oat physics

What’s Tested on OAT Physics?

The Physics section of the OAT requires strong knowledge of testable concepts and equations, as well as strategic testing skills. You’ll need to comfortable with the following topics:

  • Units, Vectors, and Kinematics
  • Newtonian Mechanics
  • Energy and Momentum
  • Thermodynamics
  • Fluid Statics
  • Electrostatics
  • Circuits
  • Simple Harmonic Motion and Waves
  • Light and Optics
  • Modern Physics

[ RELATED: Creating an OAT Study Schedule ]

Test your readiness for OAT Physics with the practice questions below:

Question 1

At a place where g is 9.8 m/s², an object is thrown vertically downward with a speed of 10 m/s. Meanwhile, a different object is thrown vertically upward with a speed of 20 m/s. Which object undergoes a greater change in speed in a time of two seconds?

Answer 1

Each object undergoes the same change in speed. Each object experiences an acceleration of 9.8 m/s², which means that each object’s speed changes by 9.8 m/s each second. Both objects thus experience the same change in speed over the two-second period, i.e., 19.6 m/s.

Question 2

A 2,000 kg experimental car can accelerate from zero to 30 m/s in 6.3 s. What is the average power of the engine needed to achieve this acceleration?

Answer 2

143 kW. The work done by the engine is equal to the change in kinetic energy of the car:

oat physics

The average power supplied is therefore equal to:

oat physics

Question 3

Is there any difference between the temperature of an object and the total random kinetic energy of the molecules of the object?

Answer 3

Yes. The temperature measures the average random kinetic energy of each of the molecules of the substance but tells nothing about the total random kinetic energy of an object, which depends on the total number of particles in the object. Thus, it’s possible to have an object (say a dilute gas) with a high temperature and low thermal energy content. However, if the number of gas particles is known, the total kinetic energy of the gas can then be calculated.

Question 4

A 10.0 Ω resistor carries a current that varies as a function of time as shown below. What energy has been dissipated by the resistor after 5 s?

oat physics

Answer 4

120 J. 

Power is energy dissipated per unit time, therefore energy dissipated is:

E = Pt

The power dissipated in a resistor R carrying a current I is:

P = I²R

Therefore, the energy dissipated in the first 2 s is:

E = I²Rt

= (2)²(10)(2)

= 80 J

The energy dissipated in the next 2 s is zero, since there is no current, and therefore no power is dissipated. The energy dissipated during the one-second interval from t = 4 s to t = 5 s is:

E = I²Rt

= (2)²(10)(1)

= 40 J

The total energy dissipated is therefore:

80 + 40 = 120 J

Question 5

A camera is a device that can take a large object and reduce it to a much smaller image which fits on the film. Assume that the camera uses a single lens to produce a real image of an object, such that the object distance is much greater than the image distance. What type of lens, converging or diverging, should be used?

Answer 5

Converging. To evaluate whether the lens is converging or diverging, determine the sign of the focal length. Positive focal lengths mean converging lenses, and negative focal lengths mean diverging lenses. Given that both i and o are positive, you have that 1/f = 1/o + 1/i is positive, which says that f is positive. Thus, the lens is converging.

 

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