<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>
Answer:
P(x < 5) = 0.70
Step-by-step explanation:
Note: The area under a probability "curve" must be = to 1.
Finding the sub-area representing x < 5 immediately yields the desired probability.
Draw a dashed, vertical line through x = 5. The resulting area, on the left, is a trapezoid. The area of a trapezoid is equal to:
(average length)·(width, which here is:
2 + 5
----------- · 0.02 = (7/2)(0.2) = 0.70
2
Thus, P(x < 5) = 0.70
Answer:
14. congruent means "THE SAME" so it will be "A" because there all the same letter.
15. D
Answer:
∠x = 67°
Step-by-step explanation:
∠x = 67° because they both are inscribed angles of the same arc
