There. That is better.
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Refer to the diagram shown below.
The suspended wire is in the shape of a parabola defined by the equation
y = ax²
where a = a positive constant.
The derivative of y with respect to x is y' = 2ax.
The vertex is at (0,0) and the line of symmetry is x = 0.
The suspended length is 41 ft, therefore half the suspended length is 20.5 ft.
The length between x = 0 and x = 20 is given by
![\int _{0}^{20} \sqrt{1+[y'(x)]^{2}} \, dx = \int_{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx =20.5](https://tex.z-dn.net/?f=%5Cint%20_%7B0%7D%5E%7B20%7D%20%5Csqrt%7B1%2B%5By%27%28x%29%5D%5E%7B2%7D%7D%20%5C%2C%20dx%20%3D%20%5Cint_%7B0%7D%5E%7B20%7D%20%5Csqrt%7B1%2B4a%5E%7B2%7Dx%5E%7B2%7D%7D%20%5C%2C%20dx%20%3D20.5)
Because we do not know the value of a, we shall find it numerically.
Define the function
The plot for f(a) versus a yields an approximate solution (from Matlab) of a = 0.01 (shown in the figure).
Therefore
y = 0.01x²
When x = 20 ft, h = 0.01(400) = 4 ft
Because the vertex of the parabola is 19 ft above ground, the support points for the wire are 19 + h = 23 ft above ground.
Answer: 23.00 ft
The suspended wire is in the shape of a parabola defined by the equation
y = ax²
where a = a positive constant.
The derivative of y with respect to x is y' = 2ax.
The vertex is at (0,0) and the line of symmetry is x = 0.
The suspended length is 41 ft, therefore half the suspended length is 20.5 ft.
The length between x = 0 and x = 20 is given by
Because we do not know the value of a, we shall find it numerically.
Define the function
The plot for f(a) versus a yields an approximate solution (from Matlab) of a = 0.01 (shown in the figure).
Therefore
y = 0.01x²
When x = 20 ft, h = 0.01(400) = 4 ft
Because the vertex of the parabola is 19 ft above ground, the support points for the wire are 19 + h = 23 ft above ground.
Answer: 23.00 ft
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;
where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;