Answer:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings
Answer:
t = 5 hr
Explanation:
Let kali moves toward east with velocity= V₁= 40 km/ h
Mat moves toward west with velocity = V₂= 50 km/hr
As Klai left one hour earlier = t₁= 1 hr
distance traveled in 1st hour = s₁ = v * t = 40 * 1 = 40 km
Remaining distance = 400 - 40 = 360 km
As they move in the opposite directions:
Relative speed= 40 + 50 = 90 km/ h
s = v * t
⇒ t = s / v
⇒ t₂ = 360 / 90
⇒ t₂ = 4 hr
Total time = t = t₁ + t₂
t = 1 hr + 4 hr
t = 5 hr
<span>A. Rocket A will travel farther horizontally than rocket B.
This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>
