Use your line graph from Part II to interpolate, or estimate, atomic radius of Tin (Sn). (1 point) Answer: By using the given line to interpolate the atomic radius of tin, it can be found that its radius is around 149pm. 5. Tin’s actual atomic radius is 140 pm. . Use the math equation below to determine the percent error of your estimated value. (2 points) Percent error = × 100 This equation means that you subtract to find the difference between the actual value (given in this problem) and the experimental value (estimated from the graph).
x and c-x, then the diagram looks as follows: The perpendicular has divided the triangle into two right-angled triangles. Now for any right-angle triangle, according to Pythagorean Theorem, [pic] = [pic] + [pic] If Pythagorean is applied to the right-angled triangles in the above triangle, then in the case of left right-angle triangle in the above diagram, it would give us the equation [pic] = [pic] + [pic] where ‘a’ = hypotenuse and ‘h’ = height/perpendicular and ‘x’ = base. Re-writing it, the equation would become which we will call Eq. A [pic] = [pic] - [pic] ---------------------( Eq. A Similarly, for the right angle triangle on the right half to triangle ABC, [pic] = [pic] + [pic] where ‘b’ = hypotenuse, ‘h’ = height/perpendicular and ‘c-x’ = base.
It then has a right triangle, an triangle which has a one 90-degree angle. An example of an right angle could be seen as the corner of a piece of paper. The hypotenuse will be the longest side of the triangle. “The Pythagorean Theorem states that the square of the hypotenuse in right triangle is equal to the sum of the squares of the other two sides. This is expressed as a^2 + b^2 = c^2.” (Weisstien, 2011) 13 40 Pythagorean triangles are right triangles in which all three sides are integers.
Week 5 Law of Sines and Cosines 1. Law of Sines: An equation to the length of all sides on a triangle and to the sines of its angles. 2. Law of Cosines: Related to the lengths of the sides of a plane triangle to the cosine of one of the angles. 3.
The circles located on the inside are called inscribed circles, inscribed circles should touch each side of the triangle at a single point. The circles on the out sides are called circumscribed circles, the circumscribed circle touches each vertex of the triangle. There are also things called the incenter and the circumcenter. The incenter is the center of the inscribed circle and the circumcenter is the center of a circumscribed circle,the relation between a circle and a triangle is that circles can be located inside and outside a triangle. (5 points) |Score | | | 2.
Current as a function of time: Based upon the given information, we will be determining the Amplitude (A), period (and thus, B), and the phase shift (C). Given information states there is not a phase shift, therefore the value of D is zero and does not need to be recorded throughout the problem. The Amplitude (A) or maximum for the current has been given: 5 Amps. This indicates that the range will span from -5 to 5 on the y-axis. It is also given that at t=zero (time=zero) the current is equal to 5 Amps, meaning there is a negative phase shift (or phase shift to the left).
Notice that the angle of incidence at the first surface is equal to the angle of refraction at the second surface. Prove algebraically (using Snell’s Law twice) why this is true. n1sin θ1=n2sin θ2 n2sin θ2=n1sin θ3 n1sin θ1=n1sin θ3 sin θ1=sin θ3 5. Do different colors refract at different angles? No, difference colors do not refract at different angles.
5.Using complete sentences, describe to the Martians how to find the inverse of your function. Q5 answer: To find the inverse is just switching out 'y' and 'x', then solving for 'y'. So you would take the function f(x) = 2x + 5, put the 'y' in place of where the f(x) is. You now have y = 2x + 5. Switch out 'y' and 'x'.
We calculated the average time and standard deviation using Excel. We used Newton’s 2nd law to calculate the period, the stopper mass, the radius, and the acceleration due to gravity. We drew a confidence interval to show that our measurements were inconsistent. | Time (T) [s] | Texp (T/# of revolutions(10)) [s] | 1 | 5.42 s | 0.547 s | 2 | 5.27 s | 0.527 s | 3 | 4.86 s | 0.486 s | 4 | 5.28 s | 0.528 s | 5 | 5.05 s | 0.505 s | 6 | 5.33 s | 0.533 s | 7 | 4.92 s | 0.492 s | 8 | 5.29 s | 0.529 s | 9 | 5.16 s | 0.516 s | 10 | 5.11 s | 0.511 s | 11 | 5.16 s | 0.516 s | 12 | 5.20 s | 0.520 s | 13 | 5.21 s | 0.521 s | 14 | 5.13 s | 0.513 s | 15 | 5.31 s | 0.531 s | | | | Rubber stopper mass [mr] = 8.2 gRadius of circle [r] = 0.4633 mWeight mass = [m] = 50.1 g | | Tavg = 0.518 sσT = .0152 sTexpt = 0.518 s ± .0152 s | | | Tthy = .5541 s V = 5.255 m/s2 | Our average time was approximately 2.374 standard deviations from the theoretical time. In analysis, we consider our results not consistent with the theoretical time.
The first class in a relative frequency table is 50–59 and the corresponding relative frequency is 0.2. What does the 0.2 value indicate? Answer: 0.2 is equal to 1/5 or 20%, 0.2 indicates 20% of the data values are in this particular interval. 3. When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6.