Answer:
[tex]0.00015\pi m/s[/tex] 0r 0.0004.71m/s
Explanation:
The speed v can be obtained through the following relationships;
[tex]v=\omega l.................(1)[/tex]
where [tex]\omega[/tex] is its angular speed and [tex]l[/tex] is the length which is also equivalent to the amplitude of motion.
Also;
[tex]\omega=\frac{2\pi}{T}..............(2)[/tex]
where T is the period of motion which is the time it takes for the second hand to make one complete revolution.
Substituting equation (2) into equation (1), we obtain the following;
[tex]v=\frac{2\pi l}{T}..............(3)[/tex]
Given;
[tex]l=4.5mm=0.0045m\\T=60.00s\\hence\\v=\frac{2\pi *0.0045}{60}..............(2)\\v=0.00015\pi m/s[/tex]
Taking [tex]\pi=3.142[/tex]
[tex]v=0.00015*3.142[/tex]
[tex]v=0,0004.71m/s[/tex]
The speed of the end of the second hand as it moves in uniform circular motion is 4.7 × 10⁻⁴m/s.
Given the data in the question;
Speed of the end of the second hand; [tex]v =\ ?[/tex]
First we determine the angular velocity of the second hand using the expression:
[tex]w = \frac{2\pi }{t}[/tex]
Where ω is angular velocity and t is time taken.
We substitute in our given value
[tex]w = \frac{2\pi }{60.00s}\\\\w = 0.1047rad/s[/tex]
Now, from Rotational kinematics:
[tex]w = \frac{v}{r}[/tex]
Where ω is the angular velocity, r is the radius and v is the velocity
We substitute in our values and solve for "v"
[tex]0.1047rad/s = \frac{v}{0.0045m} \\\\v = 0.1047rad/s * 0.0045m\\\\v = 0.00047 m/s\\\\v = 4.7*10^{-4}m/s[/tex]
Therefore, the speed of the end of the second hand as it moves in uniform circular motion is 4.7 × 10⁻⁴m/s.
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