The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

Here,
= Change in height
m = mass of super heroine
g = Acceleration due to gravity
The change in height will be,

The final position of the heroin is below the ground level,

The initial height will be the zero point of our system of reference,


Replacing all this values we have,



Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J
Answer:
hydrogen chloride...........
THE ANSWER IS 16 ohms or however its spelled
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E







Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
brainly.com/question/1452612