What is the common denmoniator of 3/4 2/3
1 answer:
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We have that the workdone in stretching natural length to is mathematically given as
W=9/2lbft
<h3>Workdone in stretching natural length</h3>Question Parameters:
- A force of 128 lb is required to hold a spring stretched 2 ft <em>beyond </em>its natural length.
- Stretching it from its natural length to 9 inches beyond its natural length
Generally the Hookes equation for the Force is mathematically given as
F=Kx
Where
3.2lb=2ft*k
Thereofore
k=16
Force becomes
f=kx
f=16x
Hence
W=9/2lbft
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Answer:
A = 0.65
B = 3.304
C = 6.07
Step-by-step explanation:
You have to subtract to the final result the initial one and divide it for the number of segments. Then, for every segment you add the number you found, for example:
Value of a segment:
Value of A: