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3 years ago

The moon is about 3.8 X 10^5 km from the earth. Show that its average orbital speed about earth is 1026 m/s

1 answer:

3 0

Orbital speed = orbital path distance/time period = 2*pi*r/T
T=27.3days*24*3600=2358720s
2*pi*3.8*10^8m/2358720s = 1020m/s

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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

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

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We need to calculate the magnetic field

Using formula of magnetic field

$B'=\dfrac{\mu_{0}I}{2\pi r}$

Where, r = distance

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Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

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We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

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3 years ago

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