Respuesta :
distance = 523 cm
Explanation:
( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) The definition of frequency is the number of rotations per second .
Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz
( c ) The tangential speed is v = angular velocity x radius of rotation
The angular velocity ω = 2π x n , where n is the number of rotations per second
Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec
The linear velocity = angular velocity x distance from center of record
Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec
Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad
Linear displacement = angular displacement x distance from center of record
= 50π/3 x 10 = 500π/3 = 523 cm
Frequency is defined as the number of rotations per second. The rotational speed of the lady bug is 5/12 revolutions per minute.
( a) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s
= 5/12 rev/sec
( b ) frequency is defined as the number of rotations per second.
The number of rotations per second is 5/12 .
Hence, frequency = 5/12 Hz
( c ) The tangential speed is
[tex]\bold {v_{t}=\omega r}[/tex]
Where,
[tex]\bold {v_{t}}[/tex] = tangential velocity
[tex]\omega[/tex] = angular velocity
r = wheel radius
The angular velocity
[tex]\bold {\omega = 2\pi \times n}[/tex]
where,
n = number of rotations per second
So, angular velocity
[tex]\bold {\omega = 2\pi \dfrac {5}{12}}\\\\\bold {\omega = {5\pi}{6}}[/tex]
Hence, tangential speed
[tex]\bold {V_t = \dfrac {5\pi}{6\times 10} }\\\bold {V_t = \dfrac {25\pi}{3} }[/tex]
(D) Angular displacement in 20 sec
[tex]\bold {\omega \times t = \dfrac {5\pi}{6} \times 20 = \dfrac {50\pi}{3} }[/tex]
Linear displacement is the product of angular displacement and distance from center of record.
So, Linear displacement
[tex]\bold { = \dfrac {50\pi}{3} \times 10 = \dfrac {500\pi}{3} = 523 cm }[/tex]
Therefore, the linear displacement of the ladybug is 523 cm.
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