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
There is no change in the weight of either
Air has no influence on the weight of an object unless the object is accelerating down through an air column. The air resistance to the free fall of the object may have an influence on its instantaneous weight of the object.
Explanation:
The weight of an object is influenced by the mass of the object and gravity. Gravity also has influence in a vacuum. Therefore the wooden and iron piece will weight the same in a vacuum as in the air.
Learn More:
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Kinda but a football field is 100 yds. (91.4m), so not really because a Wide Receiver doesn't usually run down the entire field in one play. Also they have to pace themselves because they have to run more than one play in a whole game.