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lyudmila [28]
4 years ago
14

The sun's rays are directly overhead at the ________ on or about december 21.

Physics
1 answer:
frez [133]4 years ago
4 0
A latitude called the Tropic of Capricorn, you could also put the southern hemisphere

hope this helped :)
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The saturnian moon io probably collided with
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The Voyager and Pioneer flybys of the 1970s and 1980s provided rough sketches of Saturn’s moons. But during its many years in Saturn orbit, Cassini discovered previously unknown moons, solved mysteries about known ones, studied their interactions with the rings and revealed how sharply different the moons are from one another.
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Why should I care about children in Haiti that are being targeted by Haitian gangs?
Colt1911 [192]

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you don't have to its your choice whether you want to or not.

Explanation:

but you can not leave the fact that there dieing

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Which to particles in an atom are equal in number?
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 the neutron  I hope i could help :)

8 0
4 years ago
g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes ar
zhenek [66]

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

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5 0
3 years ago
Someone please help with this
SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

Substituting the original force:

F'=\frac{2}{3}F

The new force is 2/3 of the original force

7 0
3 years ago
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