Question
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.
Which statement best describes what Rutherford concluded from the motion of the particles?
A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.
B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.
C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.
D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.
Answer:
The right answer is C)
Explanation:
In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.
So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".
Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.
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0.4 N-s is the "impulse" acted on the "beach ball".
Option: C
Explanation:
Given that,
Mass of the "beach ball" is 0.1 kg.
The speed of the ball hits is 4 m/s.
We know that,
Whenever an object is collide with other object then an impulse is acted on object, this "impulse" causes "change in momentum".
Impulse acted on the beach ball is "mass" times "velocity".
Impulse = mass × velocity
Impulse = 0.1 × 4
Impulse = 0.4 kg m/s
Impulse = 0.4 N-s
Therefore, the "impulse" acted on the ball is 0.4 N-s.
Answer:
2.5 m/s
Explanation:
The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground
v = 0 m/s + 2.5 m/s
v = 2.5 m/s
At the moment it is released, the package is rising at 2.5 m/s.
