The buoyant force on an object fully submerged in a liquid depends on (select all that apply) answer 1. the object’s mass. 2. the object’s volume. 3. the density of the liquid. 4. the mass of the liquid

2. The object's volume.
 3. The density of the liquid.  
Remember what the buoyant force is. It's the lifting force caused by the displacement of a fluid. I'm using the word fluid because it can be either a liquid or gas. For instance a helium balloon floats due to the buoyant force exceeding the mass of the balloon. So let's look at the options and see what's correct. 
 1. Object's mass
 * This doesn't affect the buoyant force directly. It can have an effect if the object's mass is lower than the buoyant force being exerted. Think of a boat as an example. The boat is floating on the top of the water. If cargo is loaded into the boat, the boat sinks further into the water until the increased buoyant force matches the increased mass of the boat. But if the density of the object exceeds the density of the fluid, then increasing the mass of the object will not affect the buoyant force. So this is a bad choice. 
 2. The object's volume.
 * Yes, this directly affects the buoyant force. So this is a good choice. 
 3. The density of the liquid.
 * Yes, this directly affects the buoyant force. You can drop a piece of iron into water and it will sink. You could also drop that same piece of iron into mercury and it will float. The reason is that mercury has a much higher density than water. So this is a good choice. 
 4. Mass of the liquid
 * No. Do not mistake mass for density. As a mental exercise, imagine the buoyant force on a small piece of metal dropped into a swimming pool. Now imagine the buoyant force on that same piece of metal dropped into a lake. In both cases, the buoyant force is the same, yet the lake has a far greater mass of water than the swimming pool. So this is a bad choice.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j