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zlopas [31]
3 years ago
7

Scientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below. Overl

apping red and blue balls in the center with a circular fuzzy green cloud outside them. What experimental evidence led to the development of this atomic model from the one before it? A few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil. Experiments with water vapor showed that elements combine in specific proportions. Equations were used to identify regions around the nucleus where electrons would likely be. Cathode rays were bent in the same way whenever a magnet was brought near them.
Physics
2 answers:
olga55 [171]3 years ago
8 0

Answer:

A). A few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.

Explanation:

Scientists decided to change the model of the atom when they discovered new evidence that showed 'few of the positive particles aimed at a gold foil seemed to bounce back off of the thin metallic foil.' On this ground, <u>Rutherford concluded that atom is mostly made up of empty space and thus, he proposed a nucleus model of atom in which the atom comprises of the tiny and positively charged nucleus is surrounded by electrons with a negative charge</u>. Thus, <u>option A</u> is the correct answer.

Anastaziya [24]3 years ago
7 0

Answer: C. Equations were used to identify regions around the nucleus where electrons would likely be.

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Well, first of all, one who is sufficiently educated to deal with solving
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If the mass of the man standing on the weighing machine is 60kg, then
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simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
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That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
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