Answer:
No real solutions.
Step-by-step explanation:
729s^-5 = 729/s^5
729/s^5 = 3^2(1-s) = 9(1-s)
729=9(1-s)(s^5)
(1-s)(s^5)=81
s^5-s^6=81
Because 81 > 0 this means s^6 must be less than s^5. This is never possible as 6 is even so is always positive and is always greater in magnitude than s^5. The only time this isnt the case is when s is a decimal less than 1, but then the difference would be also less than 1 and not 81.
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:
he grows by 5 cm every year between 1999 and 2006
Step-by-step explanation:
This is a arithmetic progression problem with the formula;
T_n = a + (n - 1)d
We are told that In 1999 Daniel was 146 cm tall. He grew to be 176 cm by the year 2006.
Thus;
a = 146
d = 2006 - 1999 = 7
Thus;
176 = 146 + (7 - 1)d
176 - 146 = 6d
30 = 6d
d = 30/6
d = 5 cm
Thus, he grows by 5 cm every year between 1999 and 2006
