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FTCE Physics (032) Practice Tests & Test Prep by Exam Edge - Topics



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Understanding what is on the FTCE Physics exam is crucial step in preparing for the exam. You will need to have an understanding of the testing domain (topics covered) to be sure you are studying the correct information.

  • Directs your study efforts toward the most relevant areas.
  • Ensures efficient and adequate preparation.
  • Helps identify strengths and weaknesses.
  • Allows for a focused approach to address gaps in understanding.
  • Aligns your preparation with the exam's expectations.
  • Increases the likelihood of success.
  • Keeps you informed about your field's current demands and standards.
There is no doubt that this is a strategic step in achieving certification and advancing your career.

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Understanding the exact breakdown of the FTCE Physics 6-12 test will help you know what to expect and how to most effectively prepare. The FTCE Physics 6-12 has 75 multiple-choice questions The exam will be broken down into the sections below:

FTCE Physics 6-12 Exam Blueprint
Domain Name % Number of
Questions
Knowledge of the nature of scientific investigation and instruction in physics 7% 5
Knowledge of the mathematics of physics 8% 6
Knowledge of thermodynamics 10% 8
Knowledge of mechanics 27% 20
Knowledge of waves and optics 18% 14
Knowledge of electricity and magnetism 20% 15
Knowledge of modern physics 10% 8


FTCE Physics 6-12 - Exam Topics Sample Questions


The figure shows three cells of 1.5 V each and a resistor R of 3 ohm. What is the current flowing through R?





Correct Answer:
0.5 a


to solve the problem of finding the current flowing through the resistor r, we need to first understand the arrangement and specifications of the circuit components provided in the figure. the figure shows three cells, each with a voltage of 1.5 v, connected in parallel. alongside these cells, there is a resistor r with a resistance of 3 ohms.

when cells are connected in parallel, the voltage across each cell remains the same across the entire parallel network. therefore, the effective voltage across the resistor r in this setup is also 1.5 v, which is the voltage of each individual cell.

to find the current flowing through the resistor r, we use ohm's law, which states that the current (i) through a conductor between two points is directly proportional to the voltage (v) across the two points and inversely proportional to the resistance (r) between them. ohm's law is mathematically expressed as i = v/r.

by substituting the values from our circuit, where v = 1.5 v and r = 3 ohms, the calculation for the current becomes: i = 1.5 v / 3 ohms = 0.5 a.

therefore, the current flowing through the resistor r is 0.5 a. this value is consistent with the principles of electrical circuits and confirms that the arrangement of the cells and the resistor in the circuit has been correctly interpreted.