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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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When temperature and pressure are applied to sedimentary and igneous rock?
spin [16.1K]
Answer:

Igneous rock

Explanation:

Igneous rocks are formed through the cooling and solidification of magma. It undergoes changes in temperature and pressure that causes it to cool, solidify, and crystallize.
6 0
4 years ago
A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re
Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

P_1 = Pressure at reservoir = 10 atm

T_1 = Temperature at reservoir = 300 K

P_2 = Pressure at exit = 1 atm

T_2 = Temperature at exit

R_s = Mass-specific gas constant = 287 J/kgK

\gamma = Specific heat ratio = 1.4 for air

For isentropic flow

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3

The density of the flow at the exit is 2.2721 kg/m³

4 0
3 years ago
Galileo held that measurements of time were _____ and that measurements of distance were ______. absolute; relative
DanielleElmas [232]

Answer:

option A

Explanation:

the correct answer is option A

The given blanks in the question will be filled with Absolute and relative receptively.

time is absolute means the numerical representation of time example 12:00 pm is absolute time.

Measurement of distance is relative. A relative distance is one which is represented with the reference of some other place, it is not an exact value.

6 0
3 years ago
(b) The output current of the generator is 2.0A. The output potential difference (p.d.) of the
Luden [163]

Answer:

P = 22 watts

Explanation:

Given that,

The output power of the generator = 2 a

The output potential difference = 11 V

We need to find the output power of the generator. The formula for the output power is given by :

P=V\times I\\\\P=11\times 2\\\\P=22\ W

So, the output power of the generator is equal to 22 Watts.

8 0
3 years ago
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
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