6. Calculate the total magnification of an object viewed under a) the low power objective and b) the high power objective?
Supply check Attendance Remember to sign up for UT Quest and submit your “Test Paper” to turnitin.com DUE TOMORROW THIS IS THE EASIEST 100 YOU WILL EVER GET!
August 26, 2010
A free body diagram is the visual representation of force vectors W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = ma net F net = ma net
#1 & 2 together 10 minutes to complete #3 & 6 Groups 1, 4, 7 #4 & 7 Groups 2, 5, 8 #5 & 8 Groups 3, 6, 9
W OE = — mg N OS = + mg ∑ F y = N OS - W OE = 0 ∑ F y = mg - mg = 0
W OE = — mg N OS ∑ F ynet = N OS - W OE = ma net
W OE = — mg N OS = + mg F OA = ma ∑ F y = N OS - W OE = 0 = mg - mg = 0 ∑ F x = ma net
W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = ma net F net = ma net
W OE = — mg N OS = + mg F OA = maf OS = µN ∑ F y = mg - mg = 0 ∑ F x = ma - µN = — ma net F net = — ma net
T OR = + mg W OE = — mg
T OR = + mg W OE = — mg F OA = ma
T OR = + mg W OE = — mg Can we directly measure the Tension? Is it still equal to +mg? Why or why not? -x +y The x and y are just the components of the actual forces. This is why they’re drawn with dotted lines. You must ALWAYS draw components with dotted lines. F OA = ma
Trigonometry: deals with angles and sides of triangles Sine (sin ) Opposite over Hypoteneuse C osine (cos ) Adjacent over Hypoteneuse T angent (tan ) Opposite over A djacent Y = Opp. X =Adj. Hyp.
x adj y opp T S O HC A HT O A i p y o d y a p d n p p s j pn p j sin Ѳ = opp = y hyp T cos Ѳ = adj = x hyp T tan Ѳ = opp = y adj x Ѳ
X T cos Ѳ =x adj Y T sin Ѳ = y opp T S O HC A HT O A i p y o d y a p d n p p s j pn p j T sin Ѳ = y T cos Ѳ = x Ѳ
T OR = + mg W OE = — mg -x T cos Ѳ =x +y T sin Ѳ = +y ∑ F y = T sin Ѳ - mg = 0 ∑ F x = ma - T cos Ѳ = 0 F OA = ma
W OE = — mg Ѳ Ѳ T OR2 T OR1 In order for this box to just be hanging, what does the upward y component of the tension HAVE to be??