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klasskru [66]
3 years ago
14

Which one of the following statements accurately describes a characteristic of adolescence?

Physics
1 answer:
Yuliya22 [10]3 years ago
7 0

Hi!


Answer D would be the best option.


During the period of adolescence, which is generally between 13-19, the individual goes through a transitional period which is characterized by changes in emotional, physical and mental capacities.

These changes consequently result in the adolescents' desire to be more independent, and to govern their lives as how they deem more fitting, which results in the rebellious nature in the individual.

The other options include characteristics that MAY be seen in individuals, but is not a defining characteristic of adolescence and is not very common.


Hope this helps.

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Suppose that the speed of an electron traveling 2.0 km/s is known to an accuracy of 1 part in 105 (i.e., within 0.0010%). What i
OverLord2011 [107]

Answer: 2.89(10)^{-3} m

Explanation:

The <u>Heisenberg uncertainty principle</u> postulates that the fact each particle has a wave associated with it, imposes restrictions on the ability to determine its position and speed at the same time.  

In other words:  

It is impossible to measure simultaneously (according to quantum physics), and with absolute precision, the value of the position and the momentum (linear momentum) of a particle. Thus, in general, the greater the precision in the measurement of one of these magnitudes, the greater the uncertainty in the measure of the other complementary variable.

Mathematically this principle is written as:

\Delta x \geq \frac{h}{4 \pi m \Delta V} (1)

Where:

\Delta x is the uncertainty in the position of the electron

h=6.626(10)^{-34}J.s is the Planck constant

m=9.11(10)^{-31}kg is the mass of the electron

\Delta V is the uncertainty in the velocity of the electron.

If we know the accuracy of the velocity is 0.001\% of the velocity of the electron V=2 km/s=2000 m/s, then \Delta V is:

\Delta V=2000 m/s(0.001\%)

\Delta V=2000 m/s(\frac{0.001}{100})

\Delta V=2(10)^{-2} m/s (2)

Now, the least possible uncertainty in position \Delta x_{min} is:

\Delta x_{min}=\frac{h}{4 \pi m \Delta V} (3)

\Delta x_{min}=\frac{6.626(10)^{-34}J.s}{4 \pi (9.11(10)^{-31}kg) (2(10)^{-2} m/s)} (4)

Finally:

\Delta x_{min}=2.89(10)^{-3} m

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Hope this helps :) 

can you please make this the brainliest answer it would really help . Thanks
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