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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15x + 4 would be the algebraic expression
Answer:
f = 10, g = 4.8 cm
Step-by-step explanation:
Area of ABCE = 60 cm²
Area of ABCD = 48 cm²
So,
Area of ADE = 60-48
=> 12 cm²
Area of ADE = 
<u><em>Where Area = 12 cm², Base = 4 cm</em></u>
12 = 
Height = 12-2
Height = 10 cm
Where Height is AD
So, AD = 10 cm
Also, <u><em>AD is parallel and equal to BC</em></u>
<u><em></em></u>
So,
f = 10 cm
<u><em>Now, Finding g</em></u>
Area of ABCD = 
<u><em>Where Area = 48 cm², Base = 10 cm</em></u>
48 = 10 * Height
Height = 48/10
Height = 4.8 cm
Whereas, Height is g
So, g = 4.8 cm
Answer:
11:52...
Step-by-step explanation:
11:52. It could be any time except the time 11:37 from now, since its NOT going to be.
Hope this helps!
