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weqwewe [10]
3 years ago
7

5−16(−x+7)=1+3(2−4x) What is the value of x?

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
6 0
<span>5−16(−x+7)=1+3(2−4x)
*Distribute what's in the parantheses*
5+16x-112 = 1+6-12x
* Combine like terms*
  16x-117 = -12x+7
*Put the variable on one side*
</span>  16x-117 = -12x+7
+12x          +12x<span>
28x-117 = 7
     +117 +117
28x = 124
/28      /28
x = 4.42 *decimals*</span>
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a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
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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): 
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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: 
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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: 
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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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