as a pendulum swings the angle theta that it makes with the vertical changes through its swing. The force of gravity pulling on the bob is given by F=mg sin theta, where g is equal to 9.8 m/s^2. If the mass of the pendulum is 0.01kg, what is the force of pulling on the pendulum when it makes a 22.5 degree angle with the vertical?
a. sqrt2-sqrt2/2 N
b.0.049sqrt-sqrt2N
c.sqrt2-sqrt2/2 N
d.0.049sqrt2-sqrt2/2 N

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ichmmmm ichmmmm · Answer 1 · 2017-03-24T18:27:38+01:00

b. 0.049sqrt 2-sqrt2 N

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stokholm stokholm · Answer 2 · 2019-03-10T05:43:01+01:00

Answer:

Step-by-step explanation:

Alright, lets get started.

Equation is : F = mg sinΘ

We have given,

[tex]m =0.01[/tex]

[tex]g=9.8[/tex]

Θ [tex]= 25[/tex]

While plugging these above values in equation:

[tex]F= 0.01 *9.8 * sin (22.5)[/tex]

[tex]F = 0.098 * sin (22.5)[/tex]

Now, if we look unit circle, we don't have angle 22.5, so we will use half angle identity

[tex]F = 0.098 * sin (\frac{45}{2} )[/tex]

[tex]F = 0.098 * \sqrt{\frac{1-cos45}{2} }[/tex]

[tex]F = 0.098 * \sqrt{\frac{1-\frac{\sqrt{2} }{2} }{2} }[/tex]

[tex]F = 0.098 * \sqrt{\frac{2-\sqrt{2}}{4} }[/tex]

[tex]F = 0.049 * \sqrt{{2-\sqrt{2} }[/tex] : Answer

Hope it will help :)