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Monica [59]
3 years ago
8

What is the main difference between the Schrödinger model and the Bohr atomic model?

Physics
1 answer:
netineya [11]3 years ago
4 0

Answer:

Schrödinger believed that electrons could only exist in orbits, but Bohr stated that electrons could be found anywhere in the atom

Explanation: I got it right on test

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A glider moves along an air track with constant acceleration a. It is projected from the start of the track (x = 0 m) with an in
Pie

Answer:

vo = 0.175m/s

a = -0.040625 m/s^2

Explanation:

To solve this problem, you will need to use the equations for constant acceleration motion:

x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a

In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:

1. x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o

2. v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2

Replacing in the first equation:

0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0

v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s

But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:

a = 0.1125 - 5v_o^2\\a = -0.040625  m/s^2

3 0
4 years ago
A seagull is flying horizontally 8.00 m above the ground at 6.00 m/s.
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A) 6m cause it is at its highest speed
B)9.81*cos(36.87)+6*sin36.87)
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