Torque Lab Report Essay

Finally we got all our number and determine the slope, and the intercept in order to find out the forecast for the next

The slope of the linear fit of the data is 1.0049. What this tells me about the water is that it is increasing at a close to constant rate – while my results were not completely accurate because the slope of the line was not one it was fairly close to the target

To begin the experiment, we measured the masses of the two stoppers and the eye bolt used to secure the stoppers that we were using in our apparatus. The mass of the first stopper was 18.8 grams and the mass of the second stopper was 50.5 grams. The mass of the eye bolt was 11.6 grams. The mass of the screw and bolt that secured our hanging mass was given to us as 25 grams. After, we chose six different hanging masses based on stopper mass. We made sure that the hanging mass was always larger than the stopper mass or else we would not be able to get the stopper to spin at a constant velocity. The first three mass ratios we chose was using the stopper with the mass of 18.8 grams and then we used a hanging mass (the mass of the screw and bolt is included) of 65 grams, 85 grams, and 105 grams. This gave the three mass

4) Use cubic regression to determine an equation for the data (or lwh where (12 – x) represents the sides and (x) represents the height of the box).

All i did for this one was basically calculate the slopes because I didn't want to over think the question. Based off the figures let's calculate the following slopes...

The average rate of change is an average slope from the initial point to the final point. We would have to use A(x)=f(x)-f(a) divided by x-a to find the average rate of change. Sometimes we would want to know what the average rate of change is in the middle of the graph. The average rate of change is used in page 10, “How many more people?.” The concept of slope comes from the idea of a constant rate of change. To find the slope, you have to calculate y1-y2 divided by x1-x2. Slope is often denoted by the letter “m” which means that m+ equals the slope. Its problem is being used in page 21, “Rates, Graph, Slopes, and Equations.” Y=mx+b is the equation of the line that you can find using the two points. Variable “m” is the slope and “b” is the y-intercept.

5. Tie one end of the 60-cm string to the mass. Place the mass on a table below

Initially, 8 pennies were added to the cup, followed by the addition of 7 pennies and 1 dime, then 4 pennies and 4 dimes, and finally 8 more pennies. There were therefore a total of 27 pennies and 5 dimes added to the cup. Table 2 demonstrates that the force (N) for dimes and pennies went up by almost 0.20 N at each interval. Therefore, the force (N) in Table 2 did not deviate much from the force (N) seen in Table 1 where all pennies were used. The reason little variation in force was seen in Table 2 was due to mostly pennies being added to the cup. Due to so many pennies being added, the dimes had little impact on the overall force (N). If roughly an equal ratio of pennies to dimes had been added to the cup, a more distinct variation between the slope’s in Figure 1 and Figure 2 would have been seen. However, the slope or the average weight of the coins, as represented in Figure 2, was 0.0249 N. The slope can be calculated by dividing the change in the force by the change in the # of pennies and dimes. The x-values represent the number of pennies and dimes, while the y-values represent force (N). The y-intercept value is equal to 0, therefore, the linear equation is y=0.0249x+0. After plotting the line on a manual curve fit, as can be seen in Figure 2, the R2 value was 0.99884. This R2 value is very close to 1, meaning that the match of the linear model to the data fits. After running a

A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single rope. It has a mass of m2 = 40.0 kg attached at one end and an unknown mass m1 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal, as shown in the figure above. The tension in the rope is T = 637 N.

the arm on the catapult. My independant variables are the items and their different weights

This regression equation can be graphed as follows assuming β0 as the intercept and β1 as the slope:

Step 10: Draw a scatter plot of T2 on the Y-axis against L on the X-axis.

As a result of trail and error method we find the angle for the third pulley and the mass that should had be suspended from it. This will balance the forces deployed on the strings due to the other two masses. While the third force is defined as the equilibrant (������⃗������) Since it is the force that establishes the equilibrium. It is also the negative of the resultant -������⃗������ = ������⃗������ = ������⃗ ������ + ������⃗������. We gathered and recorded the mass and the angle required for the third pulley to enable to put the system into the equilibrium in table 1.

Below are two tables in which I have recorded the data which I obtained during the experiment. The first table reflects the Relationship between the deflection/flexion of the cantilever and the mass of the load and the second table reflects the relationship between the flexion of the cantilever and the length of the cantilever.

Table 1 & 2: First, find the mass of the wooden block and record the data. Then place the wooden block on the inclined plane (at 0o) with the wide side down. The height of the pulley should be the same height as the screw location on the wooden block. Then hang a weight on the opposite side of the hanger and add weights until block starts to move with a constant velocity (push block to overcome fs¬). Then record the resulted weight of the hanger in Table 1 (as F). Add 500 g to the wooden block and repeat the process. Replace 500 g with 1 kg on the wooden block. Repeat the process described above.