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Natasha_Volkova [10]
3 years ago
11

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles trave

led in a straight line and some were deflected at different angles.
Physics
1 answer:
jeyben [28]3 years ago
4 0

The question is incomplete, the complete question is;

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles traveled in a straight line and some were deflected at different angles. Which statement best describes what Rutherford concluded from the motion of the particles? Some particles traveled through empty spaces between atoms and some particles were deflected by electrons. Some particles traveled through empty parts of the atom and some particles were deflected by electrons. Some particles traveled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms. Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Explanation:

Rutherford first proposed the nuclear model of the atom after his landmark experiment.

In this experiment, alpha particles from a source was focused on a thin gold foil. Some of the particles passed through empty spaces within the atom but were deflected at different angles by a small area of high-density positive charge within an atom which Rutherford later called the atomic nucleus.

Hence the answer above.

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medium

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A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
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Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

X = 5.676*10^{-3} \ mm

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

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