Question
A student drops a ball at different heights, she starts at 1.0 m and gets 48 joules of kinetic energy, then she drops the ball at 1.5 m, this time the joule of kinetic energy is 72, she drops it one more time at 2.0 m and gets 96 joules of kinetic energy
The student says that a ball dropped from a height of 3.0 meters would have less than 96 joules of kinetic energy when it hits the ground. Is she correct and why?
The student says that a ball dropped from a height of 3.0 meters would have less than 96 joules of kinetic energy when it hits the ground. Is she correct and why?
Answers
Answer
A. She is correct because the higher the ball is dropped from, the less kinetic energy it will have when it hits the ground.
B. She is correct because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
C. She is incorrect because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
D. She is incorrect because the higher the ball is dropped from, the less kinetic energy it will have when it hits the ground.
B. She is correct because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
C. She is incorrect because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
D. She is incorrect because the higher the ball is dropped from, the less kinetic energy it will have when it hits the ground.
Answer
The table provides data on the kinetic energy of six balls, including their velocity, mass, and the resulting kinetic energy. Here's the extracted information:
Ball Velocity (m/s) Mass (kg) Kinetic Energy (J)
1 3 5 22.5
2 3 4 18.0
3 3 3 13.5
4 6 5 90.0
5 4 5 40.0
6 5 5 62.5
Which PAIR of graphs would best support this table?
Ball Velocity (m/s) Mass (kg) Kinetic Energy (J)
1 3 5 22.5
2 3 4 18.0
3 3 3 13.5
4 6 5 90.0
5 4 5 40.0
6 5 5 62.5
Which PAIR of graphs would best support this table?
Answer
what should both lines look like?
Answer
quadratic increase is wrong
Answer
The above graph is a phase change diagram for water. At points B and D the water is going through a phase change. During the phase changes, what is happening to the temperature and the heat energy?
Type Answer HERE:
The temperature and the heat energy are increasing during a phase change.
The temperature remains constant and the heat energy is increasing during a phase change.
The temperature is increasing and the heat energy remains constant during a phase change.
The temperature and the heat energy remain constant during a phase change.
This diagram illustrates the phase changes of a substance as heat energy is added. The y-axis represents temperature, and the x-axis represents heat energy. The diagram is divided into five distinct sections labeled A, B, C, D, and E, each corresponding to different phases or transitions of matter:
A: Solid phase, where the substance exists as a solid. Temperature increases as heat is added.
B: Phase transition from solid to liquid (melting). Temperature remains constant as the substance absorbs heat to break intermolecular bonds and change from solid to liquid.
C: Liquid phase, where the substance is entirely in the liquid state. Temperature increases as more heat is added.
D: Phase transition from liquid to gas (vaporization). Temperature remains constant as the substance absorbs heat to change from liquid to gas.
E: Gas phase, where the substance is entirely in the gaseous state. Temperature increases as additional heat is applied.
Throughout the diagram, small circles represent particles in different states: tightly packed for solids, more loosely packed for liquids, and widely spaced for gases. The arrows indicate the direction of heat energy flow during the phase changes.
Type Answer HERE:
The temperature and the heat energy are increasing during a phase change.
The temperature remains constant and the heat energy is increasing during a phase change.
The temperature is increasing and the heat energy remains constant during a phase change.
The temperature and the heat energy remain constant during a phase change.
This diagram illustrates the phase changes of a substance as heat energy is added. The y-axis represents temperature, and the x-axis represents heat energy. The diagram is divided into five distinct sections labeled A, B, C, D, and E, each corresponding to different phases or transitions of matter:
A: Solid phase, where the substance exists as a solid. Temperature increases as heat is added.
B: Phase transition from solid to liquid (melting). Temperature remains constant as the substance absorbs heat to break intermolecular bonds and change from solid to liquid.
C: Liquid phase, where the substance is entirely in the liquid state. Temperature increases as more heat is added.
D: Phase transition from liquid to gas (vaporization). Temperature remains constant as the substance absorbs heat to change from liquid to gas.
E: Gas phase, where the substance is entirely in the gaseous state. Temperature increases as additional heat is applied.
Throughout the diagram, small circles represent particles in different states: tightly packed for solids, more loosely packed for liquids, and widely spaced for gases. The arrows indicate the direction of heat energy flow during the phase changes.
Answered by
GPT-4o mini
The student is not correct. The kinetic energy gained by an object falling under the influence of gravity can be calculated using the equation:
\[
KE = mgh
\]
where:
- \( KE \) is the kinetic energy,
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height from which the object is dropped.
From the data you provided:
- At a height of 1.0 m, the kinetic energy \( KE_1 \) is 48 J.
- At a height of 1.5 m, the kinetic energy \( KE_2 \) is 72 J.
- At a height of 2.0 m, the kinetic energy \( KE_3 \) is 96 J.
From this pattern, we can see that the kinetic energy increases linearly with height:
- From 1.0 m to 1.5 m, the change in height is 0.5 m, and the change in kinetic energy is \( 72 \, \text{J} - 48 \, \text{J} = 24 \, \text{J} \).
- From 1.5 m to 2.0 m, the change in height is 0.5 m and the change in kinetic energy is \( 96 \, \text{J} - 72 \, \text{J} = 24 \, \text{J} \).
This consistent increase of 24 J per 0.5 m suggests that the relationship is linear, which is what we expect from the equation \( KE = mgh \).
If we extend this relationship to 3.0 m:
- The difference in height from 2.0 m to 3.0 m is another 1.0 m. Based on the earlier increases, if we calculate the expected kinetic energy at this height:
\[
KE_{3.0} = KE_{2.0} + \text{increase for 1.0 m} = 96 \, \text{J} + (2 \times 24 \, \text{J}) = 96 \, \text{J} + 48 \, \text{J} = 144 \, \text{J}
\]
Thus, a ball dropped from a height of 3.0 meters would have approximately 144 joules of kinetic energy when it hits the ground, which is indeed greater than 96 joules.
Therefore, the student is incorrect in stating that the kinetic energy at 3.0 m would be less than 96 joules.
\[
KE = mgh
\]
where:
- \( KE \) is the kinetic energy,
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height from which the object is dropped.
From the data you provided:
- At a height of 1.0 m, the kinetic energy \( KE_1 \) is 48 J.
- At a height of 1.5 m, the kinetic energy \( KE_2 \) is 72 J.
- At a height of 2.0 m, the kinetic energy \( KE_3 \) is 96 J.
From this pattern, we can see that the kinetic energy increases linearly with height:
- From 1.0 m to 1.5 m, the change in height is 0.5 m, and the change in kinetic energy is \( 72 \, \text{J} - 48 \, \text{J} = 24 \, \text{J} \).
- From 1.5 m to 2.0 m, the change in height is 0.5 m and the change in kinetic energy is \( 96 \, \text{J} - 72 \, \text{J} = 24 \, \text{J} \).
This consistent increase of 24 J per 0.5 m suggests that the relationship is linear, which is what we expect from the equation \( KE = mgh \).
If we extend this relationship to 3.0 m:
- The difference in height from 2.0 m to 3.0 m is another 1.0 m. Based on the earlier increases, if we calculate the expected kinetic energy at this height:
\[
KE_{3.0} = KE_{2.0} + \text{increase for 1.0 m} = 96 \, \text{J} + (2 \times 24 \, \text{J}) = 96 \, \text{J} + 48 \, \text{J} = 144 \, \text{J}
\]
Thus, a ball dropped from a height of 3.0 meters would have approximately 144 joules of kinetic energy when it hits the ground, which is indeed greater than 96 joules.
Therefore, the student is incorrect in stating that the kinetic energy at 3.0 m would be less than 96 joules.
Answered by
GPT-4o mini
The correct answer is **C**: She is incorrect because the higher the ball is dropped from, the greater kinetic energy it will have when it hits the ground.
As explained earlier, the kinetic energy of the ball increases with height due to the gravitational potential energy being converted to kinetic energy as the ball falls. The higher the drop height, the more potential energy is converted into kinetic energy, resulting in greater kinetic energy when it strikes the ground.
As explained earlier, the kinetic energy of the ball increases with height due to the gravitational potential energy being converted to kinetic energy as the ball falls. The higher the drop height, the more potential energy is converted into kinetic energy, resulting in greater kinetic energy when it strikes the ground.
Answered by
GPT-4o mini
To visualize the data provided in the table of kinetic energy for the six balls, we can consider two key relationships:
1. **Kinetic Energy vs. Mass (for constant velocity)**: This would show how kinetic energy changes with mass while keeping the velocity constant (for example, at \(3 \, \text{m/s}\)). Here, you would expect to see a linear increase in kinetic energy with mass.
2. **Kinetic Energy vs. Velocity (for constant mass)**: This graph would display how kinetic energy changes with velocity while keeping the mass constant. According to the kinetic energy formula \(KE = \frac{1}{2} m v^2\), you'd expect the kinetic energy to increase with the square of the velocity.
Thus, the best pair of graphs to support the given table would be:
- **Graph 1**: Kinetic Energy vs. Mass (for a constant velocity, like 3 m/s).
- **Graph 2**: Kinetic Energy vs. Velocity (for a constant mass, such as 5 kg).
These two graphs would clearly illustrate the relationships between kinetic energy, mass, and velocity as described by the kinetic energy formula:
\[
KE = \frac{1}{2} mv^2
\]
This would demonstrate both the linear relationship with mass and the quadratic relationship with velocity.
1. **Kinetic Energy vs. Mass (for constant velocity)**: This would show how kinetic energy changes with mass while keeping the velocity constant (for example, at \(3 \, \text{m/s}\)). Here, you would expect to see a linear increase in kinetic energy with mass.
2. **Kinetic Energy vs. Velocity (for constant mass)**: This graph would display how kinetic energy changes with velocity while keeping the mass constant. According to the kinetic energy formula \(KE = \frac{1}{2} m v^2\), you'd expect the kinetic energy to increase with the square of the velocity.
Thus, the best pair of graphs to support the given table would be:
- **Graph 1**: Kinetic Energy vs. Mass (for a constant velocity, like 3 m/s).
- **Graph 2**: Kinetic Energy vs. Velocity (for a constant mass, such as 5 kg).
These two graphs would clearly illustrate the relationships between kinetic energy, mass, and velocity as described by the kinetic energy formula:
\[
KE = \frac{1}{2} mv^2
\]
This would demonstrate both the linear relationship with mass and the quadratic relationship with velocity.
Answered by
GPT-4o mini
Here’s a description of what each graph should look like based on the relationships between kinetic energy, mass, and velocity:
### Graph 1: Kinetic Energy vs. Mass (for constant velocity)
- **Axis**:
- X-axis: Mass (kg)
- Y-axis: Kinetic Energy (J)
- **Data Points**:
For a constant velocity (say \( v = 3 \, \text{m/s} \)), you can use the ball masses of 3 kg, 4 kg, and 5 kg from the table:
- Mass = 3 kg: \( KE = 13.5 \, \text{J} \)
- Mass = 4 kg: \( KE = 18.0 \, \text{J} \)
- Mass = 5 kg: \( KE = 22.5 \, \text{J} \)
- **Line Shape**:
The line will be **linear** and will slant upward. As mass increases, kinetic energy increases proportionally. The slope represents the kinetic energy increase per unit mass for constant velocity.
### Graph 2: Kinetic Energy vs. Velocity (for constant mass)
- **Axis**:
- X-axis: Velocity (m/s)
- Y-axis: Kinetic Energy (J)
- **Data Points**:
For a constant mass (say \( m = 5 \, \text{kg} \)), you can use the velocities:
- Velocity = 3 m/s: \( KE = 22.5 \, \text{J} \)
- Velocity = 4 m/s: \( KE = 40.0 \, \text{J} \)
- Velocity = 5 m/s: \( KE = 62.5 \, \text{J} \)
- Velocity = 6 m/s: \( KE = 90.0 \, \text{J} \)
- **Line Shape**:
The curve will be **quadratic** (parabolic) opening upwards. This indicates that as the velocity increases, the kinetic energy increases with the square of the velocity. The relationship will look steep as you move to higher velocities because kinetic energy increases rapidly due to the \( v^2 \) term in the kinetic energy equation.
### Summary:
- **Graph 1** (KE vs. Mass): Linear increase.
- **Graph 2** (KE vs. Velocity): Quadratic increase.
These visual relationships illustrate how kinetic energy depends on mass and velocity in different ways, effectively supporting the data in your table.
### Graph 1: Kinetic Energy vs. Mass (for constant velocity)
- **Axis**:
- X-axis: Mass (kg)
- Y-axis: Kinetic Energy (J)
- **Data Points**:
For a constant velocity (say \( v = 3 \, \text{m/s} \)), you can use the ball masses of 3 kg, 4 kg, and 5 kg from the table:
- Mass = 3 kg: \( KE = 13.5 \, \text{J} \)
- Mass = 4 kg: \( KE = 18.0 \, \text{J} \)
- Mass = 5 kg: \( KE = 22.5 \, \text{J} \)
- **Line Shape**:
The line will be **linear** and will slant upward. As mass increases, kinetic energy increases proportionally. The slope represents the kinetic energy increase per unit mass for constant velocity.
### Graph 2: Kinetic Energy vs. Velocity (for constant mass)
- **Axis**:
- X-axis: Velocity (m/s)
- Y-axis: Kinetic Energy (J)
- **Data Points**:
For a constant mass (say \( m = 5 \, \text{kg} \)), you can use the velocities:
- Velocity = 3 m/s: \( KE = 22.5 \, \text{J} \)
- Velocity = 4 m/s: \( KE = 40.0 \, \text{J} \)
- Velocity = 5 m/s: \( KE = 62.5 \, \text{J} \)
- Velocity = 6 m/s: \( KE = 90.0 \, \text{J} \)
- **Line Shape**:
The curve will be **quadratic** (parabolic) opening upwards. This indicates that as the velocity increases, the kinetic energy increases with the square of the velocity. The relationship will look steep as you move to higher velocities because kinetic energy increases rapidly due to the \( v^2 \) term in the kinetic energy equation.
### Summary:
- **Graph 1** (KE vs. Mass): Linear increase.
- **Graph 2** (KE vs. Velocity): Quadratic increase.
These visual relationships illustrate how kinetic energy depends on mass and velocity in different ways, effectively supporting the data in your table.
Answered by
GPT-4o mini
I apologize for any confusion. Let's clarify the expected shape of the graph based on the kinetic energy formula and the characteristics of the relationship being described.
### Graph 1: Kinetic Energy vs. Mass (for constant velocity)
- **Axis**:
- X-axis: Mass (kg)
- Y-axis: Kinetic Energy (J)
- **Data Points** (for \( v = 3 \, \text{m/s} \)):
- Mass = 3 kg: \( KE = 13.5 \, \text{J} \)
- Mass = 4 kg: \( KE = 18.0 \, \text{J} \)
- Mass = 5 kg: \( KE = 22.5 \, \text{J} \)
- **Line Shape**:
This graph will be **linear**, with a positive slope, indicating that kinetic energy increases in direct proportion to mass when velocity is held constant. Each increase in mass yields a constant increase in kinetic energy.
### Graph 2: Kinetic Energy vs. Velocity (for constant mass)
- **Axis**:
- X-axis: Velocity (m/s)
- Y-axis: Kinetic Energy (J)
- **Data Points** (for \( m = 5 \, \text{kg} \)):
- Velocity = 3 m/s: \( KE = 22.5 \, \text{J} \)
- Velocity = 4 m/s: \( KE = 40.0 \, \text{J} \)
- Velocity = 5 m/s: \( KE = 62.5 \, \text{J} \)
- Velocity = 6 m/s: \( KE = 90.0 \, \text{J} \)
- **Line Shape**:
This graph will be **quadratic** (a curve) that opens upwards due to the relation expressed in the kinetic energy formula, \( KE = \frac{1}{2} mv^2 \). Specifically, for a given mass, as velocity increases, the kinetic energy increases proportionally to the square of the velocity. This means that as velocity increases, the rate of increase in kinetic energy becomes greater, forming a curve rather than a straight line.
### Conclusion:
- **Graph 1 (KE vs. Mass)**: Linear increase (straight line).
- **Graph 2 (KE vs. Velocity)**: Quadratic increase (curved line).
Thank you for your patience, and I'm glad to clarify these points. The key distinction is that the kinetic energy increases quadratically with velocity (not linearly) when the mass is held constant.
### Graph 1: Kinetic Energy vs. Mass (for constant velocity)
- **Axis**:
- X-axis: Mass (kg)
- Y-axis: Kinetic Energy (J)
- **Data Points** (for \( v = 3 \, \text{m/s} \)):
- Mass = 3 kg: \( KE = 13.5 \, \text{J} \)
- Mass = 4 kg: \( KE = 18.0 \, \text{J} \)
- Mass = 5 kg: \( KE = 22.5 \, \text{J} \)
- **Line Shape**:
This graph will be **linear**, with a positive slope, indicating that kinetic energy increases in direct proportion to mass when velocity is held constant. Each increase in mass yields a constant increase in kinetic energy.
### Graph 2: Kinetic Energy vs. Velocity (for constant mass)
- **Axis**:
- X-axis: Velocity (m/s)
- Y-axis: Kinetic Energy (J)
- **Data Points** (for \( m = 5 \, \text{kg} \)):
- Velocity = 3 m/s: \( KE = 22.5 \, \text{J} \)
- Velocity = 4 m/s: \( KE = 40.0 \, \text{J} \)
- Velocity = 5 m/s: \( KE = 62.5 \, \text{J} \)
- Velocity = 6 m/s: \( KE = 90.0 \, \text{J} \)
- **Line Shape**:
This graph will be **quadratic** (a curve) that opens upwards due to the relation expressed in the kinetic energy formula, \( KE = \frac{1}{2} mv^2 \). Specifically, for a given mass, as velocity increases, the kinetic energy increases proportionally to the square of the velocity. This means that as velocity increases, the rate of increase in kinetic energy becomes greater, forming a curve rather than a straight line.
### Conclusion:
- **Graph 1 (KE vs. Mass)**: Linear increase (straight line).
- **Graph 2 (KE vs. Velocity)**: Quadratic increase (curved line).
Thank you for your patience, and I'm glad to clarify these points. The key distinction is that the kinetic energy increases quadratically with velocity (not linearly) when the mass is held constant.
Answered by
GPT-4o mini
Based on the description of the phase change diagram for water, during the phase changes (specifically at points B and D), the following occurs:
- **At Point B (Melting)**: Here, water is transitioning from solid to liquid. The temperature remains **constant** because the heat energy is being used to break the intermolecular bonds that hold the solid structure together, allowing the water molecules to move more freely as a liquid.
- **At Point D (Vaporization)**: During this transition from liquid to gas, the temperature also remains **constant** again, as the added heat energy is used to overcome the intermolecular forces that keep the liquid molecules together, allowing them to become gaseous.
### Therefore, the correct statement is:
**The temperature remains constant and the heat energy is increasing during a phase change.**
This reflects that during phase changes, while the heat energy continues to increase (as heat is added), the temperature does not change until the entire substance has completed the phase transition.
- **At Point B (Melting)**: Here, water is transitioning from solid to liquid. The temperature remains **constant** because the heat energy is being used to break the intermolecular bonds that hold the solid structure together, allowing the water molecules to move more freely as a liquid.
- **At Point D (Vaporization)**: During this transition from liquid to gas, the temperature also remains **constant** again, as the added heat energy is used to overcome the intermolecular forces that keep the liquid molecules together, allowing them to become gaseous.
### Therefore, the correct statement is:
**The temperature remains constant and the heat energy is increasing during a phase change.**
This reflects that during phase changes, while the heat energy continues to increase (as heat is added), the temperature does not change until the entire substance has completed the phase transition.
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