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:
84
Step-by-step explanation:
Since we are given line OP perpendicular to line DR, then angles PDR and ODR are right angles.
Angles PDA and ADR are complementary.
Angles ODU and UDR are complementary.
Angles ADR and UDR are given as congruent.
We can conclude that angles PDA and ODU are congruent.
By AA Similarity, triangles APD and UMD are similar.
DP/DM = PA/MU
3.75/10 = 4.5/MU
3.75MU = 10 × 4.5
MU = 12 (altitude of triangle DUO)
OD = OM + MD = 4 + 10 = 14 (base of triangle DUO)
area = base × height / 2
area = 14 × 12 / 2
area = 84
Answer:
The final cost is 28249*(1+3.49/100)^5=33534.74
Answer:
FUNCTIONNNNN
Step-by-step explanation:
X+6 less than or equal to 10
X+14 greater than or equal to 18
