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Stels [109]
4 years ago
6

Help me please please help

Physics
1 answer:
andrew-mc [135]4 years ago
7 0

Answer:

The answer is C.

Explanation:

If you have ever tried doing so, hair stick to the ballon. Opposites attract as well, so the answer is C.

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What gas makes up most of our atmosphere in the troposphere?
Agata [3.3K]
78 nitrogen I think
4 0
3 years ago
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Ordinarily, material presented ________ is usually the most difficult to remember due to interference
Brums [2.3K]

Memorizing items on a list are enabled by the effects of primacy and

recency.

Ordinarily, material presented <u>in the middle of a list </u> is usually the most

difficult to remember due to interference.

Reasons:

When memorizing a list of items, due to the primacy effect, the information

at the beginning of the list are given importance, in the brain, due to their

many attributes, which may be due to them being;

  • The first item
  • Definition of the task (assignment of importance level)
  • Idea of what to expect (preparation of the mind)

Also the last items on the list are recalled by the effect known as recency,

which is probably due to the tendency to store items in the short term

memory.

Therefore, in the course of memorizing a list,  the first items on the list can

be stored in the long-term memory. The items that comes last in the

list are stored in the short-term memory, while the material presented in

the middle of the list are not allocated to a storage, and are usually

the most difficult to remember due to interference, by other items that

tend to come up during the retrieval process, from the long-term memory.

<em>(Rearranging the list can aid total recall)</em>

<em />

<em />

Learn more here:

brainly.com/question/15180534

<em>The possible question options as obtained from a similar question online, are;</em>

  • <em>At the beginning</em>
  • <em>Last</em>
  • <em>At the end</em>
  • <em>In the middle of a list</em>
5 0
2 years ago
Which of the following accurately describes junction diodes?
tatiyna

Correct answer choice is :


D) In a forward-biased setup, large numbers of charge carriers will be pulled across the junction and result in a large current.


Explanation:


P-n junction diode is the easiest mode of all the semiconductor tools. Despite, diodes act a significant role in many electronic appliances. A p-n junction diode can be utilized to change the alternating current to the direct current. These diodes are applied in power supply accessories. Battery attached across the p-n junction causes the diode forward biased, pushing particles from the n-type to the p-type and starting holes in a different way. Electrons and holes cross the junction and join. Photons are carried off as the electrons and holes recombine.

8 0
3 years ago
Read 2 more answers
Suppose our Sun is about to explode. In an effort to escape, we depart in a spacecraft at v=0.800 c and head toward the star Tau
Levart [38]

Considering a scenario in which the Sun is going to explode.

In order to escape the explosion, we depart in a spacecraft with a speed of v = 0.8c.

The star Tau Ceti is 12 life years away.

At the midpoint of the journey, the Sun as well as Tau Cetik explode. at the same instant.

Imagine a hermit who is immobile with respect to both the Sun and Tau Ceti and resides on an asteroid that is midway between both. He observes the blast waves of both explosions as our spaceship passes him. This observer concludes that the two stars blew up simultaneously because he believes both stars to remain stationary.

In the frame of reference, the Sun and Tau Ceti explode at the same time and will remain stationary.

Learn more about the frame of reference here:

brainly.com/question/10962551

#SPJ4

8 0
2 years ago
Rutherford used a 5.5 MeV alpha particle. Calculate the de Broglie wavelength of these alpha particles. How does this compare to
vodomira [7]

Answer:

Wavelength of alpha particle is 6.08 \times 10^{-15} m

The de broglie wavelength of 1keV is 451 times larger than size of nucleus.

Explanation:

Given:

Energy of alpha particle E= 5.5 MeV

De broglie wavelength corresponding the energy of alpha particle is given by,

   \lambda = \frac{h}{\sqrt{2mE} }

Where h = 6.6 \times 10^{-34} Js, m = 4 m_{p}, m_{p} = 1.67 \times 10^{-27} kg

   \lambda = \frac{6.6 \times 10^{-34} }{\sqrt{2 \times 4 \times 1.67 \times 10^{-27} \times 5.5 \times 10^{6 } \times 1.6 \times 10^{-19}  } }

   \lambda = 6.08 \times 10^{-15}m

Hence, wavelength of alpha particle is 6.08 \times 10^{-15} m

Size of nucleus is ≅ 10^{-15} m

Now wavelength of 1 keV is given by,

 \lambda _{1kev} = \frac{6.6 \times 10^{-34} }{\sqrt{2 \times 4 \times 1.67 \times 10^{-27} \times 1 \times 10^{3 } \times 1.6 \times 10^{-19}  } }

 \lambda _{1kev}  = 4.51 \times 10^{-13} m

Here, \frac{\lambda_{1kev}  }{\lambda } = \frac{4.51 \times 10^{-13} }{10^{-15} } ≅ 451

Therefore, the de broglie wavelength of 1keV is 451 times larger than size of nucleus.

 

4 0
4 years ago
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