Lab Data 1-11-06 Charles Roberts

Your calculations look good.
Your results indicate that
3 N resulted in a velocity of 104 ft/sec in the sisal twine. 1.2 N resulted in a velocity of 180 ft/sec in the nylon twine.
Velocities should be given in m/sec, to be consistent with the MKS units of the force. This will be an easy conversion.

Sisal Baler Twine – 42 feet in length 3 rubber bands used Unstretched length: 9.2 cm Stretched length: 13.0 cm Calibrated to the 3rd hit of the pendulum whose length is 8 cm. (The force exerted on the rubber bands is 3.0 Newton) ** The pendulum showed to have a period of .27 seconds, which was concluded by the timer program. So the vibrations traveled a total of 84 feet (down and back in the twine) in 3 x .27 seconds = .81 sec. 84 feet/ .81 sec. = 103. 70 feet/sec. Since 3 Newtons of force were exerted on the rubber bands, the rubber bands exerted the same 3 Newton force on the twine.

Nylon Twine – 45 feet in length 2 rubber bands used Unstretched length: 9.2 cm Stretched length: 11.0 cm Calibrated to the 2nd hit of a pendulum whose length is 7.3 cm. (The force exerted on the rubber bands is 1.2 Newtons of force) ** The pendulum showed to have a .25 second period, which was concluded by the timer program. So the vibrations traveled a total of 90 feet (down and back in the string) in 2 x .25 seconds = .50 sec. 90 feet/ .50 sec = 180 feet/sec. Since there was a 1.2 Newton force applied to the rubber bands, the same 1.2 Newton force was applied to the string.

Mr. Smith, Here are some examples of the quantitative ability problems that were given to me on a study guide by Christine Kiser:

Factorial Problems: 7! 6!/3!

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.
6! / 3! = (6 * 5 * 4 * 3 * 2 * 1) / ( 3 * 2 * 1) = 6 * 5 * 4 = 120.

Fractions: 1/2 / ((3/4)/5)

(a/b) / (c/d) = (a / b) * (d / c) = a * d / (b * c).
You have to apply this idea twice in this problem.
(1/2) / ( (3/4) / 5) = (1/2) * (5 / (3/4) ) = 1/2 * (5 * (4/3) ) = 1/2 * ( 20/3) = 1 * 20 / (2 * 3) = 20 / 6 = 10 / 3.

Lograritms: log 10 100

Since 10^2 = 100, log{base 10} ( 100 ) = 2.
The log is asking what power of the base does it take to give us the number.

log 1

What power of 10 is equal to 1?
Answer:
Since 10^0 = 1, the log you are looking for is 0.

Basic algebra: linear,fractional,and quadratic equations.

Holler at me after class about the remaining questions.

Mr. Smith, These are the problem types that I may have trouble with on an exam. If you could would you give me a brief over view of the way to work the specific types of problems. As well as any tips that may help me along in the test or in life. The test does not allow calculators, so I will have to work them with out any aid of that.

Also I remember last time it gave us diagrahams of graphs of equations. Linear, quadratic ... Also the sin, cosine ,tan and the invereses were qiven. I may need to see you after class for a reveiw of them. I really appreciate all of the help, I hope I am not bothering you. Thanks, Charles

Your calculations look good. Your results indicate that 3 N resulted in a velocity of 104 ft/sec in the sisal twine. 1.2 N resulted in a velocity of 180 ft/sec in the nylon twine. Velocities should be given in m/sec, to be consistent with the MKS units of the force. This will be an easy conversion.

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040. 6! / 3! = (6 * 5 * 4 * 3 * 2 * 1) / ( 3 * 2 * 1) = 6 * 5 * 4 = 120.

(a/b) / (c/d) = (a / b) * (d / c) = a * d / (b * c). You have to apply this idea twice in this problem. (1/2) / ( (3/4) / 5) = (1/2) * (5 / (3/4) ) = 1/2 * (5 * (4/3) ) = 1/2 * ( 20/3) = 1 * 20 / (2 * 3) = 20 / 6 = 10 / 3.