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Question 4 1 pts The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 seconds, whereas the standard deviation is found to be 0.24 seconds. The standard error (or standard deviation of the mean) therefore is 0.24 seconds divided by the square root of 16, yielding 0.06 seconds. Assuming that the uncertainty of this period arises solely from random error for these 16 trials, about how many additional trials would need to be completed to reduce the standard error to 0.04 seconds? Question 5 1 pts For a certain experiment, you must provide the kinetic energy of some object with its uncertainty. (The equation for kinetic energy K = 1/2mv^2, where m is the mass of the object and v is the object's speed.) To do this, you first measure the mass of the object to be 0.015 kg. Due to the limits on precision for mass using a scale, you can only measure the mass of an object within an uncertainty of 2.7 grams (so ?m=2.7x10^-3 kg). In this same experiment, you measure the speed as 10 m/s, but you can only measure the speed with a precision ?v=1.1 m/s. Hence, the mass and speed you report are m = 15±2.7 g and v = 10±1.1 m/s, respectively. How would you report the kinetic energy using the error propagation formula given in the introduction (assuming the speed and mass measurements are independent of each other)? Hint, the derivative of kinetic energy with respect to mass dK/dm = 1/2v^2 and the derivative of kinetic energy with respect to speed dK/dv = mv. (fill in the blank) K = 0.75 ± ______ J

Question 4
1 pts
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 seconds, whereas the standard deviation is found to be 0.24 seconds. The standard error (or standard deviation of the mean) therefore is 0.24 seconds divided by the square root of 16, yielding 0.06 seconds.
Assuming that the uncertainty of this period arises solely from random error for these 16 trials, about how many additional trials would need to be completed to reduce the standard error to 0.04 seconds?
Question 5
1 pts
For a certain experiment, you must provide the kinetic energy of some object with its uncertainty. (The equation for kinetic energy K = 1/2mv^2, where m is the mass of the object and v is the object's speed.) To do this, you first measure the mass of the object to be 0.015 kg. Due to the limits on precision for mass using a scale, you can only measure the mass of an object within an uncertainty of 2.7 grams (so ?m=2.7x10^-3 kg). In this same experiment, you measure the speed as 10 m/s, but you can only measure the speed with a precision ?v=1.1 m/s. Hence, the mass and speed you report are m = 15±2.7 g and v = 10±1.1 m/s, respectively. How would you report the kinetic energy using the error propagation formula given in the introduction (assuming the speed and mass measurements are independent of each other)? Hint, the derivative of kinetic energy with respect to mass dK/dm = 1/2v^2 and the derivative of kinetic energy with respect to speed dK/dv = mv.
(fill in the blank)
K = 0.75 ± ______ J

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