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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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15) A 1720 kg car accelerates at a rate of 3.0 m/s2. How much force is the car's engine producing? (use the formula F=ma
Anastasy [175]

Answer:

<h3>The answer is 5160 N</h3>

Explanation:

To find the force acting on an object given it's mass and acceleration we use the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 1720 kg

acceleration = 3.0 m/s²

We have

Force = 1720 × 3

We have the final answer as

<h3>5160 N</h3>

Hope this helps you

4 0
3 years ago
Read 2 more answers
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
3 years ago
Sam moves a box with a with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of powe
Vlad1618 [11]

Answer:

100s

Explanation:

there are many student how can not get answer on time and step by step. so there are a group of trusted physics experts who provide step by step answer. just join this wats up group.

4 0
3 years ago
Read 2 more answers
So long as there are(is) ______ there is no movement.
patriot [66]

Answer:

answer is friction. MCQ A is answer

6 0
2 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
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