The work done by a constant force in a rectilinear motion is given by:

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

Therefore, the total work done is 578.123 J and the answer is option E
Answer:True
Explanation: The mechanical waves need a medium through which to transport energy because the perturbation travels inside of the medium so it needs molecules that connect elastically the displacement of the wave. For example the sound mainly uses the air molecules to transport the energy also the sound can be traveled in solid materials like metal rods.
<em>Labels that belong in the marked ares X, Y & Z include;</em>
X: Curves outward
Y: Image may be smaller than object
Z: Image is always virtual
<u>Since the rays never meet, the images formed by convex mirrors are always virtual and smaller than the object, and since they are smaller, the images appear to be further than they actually are.</u>
What happens to has when it cools is ...
Step 1) They will start to form liquids, like condensation.
Step 2) As liquids cool, they will turn into solids.
Step 3) As solids cool, they become more stable and solid.
If you remember the formula for potential energy,
then this question is a piece-o-cake.
<em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>
-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ?
Compared to gravity on Earth, it's only 16.5 percent as much on the Moon.
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.