Answer
toy begins to slide when the plane is at an angle of 36.0°
at this point the toy has just overcome the static friction


since toy begins to slide

now on comparing




as the body is moving at constant velocity acceleration will be zero
so now,



Answer:
I2>I1
Explanation:
This problem can be solved by using the parallel axis theorem. If the axis of rotation of a rigid body (with moment of inertia I1 at its center of mass) is changed, then, the new moment of inertia is gven by:

where M is the mass of the object and d is the distance of the new axis to the axis of the center of mass.
It is clear that I2 is greater than I1 by the contribution of the term Md^2.
I2>I1
hope this helps!!
The resultant force on the object is
∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N
which has a magnitude of
<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N
By Newton's second law, the acceleration has magnitude <em>a</em> such that
<em>F</em> = <em>m a</em>
10 N = (2 kg) <em>a</em>
<em>a</em> = (10 N) / (2 kg)
<em>a</em> = 5 m/s²
so the answer is B.