Answer:
b ac power source
Explanation:
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Answer:She would need to first know the weight of the sculpture and what she is going to move it with then she will need to use newton's second law to calculate the amount of force needed to move it
Explanation: I just did the assignment on edgunity
Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
Answer : Height, h = 20.4 m
Explanation :
It is given that,
Mass of an object, m = 500 g = 0.5 kg
Gravitational potential energy, PE = 100 J
The Gravitational potential energy is the energy which is possessed due to the height and gravity of an object. It is given as :
PE = m g h
where,
h is the height of the cliff.

h = 20.40 m
So, the height of the cliff is 20.4 m.