-- reduce the length of a wire to 1/2 . . . cut the resistance in half
-- reduce the diameter to 1/4 . . . reduce the cross-section area by (1/4²) . . . increase the resistance by 16x .
-- R2 = (R1) · (1/2) · (16) = 8 · R1
<em>-- R2 / R1 = 8</em>
Answer:(a) 2.40 (b) horizontal distance. (c) 0.630. (d) 6.50
Explanation:that's all is talking about a speed and distance and time right
Force-a push or pull exerted on an object.
Answer:
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Explanation:
The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
A = Cross sectional Area of the capacitor
d = separation between the capacitor
So,
U = CV²/2
Substituting for C
U = ϵAV²/2d
Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵAV²/2d₁
U₁ = ϵAV²/(2(3d))
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Answer:
the jet i just took the test
and got 100
Explanation: