Answer:12.8 ft/s
Explanation:
Given
Speed of hoop 
height of top 
Initial energy at bottom is

Where m=mass of hoop
I=moment of inertia of hoop
=angular velocity
for pure rolling 



Energy required to reach at top


Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

Therefore



Acceleration= velocity/time
1.1/2.7=0.41
The acceleration is .41 m/s^2
Let's list the given information. The frictional force, denoted as Ff, is equal to 0.200 N. We have to find the normal force, denoted as Fn. The relationship between Ff and Fn is written as:
Ff = μFn
where μ is the coefficient of friction
If there is no given data for μ, we can't solve this problem. Suppose μ = 0.5, then the normal force would be:
Fn = Ff/μ = 0.2/0.5 = 0.4 N
As we know that fundamental frequency is given as

here we know that

here we have
m = mass of wire = 5 g
l = length of wire = 90 cm


from above formula now



now we know that tension is due to weight of the sculpture so we will have


so its mass will be 11.76 kg
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 +
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m