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Galina-37 [17]
3 years ago
13

g In 1920, Stern and Gerlach performed an experiment that first demonstrated Group of answer choices energy quantization. space

quantization. orbital angular momentum quantization. magnetic orbital quantization. that particles behave like waves.
Physics
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

B. space quantization.

Explanation:

In 1921, Otto Stern developed the idea behind this experiment, while Walther Gerlach performed the actual experiment in 1922. The Ster-Gerlach experiment provides prove to the fact that the spatial orientation of angular momentum is quantized. To demonstrate the experiment, silver atoms were made to travel through a magnetic field path.

Before they hit the screen(usually a glass slide), they were deflected because of their non-zero magnetic moment. There was an expected result for this experiment, but the actual observation on the glass slide was a continuous distribution of the silver atoms that actually hit the glass. This experiment was useful in proving that in all atomic-scale systems, there was a quantization of angular momentum.

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3 years ago
Please answer ASAP for brainliest
OlgaM077 [116]

Answer:

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Explanation: Hope this helps please mark brainliest!

4 0
3 years ago
Read 2 more answers
022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
svetoff [14.1K]

1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

initial speed of the ball upwards

v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

a = 9.76 m/s^2

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

a = 90 m/s^2

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

v_f = v_i + at

v_f = 0 + (90)(0.33)

v_f = 29.7 m/s

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}(9.81)(2.3^2)

y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

d = 2.92 \times 2.3

d = 6.72 m

6 0
3 years ago
What type of energy transfer occurs through light or electromagnetic waves?
Free_Kalibri [48]
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4 0
3 years ago
Read 2 more answers
The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

delta(P)=F*delta(t)

P_{f} -P_{i}=F*delta(t)

2m(v_{f} -v_{i})=F*delta(t)

v_{i} =0.35-0.35=0

It is neccesary calculate the force:

F=(m+m)*g*sin\theta

Here, g = gravity = 9.8 m/s²

F=(0.2+0.2)*9.8*sin37=2.3591N

v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s

6 0
3 years ago
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