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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It can be, but set also means many other things.
Answer:
Explanation:
potential energy of compressed spring
= 1/2 k d²
= 1/2 x 730 d²
= 365 d²
This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .
Kinetic energy after crossing the rough patch
= 1/2 x 1.2 x 2.3²
= 3.174 J
Loss of energy
= 365 d² - 3.174
This loss is due to negative work done by frictional force
work done by friction = friction force x width of patch
= μmg d , μ = coefficient of friction , m is mass of block , d is width of patch
= .44 x 1.2 x 9.8 x .05
= .2587 J
365 d² - 3.174 = .2587
365 d² = 3.4327
d² = 3.4327 / 365
= .0094
d = .097 m
= 9.7 cm
If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .
I want to say that they will be primarily flat but I honestly don't know
I feel like it could be A
