Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
D
Giddy UP!!!!!!!!!!!!!!!!!!!!!
Answer:
Explanation:
Using the first law of thermodynamics:
Where 
is the change in the internal energy of the system, in this case 
, 
is the heat tranferred, and 
is the work, 
with a negative sign since the work is done by the system.
From the previous equation we solve for heat, because it is the unknown variable in this problem
And replacing the known values:
The negative sign shows us that the heat is tranferred from the system into the surroundings.
They are similar because they are all colors in the spectrum and they are different because you cant seperate primary colors but you can seperate secondary
