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:
the answer is 288 because 72 would be one fourth or 25% of 288. 72 plus 72 plus 72 plus 72 equals 288 equals 100%
4a + 6b = 10
2a - 4b = 12...multiply by -2
----------------
4a + 6b = 10
-4a + 8b = - 24 (result of multiplying by -2)
------------------add
14b = - 14
b = -14/14
b = -1
2a - 4b = 12
2a - 4(-1) = 12
2a + 4 = 12
2a = 12 - 4
2a = 8
a = 8/2
a = 4
so 12a = 12(4) = 48 <==
Answer:
Step-by-step explanation:
-5x-4y=-4 ...(1)
25x+ 20y = 20 ....(2)
multiply (1) by -5 you have : 25x+ 20y = 20 ....(2) same equation , same line
so : The system has infinitely solutions.
