Asecurity light is being installed outside a loading dock. the light is mounted 20feet above the ground. the light must be placed in an angle so that it will illuminate the end of the parking lot. if the end of the parking lot is 100 feet from the loading dock, what should be the angle of depression of the light

Step-by-step explanation:

Let x represent angle of depression.

We have been given that a security light is being installed outside a loading dock. The light is mounted 20 feet above the ground. The end of the parking lot is 100 feet from the loading dock.

We can see that security light, parking lot and angle of depression forms a right triangle with respect to ground. The side that is 20 ft is the opposite side and side 100 ft is adjacent to angle of depression.  

Now, we will use inverse tangent or arctan to solve for x.

Therefore, the angle of the depression of the light is approximately 11.31 degrees.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

well if he bought them at 5 cedis each sold them at 7 cedis each then he would make a profit of 2 cedis of each book. if in total he made 150 cedis profit that means he sold 75 books. he lost 12 books at the start so that means he originally bought a total of 82 books.

step-by-step explanation: