woah Flashcards
(17 cards)
1.2
2.75 x 0.42 = T(pq)
1.3
0.170
1.4
S = ut + 0.5x at^2
a = 2s/t^2 = (2 x 0.17)/1.2^2
1.5
Resolution of the graph is 2mm, uncertainty is half the resolution, as its a reading, multiply by 2, gets uncertainty of +-2mm
1.6
Uncertainty of s = (2/170) x 100 = 1.18%
uncertainty of T = 0.46 x 2
Add together gets you 2.1
1.7
Resultant force will be lower as they have ignored the force of friction, so the value of a will be an underestimate as it has been slowed down due to friction, creating an underestimate for g
2.1
use given formula to find a value for f, should be about 3 x 10^4
2.2
Use the curve of best fit, should be from 58-64, find a 62 for safety
2.3
Desired figure is WAY off the graph, thus requires significant extrapolation, reduce the force or increase the value of D to be able to find B for lead
2.4
Digital vernier callipers
2.5
Because the percentage uncertainty of h > than that of d due to the size of it
3.2
Use of 2T - W = ma (modified NL2) and solve for a
3.3
Max Stress = max t/area = (1.6 x 10^6)/(9.6 x 10^-3) = 1.67 x 10^8
Breaking/max = (8.9 x 10^6)/(1.67 x 10^8) gets you ratio of roughly 5x larger, so yes, is safe
3.4
Attempt to find area under graph in that area.
Should lead to 250 x 3.6 x 10^6 joules
Therefore energy stored is insufficient
4.1
Constructive interference creates a bright fringe,
so path difference of 2d + phase difference due to the reflection at A must be approximately equal to the light reflected at a and the light reflected at O must combine such that the difference between them is n lambda.
Reflection at a changes it by lambda/2 so path diff must be n + 1/2 lambda
4.2
s = 5.0mm / 11
t = (lambda x L)/2s = (590 x 10^-9 x 6.0 x 10^-2)/(2 x 0.45 x 10^-3)
= 3.9 x 10^-7
4.3
As c = f x lambda, wavelength will decrease due to change in refractive index, and as s is directly proportional to lambda as s = (lambda x L)/(2t) so s will decrease
