<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>
B
I need 20 characters lol
3 1/4 ounces = 3.25 Ounces.
3.25 x 5 = 16.25 ounces (16 1/4 ounces)
Answer:
Assuming the problem asks to rotate counterclockwise: E' (0,0) F' (0,5) G' (-4,5) H' (-4,0)
If it says rotate clockwise then your points will be: E' (0,0) F' (0,-5) G' (4,-5) H' (4,0)
Step-by-step explanation:
Translate the points first by subtracting 3 from the x-coordinates of each, then subtracting 3 from the y-coordinates of each. Next, to rotate them counterclockwise about the origin 90 degrees, switch the x and y coordinates and the sign of the resulting x-coordinate. To rotate clockwise 90 degrees switch the x and y still but change the sign of the resulting y coordinate.
