air balloon is rising upward with a constant speed of 2.03 m/s. When the balloon is 8.13 m above the ground, the balloonis
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The answer is 4.2s...
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