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zvonat [6]
3 years ago
13

This image shows a stream of positively charged particles being directed at gold foil. The positively charged particles are call

ed "alpha particles” and each one is like a nucleus without any electrons.
What is the best explanation for why a particle is striking point X?

Alpha particles are mostly empty space, so they move in random directions, and one of these alpha particles can strike point X.
The gold atoms contain negative electrons, so when alpha particles strike the gold they keep moving but in various directions.
If an alpha particle hits the gold foil, a gold nucleus splits and a particle from it flies out at point X.
When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Physics
2 answers:
coldgirl [10]3 years ago
8 0

Answer:

D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Explanation:

nikklg [1K]3 years ago
3 0

Answer:

It is D, youre right. thanks for being awesome

Explanation:

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Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
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The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

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State how heat loss by radiation is minimized in a thermos flask​
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Answer:

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Explanation:

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Explanation:

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