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Stella [2.4K]
3 years ago
8

A negatively-charged rod is brought close to (but does not touch) two neutral spheres that are in contact with each other but in

sulated from the ground. If the two spheres are then separated, what kind of charge will be on the spheres
Physics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

One sphere obtains a positive charge, while the other obtains a negative charge.

Explanation:

When the negatively-charged rod is brought close to the two neutral spheres in contact with each other, the electrons in both spheres are repelled away from the rod and thus those electrons move into the second sphere farther away from the rod.

The first sphere is now short of electrons.

When the spheres are separated, the first sphere which is short of electrons now carries a net positive charge, while the other sphere which has excess electrons now carries a net negative charge.

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What is the mass of 2000 ml of an intravenous glucose solution with a density of 1.15 g/ml?
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According to the following formula, the answer is 2,300 g or 2.3 kg:

Volume (m)/Mass (m) Equals Density (p) (V)

Here, the density is 1.15 g/mL, allowing the formula described above to result in a mass of 2.00 L:

p=m/V

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<h3>How many grams of glucose are in a 1000ml bag of glucose 5?</h3>

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5 0
2 years ago
I need this right ASAP!!! plz plz plz
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Answer:

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An object is five focal lengths from a concave mirror.how do the object and image heights compare?
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An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

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