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olya-2409 [2.1K]
2 years ago
15

In Rutherford experiment some alpha particles fired at a gold foil bounced backward as a result of ... A.reflection from the sur

faces of gold atoms B.electrostatic repulsion by gold nuclei C.electrostatic repulsion by electronsD.all of the aboveE.none of the above
Chemistry
2 answers:
IRINA_888 [86]2 years ago
6 0

Answer: The correct option is B (electrostatic repulsion by gold nuclei).

Explanation:

In the Rutherford's experiment, he used positively charged particles called alpha particles to bombard an atom in order to find out what is inside the atom. Together with two other scientists, Geiger and Marsden, they used a narrow beam of alpha particles emitted from a radioactive source to bombard a thin gold foil. The scattering of the particles from the gold foil was detected by a movable zinc sulphide screen which could be rotated to various positions around the foil.

Each time an alpha particle hit the screen, a visible flash of light or scintillation was produced. This was observed by a microscope attached to the screen. It was then observed that some of the particles followed a straight path through the gold foil while a few where scattered in a backward direction. This was as a result of electrostatic repulsion by gold nuclei which occurs due to the greater part of the mass of the atom was concentrated in a minute nucleus with positive charge.

Amanda [17]2 years ago
3 0

Answer:

B.electrostatic repulsion by gold nuclei

Explanation:

According to Rutherford's experiment, a thin gold foil was bombarded with alpha particles. Some of the particles passed through the foil undeviated, some were scattered through large angles while some bounced backwards.

It follows that the particles that bounced backwards must have encountered a massive particle of like charge.

The atom is composed of a nucleus which contains positively charged particles. Some of the alpha particles which are positively charged particles bounced back when they encountered the positively charged particles in the nucleus.

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In that sense, it needs four electrons to complete its second shell.<span />
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3 years ago
What is the overall effect of adding a solute to a Solution
user100 [1]

Answer:

By adding the solute in to solution boiling point is increased while freezing point is decreased.

Explanation:

When solute in added into the solvent the boiling point of solvent increases for example,

Water is boiled at 100 °C, when sodium chloride is added its boiling point increased. Ions of salt interact with solvent and prevent the water molecules to escape from the surface and form gas molecules. In order to make it boiled solution must be heated above 100  °C.

But there is different case with freezing point. Freezing point is the state in which substance converted into the solid. At given temperature when solute is added into the solvent it prevent the formation of solid. It required time to decrease the temperature first and as the temperature is decreases solid is formed.

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3 years ago
A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

  • Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

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Answer:

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Explanation:

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