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When a stone is tied to a string and whirled in a circle at constant speed, the tension in the string varies inversely as the radius of the circle. If the radius is 90 centimeters, the tension is 60 newtons. Find the tension if the radius is 40 centimeters.
If the radius is 40 centimeters, the tension is ____ newtons.

Sagot :

jacob193

Answer:

[tex]303.75\; {\rm N}[/tex].

Explanation:

Given that tension [tex]F[/tex] varies inversely with radius [tex]r[/tex] ([tex]r \ne 0[/tex]) of the path, there exists a constant [tex]k[/tex] ([tex]k \ne 0[/tex]) such that:

[tex]\displaystyle F = \frac{k}{r^{2}}[/tex].

To find the value of [tex]k[/tex], make use of the fact that [tex]F = 60\; {\rm N}[/tex] when [tex]r = 90\; {\rm cm}[/tex]:

[tex]\displaystyle (60\; {\rm N}) = \frac{k}{(90\; {\rm cm})^{2}}[/tex].

[tex]\displaystyle k = (60\; {\rm N})\, (90\; {\rm cm})^{2}[/tex].

The value of [tex]k[/tex] is constant regardless of the value of [tex]r[/tex]. Hence, when [tex]r = 40\; {\rm cm}[/tex], [tex]k[/tex] would still be equal to [tex](60\; {\rm N})\, (90\; {\rm cm})^{2}[/tex], and the value of [tex]F[/tex] would be:

[tex]\begin{aligned} F &= \frac{k}{r^{2}} \\ &= \frac{(60\; {\rm N})\, (90\; {\rm cm})^{2}}{(40\; {\rm cm})^{2}} \\ &= 303.75\; {\rm N}\end{aligned}[/tex].

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