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:
answer is image
Step-by-step explanation:
Hi
Let me help you out
101/25 is a in proper fraction.
To change the in proper fraction to a proper fraction you have to see which number is close to 101 by multiplying numbers to 25. You can also divide 101 by 25 which is 4 remainder 1.
25*4=100
100 is close to 101
the answer would be 4&<span>1/25</span>
if you like my answer and think it is helpful then please mark mine as brainliest
answer thank you
Answer:
x = 23
y = 110
Step-by-step explanation:
70 and 3x+1 are alternate exterior angles and that means they are equal
70 =3x+1
Subtract 1 from each side
69 = 3x
Divide by 3
69/3 =3x/3
23 =x
70 and y are same side exterior angles and that means they are supplementary ( add to 180)
70+y = 180
y = 180-70
y = 110
m∠1=31
m∠2=59
m∠3=90
m∠4=31
m∠5=90
hope this helps brainliest please!
