Ok i finally got it to work so uh heres the pic thag goes with my previous question​

The surface area is 644 cm^2.

Step-by-step explanation:

The first thing I did was to find the SA of the top figure (excluding the bottom face, I'll get to that later). The dimensions of the figure has a square base with sides measuring 3 cm and a height of 8 cm. So then, I solved the the SA of the exposed faces and did the following:

Green Figure SA = (3 x 3) + 4(8 x 3)

= (9) + 4(24)

= (9) + (96)

= 105 cm^2

Now we get to the red figure. The reason why I excluded the bottom face of the green figure was because it was reducing the SA of the top face of the red figure. Knowing that the figure has a width of 10 cm and a length of 12 cm, I did the following:

Red Figure Top Face = 10(12) - 3(3)

= 120 - 9

= 111 cm^2

Then the next step is to solve for the SA of the entire red figure. Using what I've got for the top face, I did the following:

Red Figure SA = 111 + 2(7 x 10) + 2(7 x 12) + 10(12)

= 111 + 2(70) + 2(84) + (120)

= 111 + (140) + (168) + 120

= 539 cm^2

The last step is to just add the two SA's.

SA = 105 + 539

SA = 644 cm^2

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