3,89,988 cm/min is the linear velocity
Given,
Diameter of CD = 12 cm
So, Radius of CD = 6 cm
CD is spinning at 10350 rev/min
Firstly , convert rev/min into rad/min
1 rev = 2π radians
10350 rev/min = 10350 × 2π
= 64998 rad/min
Formula used,
where,
is the Linear velocity
is the radius
is the angular velocity
= 6 cm × 64998rad/min
= 3,89,988 cm/min
Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.
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Answer:
speed and time are Vf = 4.43 m/s and t = 0.45 s
Explanation:
This is a problem of free fall, we have the equations of kinematics
Vf² = Vo² + 2g x
As the object is released the initial velocity is zero, let's look at the final velocity with the equation
Vf = √( 2 g X)
Vf = √(2 9.8 1)
Vf = 4.43 m/s
This is the speed with which it reaches the ground
Having the final speed we can find the time
Vf = Vo + g t
t = Vf / g
t = 4.43 / 9.8
t = 0.45 s
This is the time of fall of the body to touch the ground
B
V= f x lambda
V= 5m/s
F = 10hz
Lambda = ?
5 = 10 x lamba
5 /10 = lambda
Wavelength =0.5
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The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
