Answer:
0.0061 J
Explanation:
Parameters given:
Number of turns, N = 111
Radius of turn, r = 2.11 cm = 0.0211 m
Resistance, R = 14.1 ohms
Time taken, t = 0.125 s
Initial magnetic field, Bin = 0.669 T
Final magnetic field, Bfin = 0 T
The energy dissipated in the resistor is given as:
E = P * t
Where P = Power dissipated in the resistor
Power, P, is given as:
P = V² / R
Hence, energy will be:
E = (V² * t) / R
To find the induced voltage (EMF), V:
EMF = [-(Bfin - Bin) * N * A] / t
A is Area of coil
EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125
EMF = 0.83 V
Hence, the energy dissipated will be:
E = (0.83² * 0.125) / 14.1
E = 0.0061 J
It means lithium ion.
HOPE I HELPED
Rutherford is credited for discovering the nucleus of the atom. This occurred after he performed his famous gold foil experiment - by exposing a source of alpha particles to a gold foil to see how the particles deflected against the atoms of the foil, he discovered that the observed pattern only made sense when he considered the idea of a nucleus in this situation.
Answer:
see below
Explanation:
B is the activation energy
A is the enthalpy change which is negative
C is the product of oxygen and ethanol which is carbon dioxide and water
hope this helps, please mark
Answer:

Explanation:
<u>Given:</u>
- Diameter of each disc, D = 2.4 cm.
- Distance between the discs, d = 1.0 mm.
- Charges on the discs, q = ±11 nC =

The surface area of each of the disc is given by

For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by

<em>where</em>,
= surface charge density of the disc =
.
= electrical permittivity of free space = [tex\rm ]9\tiimes 10^9\ Nm^2/C^2[/tex].
The electric field strength between the discs due to negative disc is given by

Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.
The electric field strength between the discs due to positive disc is given by

The direction of this electric field is away from the positive disc, i.e., towards negative disc.
The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by
