Given data:
● Angular velocity of the turntable ω = 33 rev/min
Therefore,
ω = 33 rev/min
ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s
● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m
● μs = the coefficient of static friction
Section a:
The seed is undergoes circular motion and it is been effected by centripetal acceleration.
ac = rω^2
ac = 0.078 × 3.45^2
ac = 0.9284 m/s^2
Therefore,
the centripetal acceleration of the seed is 9.274 m/s^2
Section b:
If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.
Centripetal Force = Frictional Force
mrω^2 = μsmg
μs = rω^2 /g
μs = 0.078 × 3.45^2
------------
9.81
μs = 0.09464
Thus,
the coefficient of static friction is 0.09464.
Answer:
D. Sami had low birth weight as a lamb but quickly grew into a large sheep.
Explanation:
The Sami lamb is an example of culled because it has the characteristics of quick growth. Culling is the process of separating organisms from a group of same type of organisms due to presence of desired or undesired characteristics. Sami sheep has a unique characteristic of having less weight at time of birth but having quick growth features that make to gain more weight and become large sheep so this sheep is removed from the population for its good traits.
mesophyll is the inner tissue of a leaf and it contains may chloroplasts
<span>Only the "Analysis" section includes specific data comparisons, and the only "Conclusion" section suggests further research.???</span>
