Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first 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 at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
Answer:
.
Explanation:
When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.
The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.
Let denote the mass of this ball. Let denote the mass of water that this ball has displaced.
Let denote the gravitational field strength. The weight of this ball would be . Likewise, the weight of water displaced would be .
For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:
.
At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:
.
Therefore:
.
.
In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let denote the density of water. The volume of water that this ball should displace would be:
.
Given that while :
.
In other words, for this ball to stay afloat, at least of the volume of this ball should be under water. Therefore, the volume of this ball should be at least .
When plates move they sometimes open small cracks that release heat from under earths plate, coming from the mantle or even the core at times, this heat allows scientists to determine where the plates are moving from and where they are moving to.