Answer:
Explanation:
Efficiency of a Carnot engine = 1 – (T2/T1)
;
where T1 = source temperature and T2 = sink temperature
T2 = 10⁰C = 273+10 = 283K
The initial efficiency = 40%
So 0.4 = 1 - (283/Initial T1)
Initial T1 = 283/(1-0.4) = 471.67 K
To raise efficiency to 65%
0.65 = 1 - (283/Increased T1)
Increased T1 = 283/(1-0.65) = 808.57 K
The source temperature must be increased by 808.57 - 471.67 = 336.9 K or 336.9°C.
C is the correct answer, hope it helps
Answer:
(a) Magnitude: 14.4 N
(b) Away from the +6 µC charge
Explanation:
As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.
Let's say that a force that goes toward the +6 µC charge is positive, then:


The magnitude will be:
, away from the +6 µC charge