ACT Science : How to find data representation in physics

Study concepts, example questions & explanations for ACT Science

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Example Questions

Example Question #41 : How To Find Data Representation In Physics

Experiment 1

A scientist develops the following setup, shown in Figure 1 below, to study the charges of radioactive particles. A radioactive sample is placed into a lead box that has an open column such that the particles can only exit from one direction. A detector is placed in front of the opening. A metric ruler measuring in centimeters (cm), is aligned on the detector such that zero is directly in front of the opening of the column, with positive values extending to the left and negative values extending to the right. On the left side of the experimental setup, there is a device that generates a magnetic field that attracts positively charged particles and repels negatively charged particles.

 

                 Radioactivity_fig_1

                                                    Figure 1.

 

The device detects particles in three different places: alpha, α; beta, β; and gamma, γ; as labeled in Figure 1. The paths these particles take from the source of radioactivity are shown.

 

Experiment 2

A different scientist finds the following data, shown in Table 1, about the energies of the α, β, and γ particles by observing what kinds of materials through which the particles can pass. This scientist assumes that the ability of particles to pass through thicker and denser barriers is indicative of higher energy. Table 1 summarizes whether or not each type of particle was detected when each of the following barriers is placed between the radioactivity source and the detector. The paper and aluminum foil are both 1 millimeters thick, and the concrete wall is 1 meter thick.

 

                              Radioactivity_table_1

Based on the trajectories of the particles in Experiment 1, what can be concluded about the relative charges of the particles?

Possible Answers:

The β particle must have a charge twice as large as the γ particle, as the applied magnetic field bent it twice as much.

The α particle must have a charge twice as large as the γ particle, as the applied magnetic field bent it twice as much.

The α particle must have a charge twice as large as the β particle, as the applied magnetic field bent it twice as much.

The β particle must have a charge twice as large as the α particle, as the applied magnetic field bent it twice as much.

Correct answer:

The α particle must have a charge twice as large as the β particle, as the applied magnetic field bent it twice as much.

Explanation:

We know that the alpha particle must have a positive charge since it moves toward the magnetic field. We also know that the beta particle must have a negative charge since it moves away from the magnetic field. Lastly, the gamma particle must have no charge, as its trajectory is not affected by the magnetic field. Now we can look at the magnitude of deflection from 0 using the detector's ruler. As the alpha particle was deflected two units from zero, and the beta particle was deflected one unit from zero, we know that the alpha particle must have a charge twice that of the beta particle.

Example Question #41 : How To Find Data Representation In Physics

The graph below depicts the position of three different cars over a 15-second time interval.

Pic_2

What is the total distance traveled by Car 3 from time = 0 until time = 4s?

Possible Answers:

8m

0m

12m

–2m

4m

Correct answer:

12m

Explanation:

An important consideration in this problem is the difference between total distance traveled (which is what the question asks about) and net displacement.  Total distance traveled includes every meter that Car 3 covers, even if some of that distance moved backward or toward its original position. Net displacement, often in physics referred to simply as "distance" (but not "total distance traveled") is only concerned with the beginning and end points of the object.  

 

Here you can see that Car 3's position began at -4, moved to 4, and then returned back to 0. That means that it made a positive gain of 8 and then a negative gain of 4, for a total distance covered of 12m. The net displacement is only a net gain of 4m, but as the question asks for total distance traveled the answer is 12m.

 

 

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