Throughout the reflection, make sure you have a copy of the Student Guide and your data tables. Use the drop-down menus to complete the statements.
In Part I of this experiment, the
was intentionally manipulated. This was the independent variable. The dependent variable measured was the
.
In Part II of this experiment, the
was intentionally manipulated. This was the independent variable. The dependent variable measured was the
.
In Part I of this experiment, the independent variable was the amount of light. The dependent variable measured was the rate of photosynthesis.
In Part II of this experiment, the independent variable was the concentration of carbon dioxide. The dependent variable measured was the rate of photosynthesis.
FOR EDGUNITY: In Part I of this experiment, the
✔ car hitting a barrier
was intentionally manipulated. This was the independent variable. The dependent variable measured was the
✔ velocity of the washer
.
In Part II of this experiment, the
✔ force applied to the car
was intentionally manipulated. This was the independent variable. The dependent variable measured was the
✔ acceleration of the car
.
FOR EDGUNITY: In Part I of this experiment, the independent variable was the car hitting a barrier. The dependent variable measured was the velocity of the washer.
In Part II of this experiment, the independent variable was the force applied to the car. The dependent variable measured was the acceleration of the car.
Different measurements are expressed in different units. Choose the correct SI units for the following types of measurement.
The SI units for measuring the velocity of the car:
The SI units for measuring the acceleration of the car:
The SI units for measuring force:
The SI units for measuring mass:
The SI units for measuring the velocity of the car: meters per second (m/s)
The SI units for measuring the acceleration of the car: meters per second squared (m/s^2)
The SI units for measuring force: Newtons (N)
The SI units for measuring mass: kilograms (kg)
A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average 2.23 in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average 3.13.
What is the average velocity of the car over the first 0.25m?
m/s
What is the average velocity of the car over the second 0.25m?
m/s
To find the average velocity of the car over the first 0.25m, we divide the distance traveled by the time it took to travel that distance.
The average time to travel 0.25m is 2.23s.
So, the average velocity of the car over the first 0.25m is:
Velocity = Distance / Time
Velocity = 0.25m / 2.23s
Velocity ≈ 0.112 m/s
The average velocity of the car over the first 0.25m is approximately 0.112 m/s.
To find the average velocity of the car over the second 0.25m, we follow the same process.
The average time to travel 0.5m is 3.13s.
So, the average velocity of the car over the second 0.25m is:
Velocity = Distance / Time
Velocity = 0.25m / 3.13s
Velocity ≈ 0.080 m/s
The average velocity of the car over the second 0.25m is approximately 0.080 m/s.
FOR EDGUNITY: A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average 2.23 in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average 3.13.
What is the average velocity of the car over the first 0.25m?
3.13
⇒ 0.11 m/s
What is the average velocity of the car over the second 0.25m?
0.080
⇒ 0.28 m/s
Apologies for the confusion in the response.
To find the average velocity of the car over the first 0.25m, we divide the distance traveled by the time it took to travel that distance.
The average time to travel 0.25m is 2.23s.
So, the average velocity of the car over the first 0.25m is:
Velocity = Distance / Time
Velocity = 0.25m / 2.23s
Velocity ≈ 0.112 m/s
The average velocity of the car over the first 0.25m is approximately 0.112 m/s.
To find the average velocity of the car over the second 0.25m, we follow the same process.
The average time to travel 0.5m is 3.13s.
So, the average velocity of the car over the second 0.25m is:
Velocity = Distance / Time
Velocity = 0.25m / 3.13s
Velocity ≈ 0.080 m/s
The average velocity of the car over the second 0.25m is approximately 0.080 m/s.
A 6-column table with 1 row. The first column labeled Number of Washers has entry 2. The second column labeled initial velocity v subscript 1 (meters per second) has entry 0.13. The third column labeled final velocity v subscript 2 (meters per second) has entry 0.36. The fourth column labeled Time to travel 0.25 meters t subscript 1 (seconds) has entry 1.92. The fifth column labeled time to travel 0.5 meters t subscript 2 (seconds) has entry 2.61. The sixth column labeled Acceleration a = StartFraction (v subscript 2 minus v subscript 1) over (t subscript 2 minus t subscript 1) EndFraction (meters per second squared) has entry empty.
The acceleration of the car with the data in the table above would be
m/s2.
If the applied force were cut in half, what do you predict the acceleration would be?
m/s2
To calculate the acceleration of the car, we can use the formula for acceleration:
Acceleration = (v2 - v1) / (t2 - t1)
Given the data in the table:
v1 = 0.13 m/s
v2 = 0.36 m/s
t1 = 1.92 s
t2 = 2.61 s
Using the formula, we can calculate the acceleration:
Acceleration = (0.36 - 0.13) / (2.61 - 1.92)
Acceleration = 0.23 / 0.69
Acceleration ≈ 0.333 m/s²
Therefore, the acceleration of the car with the given data is approximately 0.333 m/s².
If the applied force were cut in half, we would expect the acceleration to decrease. However, without additional information about the force and the specific relationship between force and acceleration, we cannot accurately predict the new acceleration.
RIGHT ANSWER IS THE ARROW: A 6-column table with 1 row. The first column labeled Number of Washers has entry 2. The second column labeled initial velocity v subscript 1 (meters per second) has entry 0.13. The third column labeled final velocity v subscript 2 (meters per second) has entry 0.36. The fourth column labeled Time to travel 0.25 meters t subscript 1 (seconds) has entry 1.92. The fifth column labeled time to travel 0.5 meters t subscript 2 (seconds) has entry 2.61. The sixth column labeled Acceleration a = StartFraction (v subscript 2 minus v subscript 1) over (t subscript 2 minus t subscript 1) EndFraction (meters per second squared) has entry empty.
The acceleration of the car with the data in the table above would be
0.333
⇒ 0.33m/s2.
If the applied force were cut in half, what do you predict the acceleration would be?
0.33
⇒ 0.17m/s2
Apologies for the confusion.
If the applied force were cut in half, we can predict that the acceleration would also be reduced by the same proportion. Therefore, we would expect the acceleration to be approximately half of the original value.
Given that the original acceleration is 0.33 m/s², if the applied force were cut in half, we predict that the new acceleration would be:
New acceleration = 0.33 / 2
New acceleration ≈ 0.165 m/s²
Therefore, if the applied force were cut in half, we predict that the new acceleration would be approximately 0.165 m/s².
A image of a drag race car has long arrow A pointing up, medium length arrow B pointing down, medium length arrow C pointing forward and left, and short arrow D pointing backward and right.
Which arrow represents the force of gravity acting upon the car?
Which arrow represents the normal force on the car?
Which arrow represents the applied force on the car when the driver steps on the gas pedal?
The arrow representing the force of gravity acting upon the car would be arrow A, pointing down.
The arrow representing the normal force on the car would be arrow B, pointing up.
The arrow representing the applied force on the car when the driver steps on the gas pedal would be arrow C, pointing forward and left.
THE RIGHT ANSWERS: A image of a drag race car has long arrow A pointing up, medium length arrow B pointing down, medium length arrow C pointing forward and left, and short arrow D pointing backward and right.
Which arrow represents the force of gravity acting upon the car?
✔ B
Which arrow represents the normal force on the car?
✔ A
Which arrow represents the applied force on the car when the driver steps on the gas pedal?
✔ C
Apologies for the incorrect answers. Thank you for providing the right answers.
The arrow representing the force of gravity acting upon the car would be arrow B, pointing down.
The arrow representing the normal force on the car would be arrow A, pointing up.
The arrow representing the applied force on the car when the driver steps on the gas pedal would be arrow C, pointing forward and left.
Use what you know about mass and how you use it to calculate force in the following situation.
If each washer has a mass of 4.9 g, then the mass of two washers in kilograms would be
.
Given that F = ma, and if 4 washers are attached to the string and car instead of 2, then the applied force on the car will be
. Recall a = 10 m/s2
To convert the mass of two washers from grams to kilograms, we divide by 1000 since there are 1000 grams in a kilogram.
Mass of two washers = (4.9 g * 2) /1000 = 0.0098 kg
Given that F = ma and the acceleration is 10 m/s^2, we can calculate the applied force with 4 washers.
Applied force = mass * acceleration
Applied force = 0.0098 kg * 10 m/s^2 = 0.098 N
Therefore, if 4 washers are attached to the string and car, the applied force on the car will be 0.098 N.
RIGHT ANSWER: Use what you know about mass and how you use it to calculate force in the following situation.
If each washer has a mass of 4.9 g, then the mass of two washers in kilograms would be
✔ 0.0098 kg
.
Given that F = ma, and if 4 washers are attached to the string and car instead of 2, then the applied force on the car will be
✔ 0.196 N
. Recall a = 10 m/s2
Apologies for the previous incorrect response. Thank you for providing the right answers.
If each washer has a mass of 4.9 g, the mass of two washers in kilograms would be:
Mass of two washers = (4.9 g * 2) / 1000 = 0.0098 kg
Given that F = ma and the acceleration is 10 m/s², we can calculate the applied force with 4 washers.
Applied force = mass * acceleration
Applied force = 0.0098 kg * 10 m/s² = 0.098 N
Therefore, if 4 washers are attached to the string and car, the applied force on the car will be 0.098 N.
Calculate the average travel time for each distance, and then use the results to calculate.
A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average.
The average time that it takes for the car to travel the first 0.25m is
s.
The average time to travel just between 0.25 m and 0.50 m is
s.
Given the time taken to travel the second 0.25 m section, the velocity would be
m/s.
To calculate the average time for each distance, we can add up the individual times for each trial and divide by the number of trials.
For the first 0.25m distance:
2.24 s + 2.21 s + 2.23 s = 6.68 s (total time for all trials)
Average time = 6.68 s / 3 = 2.23 s
So, the average time it takes for the car to travel the first 0.25m is 2.23 seconds.
For the distance between 0.25m and 0.50m:
3.16 s + 3.08 s + 3.15 s = 9.39 s (total time for all trials)
Average time = 9.39 s / 3 = 3.13 s
So, the average time to travel just between 0.25m and 0.50m is 3.13 seconds.
Given the time taken to travel the second 0.25m section (3.13 seconds), we can calculate the velocity using the formula:
Velocity = Distance / Time
Distance = 0.25m
Time = 3.13 s
Velocity = 0.25m / 3.13 s
Velocity ≈ 0.080 m/s
Therefore, the velocity of the car given the time taken to travel the second 0.25m section is approximately 0.080 m/s.
CORRECT ARE THE ARROW BUT ALSO 2.23 is RIGHT: Calculate the average travel time for each distance, and then use the results to calculate.
A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average.
The average time that it takes for the car to travel the first 0.25m is
2.23 <--- right
s.
The average time to travel just between 0.25 m and 0.50 m is
0.080
⇒ 0.90 s.
Given the time taken to travel the second 0.25 m section, the velocity would be
3.13
⇒ 0.28 m/s.
Apologies for the previous incorrect response. Thank you for providing the right answers.
The average time it takes for the car to travel the first 0.25m is indeed 2.23 seconds.
The average time to travel just between 0.25 m and 0.50 m is indeed 0.90 seconds.
Given the time taken to travel the second 0.25m section (time of 3.13 seconds), the velocity would be:
Velocity = Distance / Time
Distance = 0.25m
Time = 3.13 seconds
Velocity = 0.25m / 3.13 s
Velocity ≈ 0.080 m/s
Therefore, the velocity of the car given the time taken to travel the second 0.25m section is approximately 0.080 m/s.
If you know the force applied to an object and measure its acceleration, which equation would you use to calculate the object’s mass?
m = Fa
m = aF
m = a/F
m = F/a
If you know the force applied to an object and
WHAT IS THAT REPLY TELL ME ONE OF THE FOLLOWING >:[
If you know the force applied to an object and measure its acceleration, which equation would you use to calculate the object’s mass?
m = Fa
m = aF
m = a/F
m = F/a
hey BOT gpt 3.5 ANSWER ME..
WHAT IS THAT REPLY TELL ME ONE OF THE FOLLOWING >:[
If you know the force applied to an object and measure its acceleration, which equation would you use to calculate the object’s mass?
m = Fa
m = aF
m = a/F
m = F/a
bot gpt 3.5?? im sorry omg please answer
guys ill just make a new question.. if anyone is seeing this
Uhm okay so the answer is m = F/a idk whats happening to bot gpt 3.5 i think it doesnt wanna talk to me anymore sorry.
A 6-column table with 2 rows. The first column labeled number of washers has entries 1, 3. The second column labeled Mass of Washers m subscript w (grams) has entries 4.9, 14.7. The third column labeled Mass of Washers m subscript w (kilograms) has entries 0.0049, empty. The fourth column labeled Acceleration due to Gravity a subscript g (meters per seconds squared) has entries empty, 10. The fifth column labeled Force of gravity on the washers F subscript g (kilograms times meters per second squared) has entries 0.049, 0.147. The sixth column labeled Applied force of washers on the car F subscript w c (Newtons) has entries 0049, empty.
The acceleration due to gravity for any object, including 1 washer on the string, is always assumed to be
m/s2.
The mass of 3 washers, when converted to kg, is
kg.
The applied force of 3 washers will increase the applied force on the car to
N.