The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
particles directed at the foil
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms
(1) passed through the foil
(2) remained trapped in the foil
(3) were deflected by the nuclei in gold atoms
(4) were deflected by the electrons in gold atoms
2 answers:
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
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We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
