Answer:
a)ΔS₁ = - 9.9 J/K
ΔS₂ = 69 J/K
b)The entropy change for the rod = 0 J/K
c)ΔS = 59.1 J/K
Explanation:
Given that
T₁ = 699 K
T₂= 101 K
Q= 6970 J
Change in entropy given as
For 699 K:
ΔS₁ = - 9.9 J/K ( Negative because heat is leaving from the system)
For 101 K;
ΔS₂ = 69 J/K
The entropy change for the rod = 0 J/K
Entropy change for the system
ΔS = ΔS₂ + ΔS₁
ΔS = 69 -9.9 J/K
ΔS = 59.1 J/K
The magnetic force between two wires is 0.052 N which is attract each other.
We need to know about magnetic force on a current-carrying wire formula to solve this problem. The magnetic force on two wires with same direction of current is
F = μ₀ . I1 . I2 . L / ( 2π . r )
where μ₀ is vacuum permeability (4π×10‾⁷ H/m) F is the magnetic force, I is current, L is the length of wire, r is distance of 2 wires.
From the question above, we know that:
L = 25 m
r = 6 cm
I1 = I2 = 25 A
By substituting the parameter, we get
F = μ₀ . I1 . I2 . L / ( 2π . r )
F = 4π×10‾⁷ . 25 . 25 . 25 / (2π . 0.06)
F = 0.052 N
Hence, the force between two wires is 0.052 N which is attract each other.
Find more on magnetic force at: brainly.com/question/13277365
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The graph between the strength of the magnet(number of paper clips picked) and battery is approximately a straight line.
For 25 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 5 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.
For 50 coil, with increase of 1.5 V battery voltage, the electromagnet picks about 15 more clips. So, for a 7.5 V battery, it would pick about 30 paper clips.