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AEPA General Science (NT311) Practice Tests & Test Prep by Exam Edge - FAQ


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The best test prep involves studying both the subject matter and the exam itself! Read on for AEPA General Science FAQs and other test information.

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Why should I use Exam Edge to prepare for the AEPA General Science Exam?


FAQ's for Exam Edge AEPA General Science practice tests

We have ten great reasons why Exam Edge is the #1 source on the internet when it comes to preparing for AEPA General Science test:

  • Comprehensive content: Exam Edge's AEPA General Science practice tests are created specifically to prepare you for the real exam. All our AEPA General Science practice test questions parallel the topics covered on the real test. The topics themselves are covered in the same proportions as the real test too, based on outlines provided by the Arizona Educator Proficiency Assessments in their AEPA General Science test guidelines.

  • Realistic practice: Our AEPA General Science practice exams are designed to help familiarize you with the real test. With the same time limits as the real exam, our practice tests enable you to practice your pacing and time management ahead of test day.

  • Detailed explanations: As you complete your practice tests, we show you which questions you answered correctly and which ones you answered incorrectly, in addition to providing you with detailed step-by-step explanations for every single AEPA General Science practice exam question.

  • Performance insights: After you complete a practice test, we provide you with your raw score (how many you answered correctly) and our estimate of the AEPA General Science score you would have received if you had taken the real test.

  • Ease of access: Because all our practice tests are web-based, there is no software to install. You can take AEPA General Science practice exams on any device with access to the internet, at any time.

  • Flexible use: If you must pause while taking one of our practice tests, you can continue right where you left off. When you continue the test, you will start exactly where you were, and with the same amount of time you had remaining.

  • Thousands of unique questions: We offer 15 different online practice exams with 2,250 unique questions to help you prepare for your AEPA General Science !

  • Low cost: The cost of ordering 5 practice tests is less than the cost of taking the real AEPA General Science test. In other words, it would be less expensive to order 5 practice tests than to retake the real AEPA General Science exam!

  • Our trusted reputation: As a fully accredited member of the Better Business Bureau, we uphold the highest level of business standards. You can rest assured that we maintain all of the BBB Standards for Trust.

  • Additional support: If you need additional help, we offer specialized tutoring. Our tutors are trained to help prepare you for success on the AEPA General Science exam.

What score do I need to pass the AEPA General Science Exam?

To pass the AEPA General Science test you need a score of 220.

The range of possible scores is 100 to 300.

How do I know the practice tests are reflective of the actual AEPA General Science ?

At Exam Edge, we are proud to invest time and effort to make sure that our practice tests are as realistic as possible. Our practice tests help you prepare by replicating key qualities of the real test, including:

  • The topics covered
  • The level of difficulty
  • The maximum time-limit
  • The look and feel of navigating the exam
We have a team of professional writers that create our AEPA General Science practice test questions based on the official test breakdown provided by the Arizona Educator Proficiency Assessments. We continually update our practice exams to keep them in sync with the most current version of the actual certification exam, so you can be certain that your preparations are both relevant and comprehensive.

Do you offer practice tests for other Arizona Educator Proficiency Assessments subjects?

Yes! We offer practice tests for 46 different exam subjects, and there are 765 unique exams utilizing 80030 practice exam questions. Every subject has a free sample practice test you can try too!

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How do I register for the real Arizona Educator Proficiency Assessments?

For up-to-date information about registration for the Arizona Educator Proficiency Assessments, refer to the Arizona Educator Proficiency Assessments website.


AEPA General Science - FAQ Sample Questions

What does a silicon crystal doped with boron do?





Correct Answer:
it creates a p-type semiconductor in which the holes are the majority carriers.


when a silicon crystal is doped with boron, it transforms into what is known as a p-type semiconductor. in this process, boron atoms, which have three valence electrons, are introduced into the silicon crystal, which inherently has four valence electrons. this discrepancy results in the creation of "holes" or missing electrons in the crystal lattice.

these holes behave as positive charge carriers because they effectively represent the absence of an electron, which would normally contribute a negative charge. as a result, these holes become the majority carriers in the semiconductor material. in contrast, electrons, which are the natural charge carriers in an undoped silicon crystal, become the minority carriers in a boron-doped silicon crystal.

the term "p-type" comes from the positive charge of the holes. in electronics, the movement of holes within a semiconductor material can be harnessed to create electrical circuits and components that function based on the manipulation of these charge carriers. this is essential to the functionality of many electronic devices, including transistors, diodes, and solar cells.

it's important to note that doping silicon with boron does not create an n-type semiconductor. n-type semiconductors are created by doping silicon with elements that have more valence electrons than silicon, such as phosphorus or arsenic, which leads to an excess of free electrons that act as the majority carriers. thus, the process of doping with boron specifically leads to the creation of p-type semiconductors where holes dominate the charge transport mechanism.

A body of mass (m) is thrown vertically upward with a velocity (v). Identify the equation for the height (h) at which the kinetic energy of the body is half its initial value.





Correct Answer:
h=v2/4g
to solve the question of identifying the height at which the kinetic energy of a body thrown vertically upward is half its initial value, we start by considering the basic principles of mechanics involving kinetic energy (ke) and potential energy (pe).

initially, when the body is thrown upward with a velocity \( v \), the kinetic energy (\( ke \)) of the body is given by: \[ ke = \frac{1}{2}mv^2 \] where \( m \) is the mass of the body, and \( v \) is its initial velocity.

as the body rises, it loses kinetic energy due to the work done against gravity, and gains potential energy correspondingly. according to the principle of conservation of energy, the total mechanical energy (sum of kinetic and potential energy) of the body remains constant (ignoring air resistance). thus, any loss in kinetic energy is converted into an equal amount of potential energy.

at a certain height \( h \), let's say the kinetic energy of the body becomes half of its initial value. so, the new kinetic energy (\( ke' \)) at height \( h \) is: \[ ke' = \frac{1}{2} \times \frac{1}{2}mv^2 = \frac{1}{4}mv^2 \] this implies that the kinetic energy has decreased by \(\frac{1}{4}mv^2\), which must be the gain in potential energy (\( pe \)) at height \( h \): \[ pe = mgh = \frac{1}{4}mv^2 \] where \( g \) is the acceleration due to gravity.

solving the above equation for \( h \), we get: \[ mgh = \frac{1}{4}mv^2 \\ gh = \frac{v^2}{4} \\ h = \frac{v^2}{4g} \] this is the height at which the kinetic energy of the body is half of its initial value.

therefore, the correct answer is \( h = \frac{v^2}{4g} \). this equation shows us how high the body will travel before its kinetic energy is reduced to half its original value due solely to the conversion of kinetic energy to potential energy under the influence of gravity.