Answer :
Final answer:
The distance between the centers of the two 99 kg lead spheres is found to be 130.28 m. This was calculated using the Pythagorean theorem taking into account the length of the cables and the distance between the points of anchorage.
Explanation:
The question refers to two lead spheres that are hanging from cables and we need to find the distance between the centers of these spheres. This involves a little bit of trigonometric principles, particularly the Pythagorean theorem which states the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides in a right triangle. Considering the triangle formed by the line connecting the spheres’ center and the lines formed by the spheres to their respective top points of anchoring, we have:
Hypotenuse (c) = 130.3 m (Length of the cable),
Side (b) = 1.02 m / 2 = 0.51 m (Half the distance between the two points of anchoring as each sphere forms a triangle with the point of anchoring),
Side (a) = the distance between spheres
Then, by Pythagorean theorem, a= sqrt(c² - b²) = sqrt((130.3)² - (0.51)²) = 130.28 m (Distance between the centers of the two spheres)
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Answer:
the distance between the centers of the spheres is 1.72 × 10⁻⁷m
Explanation:
The mass of the two spheres ,m = 99kg
The length of the cable is L= 130.3 m
The gravitational force
[tex]F = \frac{Gm^2}{r^2} = mg\tan\theta[/tex]
[tex]\tan\theta=(\frac{Gm}{gr^2} )[/tex]
Distance change on each side is [tex]L \sin\theta[/tex]
so, the total distance is ,S = [tex]2L\sin\theta[/tex]
For small angle we have,
[tex]\sin\theta=\tan\theta=\theta[/tex]
The above equation can be written as
[tex]S = 2L\sin\theta=2l\theta[/tex]
[tex]S = 2L(\frac{Gm}{gr^2}) \\\\= 2(130.3)(\frac{(6.67\times10^-^1^1)(99)}{(9.8\times1.02)} )\\\\= 1.72\times10^-^7m[/tex]
So, the distance between the centers of the spheres is 1.72 × 10⁻⁷m
