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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Table of the graph:
x: <em>
</em>
1 2 3
y: 5 25 125
Average Rate of Change =
Section A = 25-5/2-1 =20/1 =20
Section B = 125 - 25/ 3-2 = 100/1 = 100
So, Section B is 5 times greater than A.
Section B is greater because the slope of two points is greater than points in Section A.
Answer: A and B are the legs and C is the hypotenuse
Answer:
A.2304
B.5625
C.6776
Step-by-step explanation:
The easiest way to show you is by using calculator. If you can't, try using Khan Academy if you want.
<span>12.3
Volume function: v(x) = ((18-x)(x-1)^2)/(4pi)
Since the perimeter of the piece of sheet metal is 36, the height of the tube created will be 36/2 - x = 18-x.
The volume of the tube will be the area of the cross section multiplied by the height. The area of the cross section will be pi r^2 and r will be (x-1)/(2pi). So the volume of the tube is
v(x) = (18-x)pi((x-1)/(2pi))^2
v(x) = (18-x)pi((x-1)^2/(4pi^2))
v(x) = ((18-x)(x-1)^2)/(4pi)
The maximum volume will happen when the value of the first derivative is zero. So calculate the first derivative:
v'(x) = (x-1)(3x - 37) / (4pi)
Convert to quadratic equation.
(3x^2 - 40x + 37)/(4pi) = 0
3/(4pi)x^2 - (10/pi)x + 37/(4pi) = 0
Now calculate the roots using the quadratic formula with a = 3/(4pi), b = -10/pi, and c = 37/(4pi)
The roots occur at x = 1 and x = 12 1/3. There are the points where the slope of the volume equation is zero. The root of 1 happens just as the volume of the tube is 0. So the root of 12 1/3 is the value you want where the volume of the tube is maximized. So the answer to the nearest tenth is 12.3</span>
