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olga2289 [7]
3 years ago
6

Ganymede is one of the many moons of jupiter. it is nearly a sperical in shape. it is larger than the planet mercury and slightl

y smaller than the planet mars. if it is so large compared with the bodies around it, why is it called a moon and not a planet?
Physics
1 answer:
Evgen [1.6K]3 years ago
7 0
Ganymede is the largest satellite in our solar system. It is larger
than Mercury and Pluto, and three-quarters the size of Mars.

==> If Ganymede orbited the sun instead of orbiting Jupiter, it would
easily be classified as a planet.
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___________ force can be thought of as either a push or a pull, as long as it makes the object move in a circular path.
fomenos
I believe it would be the Centripetal force.
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2 years ago
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Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values
marishachu [46]

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

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3 years ago
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put i
Margaret [11]

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t =RC

Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F

t = 56×10^-6 × 21

t = 1176×10^-6

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4 0
3 years ago
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The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
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The answer is:

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V = 86 km/ 1.3 hrs

V = 66.15 km/ hrs

I hope this helps!!
5 0
2 years ago
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