To successfully pick up the metal object, the student should get a larger magnet with a larger magnetic field. Increasing the size of the magnet will increase its magnetic field strength and the distance over which it can attract ferromagnetic materials.
A student tries to pick up a metal object with a magnet, but the magnet is too far away. What can he change in order to successfully pick up the object?(1 point)
Responses
Get a larger magnet with a larger magnetic field
Get a smaller magnet with a smaller magnetic field
Get a smaller magnet with a smaller magnetic field
Rotate the magnet so the opposite pole is facing the object
Rotate the magnet so the opposite pole is facing the object
Rotate the magnet so the center is facing the object
11 answers
A student conducts an experiment with a magnet and four objects. He puts the magnet near each of the objects, then observes and records the outcomes. Which question is the student’s experiment designed to answer?(1 point)
Responses
Can a magnetic force change the shape of an object?
Can a magnetic force change the shape of an object?
Which type of magnet can move an object?
Which type of magnet can move an object?
Which objects react to a magnetic force?
Which objects react to a magnetic force?
How does an object’s mass affect its reaction to a magnetic force?
Responses
Can a magnetic force change the shape of an object?
Can a magnetic force change the shape of an object?
Which type of magnet can move an object?
Which type of magnet can move an object?
Which objects react to a magnetic force?
Which objects react to a magnetic force?
How does an object’s mass affect its reaction to a magnetic force?
The student's experiment is designed to answer the question: Which objects react to a magnetic force? This is because he is observing the interactions between the magnet and the different objects.
A student made an electromagnet using an iron bolt, copper wire coiled three times around the bolt, and a battery. How can he increase the strength of his magnet?(1 point)
Responses
Add batteries to increase the voltage, and decrease the space between the coils of wire.
Add batteries to increase the voltage, and decrease the space between the coils of wire.
Add batteries to increase the voltage, and increase the space between the coils of wire.
Add batteries to increase the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and decrease the space between the coils of wire.
Add batteries to decrease the voltage, and decrease the space between the coils of wire.
Responses
Add batteries to increase the voltage, and decrease the space between the coils of wire.
Add batteries to increase the voltage, and decrease the space between the coils of wire.
Add batteries to increase the voltage, and increase the space between the coils of wire.
Add batteries to increase the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and increase the space between the coils of wire.
Add batteries to decrease the voltage, and decrease the space between the coils of wire.
Add batteries to decrease the voltage, and decrease the space between the coils of wire.
To increase the strength of the electromagnet, the student should add batteries to increase the voltage, and decrease the space between the coils of wire. Increasing the voltage provides more current, which enhances the magnetic field, and decreasing the space between the coils increases the density of the magnetic field lines, further strengthening the magnet.
A student wraps 10 coils of copper wire around an iron bolt, and then she connects the ends of the wire to a battery. Next, the student uses the nail to lift paper clips. She then conducts a second trial using the same wire, battery, and nail, but with 30 coils of copper wire wrapped around the bolt. What will happen in the second trial?(1 point)
Responses
The bolt will lift less paper clips, because more conductive material in the system will balance more paper clips.
The bolt will lift less paper clips, because more conductive material in the system will balance more paper clips.
The bolt will lift the same number of paper clips, because both trials used the same type of paper clips.
The bolt will lift the same number of paper clips, because both trials used the same type of paper clips.
The bolt will lift more paper clips, because more coils of wire produces a stronger magnetic field.
The bolt will lift more paper clips, because more coils of wire produces a stronger magnetic field.
The bolt will lift the same number of paper clips, because both trials used the same type of battery.
Responses
The bolt will lift less paper clips, because more conductive material in the system will balance more paper clips.
The bolt will lift less paper clips, because more conductive material in the system will balance more paper clips.
The bolt will lift the same number of paper clips, because both trials used the same type of paper clips.
The bolt will lift the same number of paper clips, because both trials used the same type of paper clips.
The bolt will lift more paper clips, because more coils of wire produces a stronger magnetic field.
The bolt will lift more paper clips, because more coils of wire produces a stronger magnetic field.
The bolt will lift the same number of paper clips, because both trials used the same type of battery.
In the second trial, the bolt will lift more paper clips, because more coils of wire produces a stronger magnetic field. Increasing the number of coils around the iron bolt enhances the magnetic field strength, allowing it to lift more paper clips.
At the bowling alley, Terry paid $22.50 and bowled 3 games while Tommy bowled 2 games and paid $15. Interpret the rate of change by identifying the cost of each game.(1 point)
$
per game
$
per game
To find the cost per game, we can divide the total amount paid by each person by the number of games they bowled.
For Terry:
- Total cost: $22.50
- Number of games: 3
Cost per game for Terry: \[ \text{Cost per game} = \frac{22.50}{3} = 7.50 \]
For Tommy:
- Total cost: $15.00
- Number of games: 2
Cost per game for Tommy: \[ \text{Cost per game} = \frac{15.00}{2} = 7.50 \]
Both Terry and Tommy paid the same rate per game.
Therefore, the cost of each game is $7.50 per game.
Use the image to answer the question.
An illustration shows three coiled wire springs stretched to varying lengths. The first spring does not have any weights pulling it downward, and the length of the spring is marked as a question mark inches. The second spring has a 2-pound weight attached to the bottom, and its length is marked as 12 inches. The third spring has a 5-pound weight attached to the bottom and its length is marked as 18 inches.
A spring has a length of 12 inches when a 2-pound weight is attached, and a length of 18 inches when a 5-pound weight is attached. Use rate of change to find the length of the spring when no weights are attached.
(1 point)
inches
An illustration shows three coiled wire springs stretched to varying lengths. The first spring does not have any weights pulling it downward, and the length of the spring is marked as a question mark inches. The second spring has a 2-pound weight attached to the bottom, and its length is marked as 12 inches. The third spring has a 5-pound weight attached to the bottom and its length is marked as 18 inches.
A spring has a length of 12 inches when a 2-pound weight is attached, and a length of 18 inches when a 5-pound weight is attached. Use rate of change to find the length of the spring when no weights are attached.
(1 point)
inches
To find the length of the spring when no weights are attached, we can use the information provided about the weights and the corresponding lengths.
We know:
- A 2-pound weight stretches the spring to 12 inches.
- A 5-pound weight stretches the spring to 18 inches.
We can first find the rate of change of length per pound of weight.
The change in length when the weight increases from 2 pounds to 5 pounds is: \[ \text{Change in length} = 18 \text{ inches} - 12 \text{ inches} = 6 \text{ inches} \]
The change in weight is: \[ \text{Change in weight} = 5 \text{ pounds} - 2 \text{ pounds} = 3 \text{ pounds} \]
Now, we find the rate of change of length per pound: \[ \text{Rate of change} = \frac{6 \text{ inches}}{3 \text{ pounds}} = 2 \text{ inches per pound} \]
Now, if we know that the spring is 12 inches with a 2-pound weight attached, we can find the length of the spring without any weights. From the 2-pound weight, we can calculate how much the spring would shorten if we removed the weight.
Since the rate is 2 inches per pound, for 2 pounds: \[ \text{Total length change with 2 pounds} = 2 \text{ inches/pound} \times 2 \text{ pounds} = 4 \text{ inches} \]
Thus, when no weights are attached, the length of the spring would be: \[ \text{Length without weight} = 12 \text{ inches} - 4 \text{ inches} = 8 \text{ inches} \]
Therefore, the length of the spring when no weights are attached is 8 inches.