To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Explanation:
(a) The given figure is a convex lens.
(b) In this figure, the object is placed between F and optical center of a lens. Convex lens is a converging lens. It converges the beam of light falling on it after reflection. The image is formed on the same side of the lens as the object.
The formed image is enlarged and it is virtual and erect.
(i) Type : virtual
(ii) Orientation : upright
(iii) Size : Enlarged
I uploaded the answer to
a file hosting. Here's link:
bit.ly/3gVQKw3
south = -(north)
Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)
Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)
Displacement = (4 - 2 + 5 - 5) km north
<u>Displacement = 2 km north </u>
