A stone is thrown vertically upward at a speed of 41.50 m/s at time t=0. A second stone is thrown upward with the same speed 3.0
1 answer:
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Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:
- where Rint = r and RL = r
- Replacing these values in I₁, we have:
- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:
- We can find the relationship between I₂, and I₁, dividing both sides, as follows:
- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.
Answer:
a)
b)
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K
Negative sign because heat is leaving.
b)
The entropy change of reservoir 101 K
c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K