Name a radioactive gas that is released from the ground
1 answer:
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Answer:
-5.9 rad/s^{2}
Explanation:
radius (r) = 30 cm = 0.3 m
mass (m) = 1.05 kg
initial speed (u) = 77 rpm
final speed (v) = 0 rpm
time (t) = 1.37 s
angular acceleration =
therefore
initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s
final speed (v) = 0 rpm = 0 rad/s
angular acceleration = = -5.9 rad/s^{2}
For this case we have that the fan spin period is given by:
Where:
C: It is the number of cycles
t: It's time
According to the data of the statement we have:
Thus, the fan spin period is
Answer:
Fan spin period is