ACT Science : Physics
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Example Question #34 : How To Find Data Representation In Physics
Sometimes scientists need to scale down experiments because collecting full-scale data may not be practical or possible. Most of the time, the data collected during a small scale-experiment is an accurate indicator of full-scale results. For example, research of craters on the Earth’s surface has indicated that the Earth has experienced a history of massive meteor collisions. It has been hypothesized that, amongst other things, these impacting meteorites caused the Earth’s oceans to vaporize. The detrimental effects associated with meteor collisions would have made the planet to be uninhabitable to humans. Due to these adverse effects, an experiment that could produce meteor-sized craters would not be possible in a full-scale setting.
Researchers have developed small-scale experiments that are exemplary of Newton’s three laws of motion. Newton’s first law states that an object in motion will stay in motion unless it is acted on by another force. This phenomenon is known as inertia. The second law comes in the form of an equation that calculates how much force is exerted on two colliding objects with respect to mass and acceleration. In other words force is equal to the mass of an object multiplied by its acceleration. The third law states that for every action there is an equal and opposite reaction. In other words, forces always occur in pairs. A counter force that is equal in magnitude and opposite in direction compliments each force in a particular direction. For example, if you push downwards on a table then the table, in theory, pushes upwards with the same force.
In a particular small-scale exercise, students were asked to design an experiment in which two “meteorites” would be dropped into a medium of sand and the subsequent crater diameters would be measured. Afterwards, the sand would be leveled and the experiment would be repeated. The students were asked to identify all of the factors that could affect the formation of the craters. This was done in order to ensure that each experiment was testing one independent and one dependent variable at a time. Otherwise, it would be unclear which variable is causing the observed effect. Possible variables that could alter the outcome of the experiment included but were not limited to the following: height of the fall, speed of the “meteorite,” angle of impact, and the mass of meteorite.
Consider an experiment that seeks to identify the mass of the meteorite analog. All other variables would have to remain constant in order to ensure accurate results. In order to do this, the height of the fall was set to one meter. The object was placed on top of a vertical meter stick. In order to keep the acceleration and height of the drop constant, it was rolled off of the stick instead of being pushed or simply dropped. Last, a protractor was used to keep the meter stick at a ninety-degree angle. In the procedure, a marble and a golf ball were used as “meteorites” of different masses.
The results of the experiment were recorded in the table and figure provided.
| Crater Diameter (cm) | ||
| Trial Number | ||
| Marble (mass = 4g) | Golf ball (mass = 46g) | |
| 1 | 4.5 | 7.75 |
| 2 | 4.5 | 6 |
| 3 | 4 | 5.5 |
| 4 | 3 | 6 |
| 5 | 3 | 6.5 |
| 6 | 3.5 | 7 |
| 7 | 3 | 5 |
| 8 | 3 | 5 |
| 9 | 3 | 7 |
| 10 | 3.5 | 6 |
| Average | Average | |
| 3.5 cm | 6.175 cm | |
How does the height affect the results of the experiment?
As the height increases, the diameter of the crater decreases.
None of these
As the height decreases, the diameter of the crater increases.
As the height increases, the diameter of the crater increases.
As the height increases, the diameter of the crater increases.
Gravity is a vertical force of acceleration, meaning that as an object draws closer to another object, the velocity will increase over time. Given this information we can theorize that if an object falls for a longer time then it should travel faster until it reaches its terminal velocity; therefore, as the height increases, the size of the crater should also increase.
Example Question #231 : Physics
Students are given an assignment to create a circuit from a given set of components. Four different students create circuits in unique manners and the professor measures the resistance of the circuit, the power output of the circuit and the heat given off by each circuit after each circuit was running for a certain amount of time. Circuits with resistors in parallel have a lower total resistance than those that have resistors in series.
Based on the information above, what is a way to increase the power output of a circuit?
Cannot be determined from the given information
Increase the resistance
Decrease the resistance
Increase the lengths of the wires of the circuits
Decrease heat output
Decrease the resistance
There is no information given on how the lengths of the wires affect the power of the circuit so this is not the correct answer. As the heat increases so does the power output of the circuit so decreasing the heat given off by the circuit would not increase the power of the circuit. In the given table it can be seen that as the resistance decreases, the power output of the circuit increases. So in order to increase the power of the circuit the resistance can be decreased.
Example Question #232 : Physics
A scientist is exploring the nature of energy and work in two experimental systems.
Experiment 1
She first sets up a system with a ball on an inclined plane, and calculates the potential and kinetic energies of the ball at three positions. She places the ball at position 1, stops the ball after rolling to position 2, and then again allows it to roll to position 3. The measured kinetic and potential energies are shown in the table, along with measurements of the force of friction acting on the ball.
|
Ball Position |
Potential Energy |
Kinetic Energy |
Force of Friction |
|
1 |
150 kJ |
X kJ |
0 N |
|
2 |
75 kJ |
75 kJ |
0 N |
|
3 |
X kJ |
150 kJ |
0 N |
Experiment 2
The same scientist then sets up another experiment with the same ball, again at three positions. The ball is provided a slight push, and then allowed to roll down three levels without any additional external input of energy.
She uses the following formulae to calculate the energy levels of the ball:
Potential Energy = Mass of the Ball x Acceleration Due to Gravity x Height of the Ball
Kinetic Energy = (1/2) x Mass of the Ball x (Velocity of the Ball)2
She also measures the internal energy of the ball, a value that she defines as the amount of energy contained in the motion of the molecules that make up the matter of the ball.
|
Position |
Potential Energy |
Velocity |
Internal Energy |
|
1 |
A |
1 m/s |
5 kJ |
|
2 |
B |
5.47 m/s |
6 kJ |
|
3 |
C |
10.94 m/s |
7 kJ |
In Experiment 2, the scientist wishes to approximate the total loss of potential energy from position 1 (PE1) to position 3 (PE3). Which of the following is true?
The potential energy of the ball at position one is going to be 20 times higher than the potential energy of the ball at position 3 because the ball is 20 m higher. Potential energy, according to the provided formula, is dependent on this height for its magnitude.
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.
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.
Based on the trajectories of the particles in Experiment 1, what can be concluded about the relative charges of the particles?
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.
The α particle must have a charge twice as large as the β particle, as the applied magnetic field bent it twice as much.
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.
What is the total distance traveled by Car 3 from time = 0 until time = 4s?
8m
0m
12m
–2m
4m
12m
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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