Experimental General Physies for Engineering I PHYS 192 10257 L0 Exam duration:

Question

Experimental General Physies for Engineering I PHYS 192 10257 L0 Exam duration: 2 hours College of Arts and Sciences Question 2 a) Calculate the ernor of the mean of the following values. (Don't forget the unit and give the result with 2 significant figures and show calculation). (2pts) [h-835.36 s, tz-835.48 s, t,-835.21 t4-9426.13 sl s, b) 1 cm and the width is 57 ± 0S cm. what is Given that, the length of your table is 135 ± the surface area of this table and its erro? A-L W (show the details of your calculations) (2pts)

Explanation / Answer

Q2. a) To find the mean error, first we will find the average. for given observations.

Mean = (t1 + t2 + t3 + t4)/4 = (835.36+835.48+835.21+9426.13)/4 = 11932.18/4 = 2983.045 s

Now, we will find the range, which is the difference between the largest and the smallest observation.

i.e. Range = 9426.13 - 835.21 = 8590.92

The error in the mean, then, is given by the half of the range i.e. 8590.92/2 = 4295.46

We will add this error in the mean to our mean i.e. 2983.05 ± 4295.46 s

b) Area of the table A = L * W

L = 135 ± 1 cm, W = 57 ± 0.5 cm, L = 1 cm, W = 0.5 cm

So, A = L *W = 135 * 57 = 7695 cm²

We have error in area A is A

A/A = ± [(L/L) + (W/W)]

          = ± [(1/135) + (0.5/57)]

          = ± [0.007 + 0.009]

So, A = ±(0.016) * A

i.e. A = ± 0.016 * 7695 = ±123.12

Therefore area A = 7695 ± 120 cm² (Answer)           

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.