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LenKa [72]
3 years ago
10

How did Rutherford's experiments demonstrate that Thomson's model of the atom was incorrect?

Physics
2 answers:
chubhunter [2.5K]3 years ago
8 0
Rutherford's Gold foil experiment involved the firing of atoms at a piece of gold foil only a few atoms thick. Rutherford expected that the particles would go straight through, but some were deflected. This experiment proved that atoms have a dense positively charged nucleus.

Answer: d. They showed that atoms have a dense nucleus.




mestny [16]3 years ago
8 0

Answer:

d) They showed that atoms have a dense nucleus.

Explanation:

Thomson's  model  predicted that the atom is composed of negatively charged electrons, which is surrounded by positively charged cloud. This was called Plum pudding model of the atom.

But Rutherford proved it to be incorrect . His gold foil experiment showed that the atom's central part is heavy and positively charged. The deflections of the alpha particles showed that the heavy central region is surrounded by negatively charged particles called electrons.

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Answer:

The answers to your questions are given below

Explanation:

22. The energy of an electromagnetic wave and it's frequency are related by the following equation:

E = hf

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E => is the energy

h => is the Planck's constant

f => is the frequency

From the equation i.e E = hf, we can conclude that the energy of a wave is directly proportional to it's frequency. This implies that an increase in the frequency of the wave will lead to an increase in the energy of the wave and also, a decrease in the frequency will lead to a decrease in the energy of the wave.

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Is the change in velocity divided by the time needed for the change to occur.
Montano1993 [528]

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Explanation:

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2 years ago
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibrati
coldgirl [10]

Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

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3 years ago
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