Answer:
Step-by-step explanation:
<u>Use the law of cosines to find the side AB:</u>
<u>Use the Heron's area formula next:</u>
, where s- semi perimeter
- s = 1/2[x + x + 3 +
) = 1/2 (2x + 3 + ) - s - a = 1/2 (2x + 3 + - 2x - 6) = 1/2 (
- 3) - s - b = 1/2 (2x + 3 + - 2x) = 1/2 ( + 3)
- s - c = 1/2 (2x + 3 + - 2) = 1/2 (2x + 3 - )
<u>Now</u>
- (s - a)(s - b) = 1/4 [(x²+3x+9) - 9] = 1/4 (x² + 3x)
- s(s - c) = 1/4 [(2x + 3)² - (x² + 3x + 9)] = 1/4 (3x²+ 9x) = 3/4(x² + 3x)
<u>Next</u>
- A² = 3/16(x² + 3x)(x² + 3x)
- 300 = 3/16(x² + 3x)²
- 1600 = (x² + 3x)²
- x² + 3x = 40
<u>Substitute this into the first equation:</u>
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
Learn more about spring constant:
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Answer:
20 because 40 is a positive and spent is negative
Step-by-step explanation:
The answer is C.) A+2/5B.
Adding equation A to 2/5B will result in canceling the x terms, i.e. 2/5x + 2/5(-x) = 0.
Answer:
The answer is x=5 and x=−5
x² - 25 = 0
(x-5)(x+5)=0
x²-5x+5x-25=0
x-5=0 x+5=0
x=5 or x=-5
