In the experiment studying Uniform Circular Motion, you were asked to plot the p
ID: 1553727 • Letter: I
Question
In the experiment studying Uniform Circular Motion, you were asked to plot the product of the mass and the radially-inward acceleration along the vertical axis versus the axis versus the radially-inward force along the horizontal axis. The slope of that plot is a. the initial acceleration times the mass b. the initial applied force c. expected to be equal to 1, since the ratio is 1 d. the mass being accelerated in a circular path e. the reciprocal of the mass being accelerated in a circular path During your analysis in the Projectile Motion lab, you plotted the x component of the velocity versus times. If the result of a linear regression to the data gives a slope more than three standard deviations less than zero, which of the following is supported by your result? a. Your experiment is a good example of "free fall" projectile motion. b. Your experiment is free from significant sources of systematic likely random error. c. Your experiment likely suffered from a large amount of random error. d. Your experiment likely suffered from significant sources of systematic error. e. both (b) and (d) In the "Newton's 2^nd Law" lab, you decide to use a 150g hanging mass to accelerate a glider (M-450g) along the air track via a string and pulley system. What will the acceleration of the system be? a. 2.45 m/s^2 b. 3.27 m/s^2 e. 9.81 m/s^2 d. 7.36 m/s^2 e. none of the above In the "Newton's 2^nd Law" lab you make a plot of a (m/s^2) vs. F(N) and perform linear regression. The results are given below. Calculate the system mass with its appropriate uncertainty and report them in the correct format. Show your work. If you measure the experimental value of some quantity to be 3.45 plusminus 0.15 and the "expected value" is 365, you should conclude that: a. The experimental value is NOT in agreement with the expected value, since they are not equivalent. b. The experimental value IS in agreement with the expected value, since the z value is less than 2. c You must have made a mistake in your measurements, since the z value is less than 2. d. You must have made a mistake in your analysis, since the z value is less than 2.Explanation / Answer
(5) According to Newton's second law of motion, producti of mass and inward acceleration to the centre is equal to the force acting inward to the centre. Hence both the values are same. While plotting in a graph, since both the values are same, the ratio is 1. Hence the slope will also be 1. ( slope =tan45o)
(7) The hanging mass of 150g will exert a tensile force of (0.15*9.81) over the glider. Hence the acceleration on the glider =Force/mass = (0.15*9.81)/0.45 = 3.27 m/s2
(8) The system mass can be determined by Force/acceleration. The coefficients say that the Force=1.82461446 N and the acceleration on the system is 0.008725028 m/s2 . Standard error says the difference between the estimated value on the regression line and the actual value. Hence the actual values of the acceleration = |0.008725028-0.008388067| = 0.000336961 and the Force of the system is |1.82461446-0.025780809|=1.798833651.Therefore the mass of the system is Force/acceleration = 1.798833651/0.000336961= 5338.403112 kg
(9) The Z= (3.45-3.65)/0.15=1.333. The Z value is less than 2. Hence the experimental value is in agreement with the expected value.