2. With regard to the pH scale, a solution with a pH
1 answer:
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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
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v =
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s