Answer:
The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.
Explanation:
- It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:
- Where ε, is the dielectric constant of the material that fills the space between plates.
- When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
- At the same time, the capacitance of a capacitor, by definition, is as follows:
- If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
- As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
- So, the change in the charge of the positive plate is +2.5 nC.
Answer:
E = h ν energy of electromagnetic particle
(b) has the greater frequency
-Surgen de una interacción.
-Nunca aparece una sola: son dos y simultáneas.
-Actúan sobre cuerpos diferentes: una en cada cuerpo.
-Nunca forman un par de fuerzas: tienen la misma línea de acción.
-Un cuerpo que experimenta una única interacción no está en equilibrio, pues sobre el aparece una fuerza unica que lo acelera. Para estar en equilibrio se requieren por lo menos dos interacciones.
Las mas importantes son la 2,3,4 característica
They are right the answer is A true
-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.
-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.
-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.
-- I bet it'll fit now.
-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.
