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Daniel [21]
3 years ago
14

Which best describes earth's magnetic field lines?​

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer:

The field lines go out of Earth near Antarctica, enter Earth in northern Canada, and are not aligned with the geographic poles.

Explanation:

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The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Colum
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"The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Column 2 would be on the y-axis."


typical plot is speed or velocity on the y-axis n time on the x-axis so the ans is Column 1 should be titled “Time,” and Column 2 should be titled “Velocity.”

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Why is the likelihood of amajor earthquake along the san andreas fault so worrisome??
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Fraternity hazing is acceptable because it is an initational rite to the brotherhood
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Answer:

that is the right answer

Explanation:

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A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present
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Answer:

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b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

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3 years ago
n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
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