The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Work done (Wd) =?
<h3>How to determine the spring constant</h3>
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Spring constant (K) =?
F = Ke
Divide both sides by e
K = F/ e
K = 3 / 0.6
K = 5 pound/foot
Thus, the spring constant of the spring is 5 pound/foot
<h3>How to determine the work done</h3>
- Spring constant (K) = 5 pound/foot
- Extention (e) = 0.7 feet
- Work done (Wd) =?
Wd = ½Ke²
Wd = ½ × 5 × 0.7²
Wd = 2.5 × 0.49
Wd = 1.23 foot-pound
Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound
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Answer:
It has to be 16in since 256 is obviously to much and 4 and 8 are too little
Step-by-step explanation:
We are asked to solve for the incremental net income from reworking the phones.
Number of items = 1000
Production Cost = 1000*$60 = $60,000
Salvage value = 1000*$30 = $30,000
Rework Cost = 1000 *80 = $80,000
Price when resold = 1000 * $120 = $120,000
Incremental Net = $120,000 + $30,000 - $60,000 -$80,000
Incremental Net = $10,000
The answer is $10,000.
Answer:
4b -5
Step-by-step explanation:
Perform the indicated multiplication first: 4b - 24 + 19. Next, combine like terms, obtaining 4b -5 (answer)
(2x-3)(x+4) —> x= 3/2 x= -4
