The answer is b fission hope this helps
Answer:
(a) 
(b) 
Solution:
As per the question:
Mass of the object, m = 1.30 kg
Length of the rod, L = 0.780 m
Angular speed, 
Now,
(a) To calculate the rotational inertia of the system about the axis of rotation:
Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

(b) To calculate the applied torque required for the system to rotate at constant speed:
Drag Force, F = 

Well the wife one is to be able to handle the nail and the sharp part is to put it in side of something like a wall
Answer:
2500 kg/m³
Explanation:
The density is ...
(107 g)/((12.2 cm)(3.7 cm)(0.95 cm)) = 107/42.883 g/cm³ ≈ 2.5 g/cm³
In kg/m³, the density is ...
(2.5 g/cm³)×(1 kg/(1000 g))×(100 cm/m)³ = 2500 kg/m³
The conversion from CGS to MKS units of density is done using the usual method of units conversion. The conversion factor multipliers each change the units without changing the value, because their numerator is equal to the denominator.
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2 significant figures are used because the block dimensions have a precision of 2 significant figures.