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Natalka [10]
4 years ago
8

I'm not sure how to do this.

Mathematics
1 answer:
Mila [183]4 years ago
5 0
Your answer would be 24 zeros. The 24 above the 10 represents how many zeros will be in the standard notation form. 
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A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
AnnZ [28]
  <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>
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4 years ago
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Alika [10]

Step-by-step explanation:

hope this helps you dear friend.

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3 years ago
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expeople1 [14]

Answer:

find your answer on math papa

Step-by-step explanation:

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4 years ago
Kiley spent $20 on rides and snacks at the state fair. If x is the amount she spent on rides, and y is the amount she spent on s
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Both linear and proportional because as x increases y decreases at the same rate
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3 years ago
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igor_vitrenko [27]

Look at the picture.

The point-slope form of a line:

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