Gravity

Graph

Analysis
VM: As the mass increases, the force increases.proportionally
MM: y=9.7045x+0.0122
Fg=(9.7045N/kg)M
Fg is force of gravity.
M is mass.
(9.7045N/kg) is g:gravitational field strength on earth. 

Slope: 9.7045
what increases> increases by 9.7045units? per every kg

Y-int: 0.0122
y-intercept is almost 0 because there is 0 force at 0 mass units??

Data Table too many decimal places....

Mass (kg)	Force (N)
0.060000 kg	0.581000 N
0.100000 kg	0.980900 N
0.150000 kg	1.473000 N
0.250000 kg	2.456000 N
0.550000 kg	5.342000 N

Conclusion/Evidence/Claims

Mass and weight differ because weight is the force of gravity while mass is the amount of stuff in something. Everyone's graphs are the same because we all got almost 9.8 as the gravitational field strength on Earth. All graphs are linear with the same slope, traveling in a positive direction.  nothing is traveling, it;s a different relationship The field is the same because the gravitational field remains the same no matter where you are on Earth. good!Our new equation was Fg=mg, which we got by multiplying the mass by the gravitational field strength, and therefore you get the total force of gravity. F is the force of attraction between the two objects. G is the universal gravitational constant. M is the mass of the objects.

Greater the mass, the more weight will be on the object. good!

Bonus:

Light and heavy objects hit the ground at the same time when dropped because 

No matter the size, a big object's Fg/mass will be the same as a small object's Fg/mass.  excellent!

The Moon Lab

To calculate the distance between the Earth and the moon, we had to follow several steps. First, we found the distance between the earth and the moon which is 238,900 miles or 384,400 km. Then, we calculated how many earths is earth away from the moon. It takes 30 earths to get to the moon. Afterwards, we calculated the proportions of our "earth" and "moon."

Earth was 16 in
Moon was 4 inches
So then we measured boards and laid them across the classroom, which will be made up of diamter "earths".
In the solution, we came to the conclusion that the distance between the Earth and the moon consisted of 30 earths, or approximately 480 inches.

Dueling Buggies Lab

Objective Statement
The problem we were trying to solve was to come up with a way where we could calculate where the two cars, with different velocities, coming from opposite directions would meet. To find out the position and the time of the collision, we needed to either come up with an algebraic equation or solve it graphically. To do so, we needed to collect our measurements and calculate the different variables that would make up the formula needed to solve our problem.

Your Plan

What we did was we found the average velocities of both of the buggies. Because one was faster than the other, we ran 2 trials.

1. Aligned 2 rulers after each other, with the overall length of 80 inches. 
2. Marked every 10 inches the rulers.
3. Placed the buggy at 0 inches. 
4. Once the buggy would start traveling, we started the stopwatch.
5. With the stopwatch, we timed how many seconds it took for the buggy to reach the markings at 10 inches. 
6. After recording down the relatively close times, we calculated the average of all of them (7-8 for each buggy) to get the most legitimate time. 
7. Plugged our results into Velocity=Distance/Time formula to calculate speed. 
8. Repeated the process for the 2nd buggy. 

To calculate the average, we added all of the time measurements together and divided by how many measurements there were recorded. To convert our sec per 10 inches to velocity, we divided the time by distance. By doing so, we found the rate for both of the buggies.
Calculations
1st buggy
10in/1.84sec= 5.43in/sec

2nd buggy
10in/0.46sec=21.74in/sec
distance/time = speed ... I need to see those calcualtions

From the data we collected, it was evident that the 2nd buggy was much faster than the 1st buggy because it took the 2nd buggy 4 times as fast to reach the 10 inches compared to the 1st buggy. 

Data Analysis

After a long time of not figuring out how to connect the information to solve our problem, we finally reached a conclusion. Since the problem included rate, time, and distance, it occurred to me that we could use the rate times time = distance formula. Therefore I tried an equation we usually see in our math books in a word problem: (rate and time of the first buggy) plus (the product of rate and time of the second buggy) which would equal the overall distance. With the lab, the equation looked a little like this: (5.43t)+(21.7t)=the given distance.  Adding these 2 equal the total distance because the amount they both traveled equal the total distance. (5.43t) is the amount of distance the first buggy traveled, and (21.7t) is the amount the second buggy traveled until the collision. You need the total distance to calculate the collision point. WHY does adding these 2 equal the total distance...  this is important to explain - 5.43 is the rate of the first, slow buggy, while 21.7 is the rate of the faster buggy. 

ok good!  :)

Buggy #1	Buggy #2
1.90 sec/10 in	0.38 sec/10 in
1.95 sec/10 in	0.41 sec/10 in
1.68 sec/10 in	0.40 sec/10 in
1.9 sec/10 in	0.48 sec/10 in
1.76 sec/10 in	0.52 sec/10 in
1.93 sec/10 in	0.55 sec/10 in
=1.83 sec/10 in	=0.46 sec/10 in

Model/Designing a Solution

With our formula, we would be solving how many seconds it would take for the two buggies to collide. With the given distance of 65, we plugged it into the equation                 (5.43t)+(21.7t)=65 to get an answer of 2.4 seconds. Then, we ran a trial run with the stopwatch and it took exactly 2.4 seconds for the cars to collide at 17.5 inches, meaning our method was successful.