ACT Science : ACT Science
Study concepts, example questions & explanations for ACT Science
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Example Questions
Example Question #1161 : Act Science
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 |
A second scientist is examining this data and wants to test if introducing an electric field around the ramp in Experiment 1 changes the measured energy values. To do this, he sets up another system, identical to that in Experiment 1, but with the addition of an electric field. He compares data generated by both the original system in Experiment 1 and the new system with the electric field. Experiment 1 is now serving as what?
A modifier
A confounder
A variable
A control group
An experimental group
A control group
A control group serves as a comparison group to an experimental group. The control group and the experimental group should be identical except in the presence of one variable that is being studied. In this case, that one variable is the presence of an electric field.
Example Question #41 : Physics
Sound waves travel through a medium by mechanically disturbing the particles of that medium. As particles in the medium are displaced by the sound wave, they in turn act upon neighboring particles. In this fashion, the wave travels through the medium through a parallel series of disturbed particles. Like in other forms of motion, the rate at which the sound wave travels can be measured by dividing the distance over which the wave travels by the time required for it to do so.
Study 1
A group of students hypothesizes that the velocity of sound is dependent upon the density of the medium through which it passes. They propose that with more matter in a given space, each particle needs to travel a shorter distance to disturb the adjacent particles. Using two microphones and a high speed recording device, the students measured the delay from the first microphone to the second. They chose a variety of media, shown in Table 1, and measured the velocity of sound through each using their two-microphone setup. The results are found in Table 1.
Study 2
The students wanted to test their hypothesis by using the same medium at different densities. To do this, they heated pure water to various temperatures and repeated the procedure described in Study 1. Their results can be found in Table 2.
Assume that density of a substance is the only contributing factor to velocity of sound through that substance. If the students' hypothesis in Study 1 is correct, what might they have predicted for the velocity of sound through lead? (Assume all other values in Table 1 remained the same.)
The students hypothesized that velocity of sound through a substance is directly dependent upon the density of that substance. In other words, as density increases, velocity of sound also increases. Because lead is the densest substance in Table 1, only velocities greater than 5,130 m/s (the velocity through iron) would satisfy the hypothesized relationship between velocity and density.
Example Question #1162 : Act Science
Sound waves travel through a medium by mechanically disturbing the particles of that medium. As particles in the medium are displaced by the sound wave, they in turn act upon neighboring particles. In this fashion, the wave travels through the medium through a parallel series of disturbed particles. Like in other forms of motion, the rate at which the sound wave travels can be measured by dividing the distance over which the wave travels by the time required for it to do so.
Study 1
A group of students hypothesizes that the velocity of sound is dependent upon the density of the medium through which it passes. They propose that with more matter in a given space, each particle needs to travel a shorter distance to disturb the adjacent particles. Using two microphones and a high speed recording device, the students measured the delay from the first microphone to the second. They chose a variety of media, shown in Table 1, and measured the velocity of sound through each using their two-microphone setup. The results are found in Table 1.
Study 2
The students wanted to test their hypothesis by using the same medium at different densities. To do this, they heated pure water to various temperatures and repeated the procedure described in Study 1. Their results can be found in Table 2.
Jackie read about this experiment and decided to recreate Study 1. She found that each of the velocities she measured are higher than those in the original experiment. What is one possible explanation for this difference?
Jackie used a shorter sample of each substance.
Jackie didn't measure the density of each substance.
Jackie used substances at a different temperature than those used in the original experiment.
Jackie used lower-quality microphones.
Jackie used substances at a different temperature than those used in the original experiment.
As study 2 shows, changes in temperature are correlated to changes in velocity. If Jackie used substances at a different temperature than those used in the original experiment, she would find different velocities of sound for those substances.
A shorter sample of each substance would not affect sound velocity, as the distance traveled is accounted for in the equation for velocity. The act of measuring the density of each substance would not affect the velocity of sound.
Example Question #1164 : Act Science
A student is doing an experiment relating force and distance to work and work and time to power. She is given the following table:
|
Force |
Distance |
Work |
Time |
|
3 Newtons |
1 meter |
3 Joules |
10 Seconds |
|
4 Newtons |
10 meters |
40 Joules |
20 Seconds |
|
1.5 Newtons |
7 meters |
10.5 Joules |
1 Second |
She is also told that:
Power = Work/ Time
Work is in Joules and Time is in seconds. Power is measured in Watts.
If the Watts of power equals the amount needed to light the light bulb attached then the bulb will be lit. If more Watts of power are present, an attached light bulb will burn brighter.
If the student were to do an experiemt with a force of 6 Newtons and a distance of 1 meter, how much work would the student be doing?
6 Joules
5.5 Joules
0.3 Watts
12 Joules
6 Joules
The table values show that the Work is calculated by multiplying the force by the distance (in this case 6x1=6 Joules of power).
Example Question #42 : Physics
A student is doing an experiment relating force and distance to work and work and time to power. She is given the following table:
|
Force |
Distance |
Work |
Time |
|
3 Newtons |
1 meter |
3 Joules |
10 Seconds |
|
4 Newtons |
10 meters |
40 Joules |
20 Seconds |
|
1.5 Newtons |
7 meters |
10.5 Joules |
1 Second |
She is also told that:
Power = Work/ Time
Work is in Joules and time is in seconds. Power is measured in Watts.
If the Watts of power equals the amount needed to light the light bulb attached then the bulb will be lit. If more Watts of power are present, an attached light bulb will burn brighter.
What is a Joule equal to according to the table?
Newton/ Watt
Newton x Meter
Newton x Watt
Newton/ Meter
Newton x Meter
From the table you can determine that the Unit for Work is a Joule and that Work is calculated by multiplying the force by the distance travelled. Since force is in Newtons and distance is shown in meters, you can infer that the unit for force multiplied by the unit for distance equals the unit for Work (Newton x Meter)
Example Question #1165 : Act Science
A student is doing an experiment relating force and distance to work and work and time to power. She is given the following table:
|
Force |
Distance |
Work |
Time |
|
3 Newtons |
1 meter |
3 Joules |
10 Seconds |
|
4 Newtons |
10 meters |
40 Joules |
20 Seconds |
|
1.5 Newtons |
7 meters |
10.5 Joules |
1 Second |
She is also told that:
Power = Work/ Time
Work is in Joules and time is in seconds. Power is measured in Watts.
If the Watts of power equals the amount needed to light the light bulb attached then the bulb will be lit. If more Watts of power are present, an attached light bulb will burn brighter.
If the student is trying to light a 2 Watt Bulb, which of the following experimental scenarios would work?
60 Joules of Work in 30 seconds
40 Joules of Work in 20 seconds
2 Joules of Work in a second
All of these
All of these
The formula Power (in Watts) = Work / Time is given in the experiment above. Since the light bulb is 2 Watts, it needs that many Watts of power to light it. All of the answers divide out to be 2 Watts of Power and therefore would all be enough to light the light bulb.
Example Question #1166 : Act Science
A student is doing an experiment relating force and distance to work and work and time to power. She is given the following table:
|
Force |
Distance |
Work |
Time |
|
3 Newtons |
1 meter |
3 Joules |
10 Seconds |
|
4 Newtons |
10 meters |
40 Joules |
20 Seconds |
|
1.5 Newtons |
7 meters |
10.5 Joules |
1 Second |
She is also told that:
Power = Work/ Time
Work is in Joules and time is in seconds. Power is measured in Watts.
If the Watts of power equals the amount needed to light the light bulb attached then the bulb will be lit. If more Watts of power are present, an attached light bulb will burn brighter.
The student tried the experiment again, but this time she moved the object over a distance of 40 meters, but in a circle so she ended up in the same place. She thinks that the distance would be 0 meters because she didn't change locations, so she cannot light a bulb because no power would be created. Is this true?
Yes because the light bulb cannot work if she moves in a circle.
Yes because the distance would be 0 meters and the Work done and Power would also be 0.
No because work has nothing to do with how far she travelled, only the force she used.
No because the experiment never discussed the paths taken, only the distance moved.
No because the experiment never discussed the paths taken, only the distance moved.
The experiment never discusses paths because that does not effect how much work took place. If that were true that no work is done when you move in a circle, running on a track would be easier than running in a straight path, which is not true. The correct answer is the only one that correctly states that the student's theory is false and gives a logical answer because force AND distance are factors in the Work done, not just the force.
Example Question #12 : How To Find Experimental Design In Physics
A student is doing an experiment relating force and distance to work and work and time to power. She is given the following table:
|
Force |
Distance |
Work |
Time |
|
3 Newtons |
1 meter |
3 Joules |
10 Seconds |
|
4 Newtons |
10 meters |
40 Joules |
20 Seconds |
|
1.5 Newtons |
7 meters |
10.5 Joules |
1 Second |
She is also told that:
Power = Work/ Time
Work is in Joules and time is in seconds. Power is measured in Watts.
If the Watts of power equals the amount needed to light the light bulb attached then the bulb will be lit. If more Watts of power are present, an attached light bulb will burn brighter.
The student repeats the experiment at home and finds out that the light bulb burns brighter if she pushes the object up an incline rather than a flat surface of the same distance in the same amount of time. What would explain this?
The student must be using less force to push the object up the incline.
Watts have nothing to do with the distance travelled or force used.
The student must be using more force to push the object up the incline.
The student must have miscalculated.
The student must be using more force to push the object up the incline.
The only variable that could be changing in the student's experiment is the force used since she stated that the distance and time are constant between the two screnarios, but we know that in the incline scenario there are more Watts of power causing the bulb to burn brighter. Knowing that the only things that effected power were those in the table (force, distance, work and time) would allow you to use process of elimination to figure out that only the force could be changing. Logically this makes sense because pushing something up an incline is much harder "work" than pushing it along a flat surface.
Example Question #1163 : Act Science
A physicist wishes to study the trajectory of a ball launched horizontally. She varies parameters such as the launching velocity, starting height, and mass of the ball. For each trajectory, she records the time of flight (in seconds) and horizontal displacement (in meters). She assumes air resistance is negligible.
Figure 1
Using all of the data she collects, she constructs the following table:
Table 1
Which of the following variables were controlled in Trial 2?
Starting height and launching velocity
Mass of the ball only
Starting height only
Mass of the ball and launching velocity
Mass of the ball and launching velocity
In Trial 2, only the starting height varies, as the mass of the ball and launching velocity remain constant.
Example Question #51 : Physics
A particle separator functions by using a magnetic field and an electric field pointing perpendicularly. When a charged particle is launched into the particle separator, a constant electric force is exerted on the particle proportional to the particle's charge. Additionally, the magnetic field exerts a force on the particle that is in the opposite direction of the electric force and that is proportional to the particle's charge and velocity. A particle will make it through the particle separator only if the opposing magnetic forces and electric forces are equal in magnitude as they will not cause a net change in the particle's direction.
According to the information of the passage, by what physical characteristic does a particle separator separate particles?
Charge
Velocity, charge, and mass
Velocity and charge
Velocity
Mass
Velocity
The correct answer is velocity. This is a difficult question because it requires critical consideration of the passage information. In the passage, we learn that both the magnetic and electric forces are proportional to the particle's charge. However, we also learn that only the magnetic force is proportional to the particle's velocity. Therefore, we know that no matter what charge a particle has, it will make it through the particle separator only if it has the right velocity such that the magnetic force produced is enough to counter the electric force on the particle. Mass is not mentioned in the passage, so we can ignore any answer including mass.
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