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:
1. Rate of change is -5 every step.
2. Rate of change is -5 feet every one hour.
3. Inital water level is 30 ft
Step-by-step explanation:
Answer:
So, cows heart best is faster than horse.
Step-by-step explanation:
Time and heartbeats are in direct proportion.
p = kx
Where p is the beats per minute and x represents time in minutes
Now, to find the value of k, substitute p = 152 and x = 4
152 = 4k
k = 152/4
k = 38
Cow: y = 65x
So, cows heart best is faster than horse
Answer:
The answer is A
Step-by-step explanation:
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