A line with a positive slope is parallel to one of the lines shown. What is its slope? Use the graph shown below.

Step-by-step explanation:

Only the blue line has a positive slope.  So the new line is parallel to the blue line.

Clearly, the blue line goes through the point (0, 3) (this is the y-intercept).  Follow this blue line upward and to the right until it intersects the horizontal line at the top of the diagram.  We must estimate the x-value of this point of intersection.  A good approximation would be 5 1/3.

Now estimate the slope of the blue line.  Starting at (0, 3), we move 5 1/3 units to the right (this is the "change in x," or "run")  We end up at (5 1/3, 3).  

Now, from the horizontal line y = 3 upward to the horizontal line at the very top of the graph, is 5 units.  This is the "rise."

Thus, the approximate value of the slope here is

                              5               5             15

m = rise / run = = = = m

                          5 1/3           16/3             16

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

x= 3^y

step-by-step explanation: