Answer:
236.3 x
C
Explanation:
Given:
B(0)=1.60T and B(t)=-1.60T
No. of turns 'N' =100
cross-sectional area 'A'= 1.2 x
m²
Resistance 'R'= 1.3Ω
According to Faraday's law, the induced emf is given by,
ℰ=-NdΦ/dt
The current given by resistance and induced emf as
I = ℰ/R
I= -NdΦ/dtR
By converting the current to differential form(the time derivative of charge), we get
= -NdΦ/dtR
dq= -N dΦ/R
The change in the flux dФ =Ф(t)-Ф(0)
therefore, dq =
(Ф(0)-Ф(t))
Also, flux is equal to the magnetic field multiplied with the area of the coil
dq = NA(B(0)-B(t))/R
dq= (100)(1.2 x
)(1.6+1.6)/1.3
dq= 236.3 x
C
Positive Work.
Negative Work.
Case of zero work done.
Displacement at an angle to the force.
Energy.
Kinetic Energy.
work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement.
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.
But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.
The Moon has nothing to do with any of this.