I believe the answer is free electrons
1. The problem statement, all variables and given/known data A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. 2. Relevant equations When a capacitor is charged, the electric field E, and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: ∮B⃗ ⋅dl⃗ =μ0(I+ϵ0dΦdt). You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero.
Answer Whooping Crane
An experimental population is a geologically depicted gathering of reintroduced plants or creatures that is segregated from other existing populaces of the species. It is done to protect the endangered species from getting extinct. Whooping crane is also one of those birds which were endangered and experimental population technique was used.
Answer:
p_{f} = 6 m / s
Explanation:
We can solve this exercise using conservation of momentum. For this we define a system formed by the two balls, so that the forces during the collision have been intense and the moment is preserved
Initial instant. Before the crash
p₀ = m v +0
Final moment. Right after the crash
= (m + m) v_{f}
how the moment is preserved
p₀ = p_{f}
m v = 2 m v_{f}
v_{f} = v / 2
we calculate
v_{f} = 12/2
p_{f} = 6 m / s
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by

where k is the elastic constant of the spring and

is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:

where the negative sign is given by the fact that

points in the opposite direction of the displacement of the cart, and where

therefore, the work done by the weight is