In moving out of a dormitory at the end of the semester, a student does1.6×10^4jof work. in the process, his internal energy decreases by4.2×10^4jdetermine each of the following quantities (including the algebraic sign): (a)w,(b)δu, and (c)q

         W = 1.6 × 10⁴ J , ΔU = - 4.2 x 10⁴ J , Q = - 2.6 x 10⁴ J

Explanation:

given,                                                

Work done by the student          

         W = 1.6 × 10⁴ J                        

decrease in internal energy        

       ΔU = - 4.2 x 10⁴ J                  

Using formula for heat                      

  Q = W + ΔU                                        

  Q = 1.6 × 10⁴ J + ( - 4.2 x 10⁴ J)      

  Q = - 2.6 x 10⁴ J                                    

Hence,                                                              

         W = 1.6 × 10⁴ J , ΔU = - 4.2 x 10⁴ J , Q = - 2.6 x 10⁴ J

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14