Answer:
E. There is not enough information to find the final temperature.
Explanation:
We do not the actual masses of ice and water involved in the question, so we cannot determine if the water freezes or the ice melts. So, there is not enough information to find the final temperature.
It shows the ray passing through the boundary.
Answer:
2.54 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the radial width of 13.1 μm, dr = 13.1 μm. r = 1.50 mm. We have a differential current dI. We remove the first integral by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ ⇒ dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr evaluate at r = 1.50 mm = 1.50 × 10⁻³ m and dr = 13.1 μm = 0.013 mm = 0.013 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.34 × 10 A/m³ × (1.50 × 10⁻³ m)² × 0.013 × 10⁻³ m = 2544.69 × 10⁻⁹ A = 2.54 × 10⁻⁶ A = 2.54 μA