.........,....,,,???........
Answer:
The correct answer is C. 45.5 lbs.
Explanation:
In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.
The formula for any problem involving a lever is:

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.
The parameter of the formula that you need is F_l:

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=45.5 lbs
Answer:
B. Chemical to Electrical to Radiant
Explanation:
To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.
The net height from the point where it begins to roll with an inclination of 30 degrees would be



In the case of Inertia would be given by

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is



Replacing in Energy conservation Equation we have that
Potential Energy = Kinetic Energy of Rolling Object




Therefore the correct answer is C.
Answer:
the correct one is 2. the equipotential lines must be closer together where the field has more intensity
Explanation:
The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.
In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by
V = k q / r
also the electric field and the electuary potential are related
E =
therefore the equipotential lines must be closer together where the field has more intensity
When checking the answers, the correct one is 2