I) if the triangles are similar then the corresponding angles are congruent. this means that the angles are of the same size.
ii) If ΔABC is similar to ΔDEF, therefore; m∠A = m∠D, m∠B=m∠E, and m∠C=m∠F,
thus, if m∠A= 52, m∠D=52, and if m∠E=65, then m∠B=65, thus to get m∠C;
180- (52+65)
= 63 , therefore; m∠C= 63
iii) if two figures are similar they have the same shape and not necessarily the same size while if two figures are congruent then they have the same shape and size
Answer:
see the attachment photo!
40 it's the same angle as angle A rotation doesn't change anything
<span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:
a = omega^2*r
omega and r in terms of given data:
omega = 2*Pi/T
r = d/2
Thus:
a = 2*Pi^2*d/T^2
What forces cause this acceleration for the passenger, at either top or bottom?
At top (acceleration is downward):
Weight (m*g): downward
Normal force (Ntop): upward
Thus Newton's 2nd law reads:
m*g - Ntop = m*a
At top (acceleration is upward):
Weight (m*g): downward
Normal force (Nbottom): upward
Thus Newton's 2nd law reads:
Nbottom - m*g = m*a
Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:
Ntop = m*(g - a)
Nbottom = m*(g + a)
Substitute a:
Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)
We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)
Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)
Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;
Results:
Ntop/(m*g) = 71.64%...she feels "light"
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
