Answer:
a) E = ρ / e0
b) E = ρ*a / (e0 * r)
c) E = 0
Explanation:
Because of the geometry, the electric field lines will all have a radial direction.
Using Gauss law
Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.
We can make this cylinder of different radii to evaluate the electric field at different points.
Then:
A = 2*π*r (area of cylinder per unit of length)
Q/e0 = 2*π*r*E
E = Q / (2*π*e0*r)
Where Q is the charge contained inside the cylinder.
Inside the cable core:
There is a uniform charge density ρ
Q(r) = ρ * 2*π*r
Then
E = ρ * 2*π*r / (2*π*e0*r)
E = ρ / e0 (electric field is constant inside the charged cylinder.
Between ther inner cilinder and the tube:
Q = ρ * 2*π*a
E = ρ * 2*π*a / (2*π*e0*r)
E = ρ*a / (e0 * r)
Outside the tube, the charges of the core cancel each other.
E=0
Answer:
The correct answer is part 'c' 160 dB
Explanation:
When noise levels of different intensities are superimposed the resultant intensity is given by the equation
where,
is the intensity of a general sound level
Since we have 10000 fans each producing sound of 80dB thus the resultant intensity is given using the above formula as
30 beats/1min = 30 beats/60 sec = .5 beat/sec = 1/2 Hz
740 Hz - 1/2 Hz = 739.5 Hz
Answer - c. 739.5 Hz
The light bulb doesn't do any "work" ... it doesn't exert force,
or maintain a force through a distance. It does, however,
dissipate energy, in the form of heat and light.
The quantity of energy that it dissipates is ...
(4 watts) x (8 hours) x (3600 sec/hour) x (1 joule/watt-sec)
= (4 x 8 x 3600) joules = 115,200 joules .
Answer:
Explanation:
Assuming we have a two hot liquid like 70°C tea and 30°C water, the tea will cool down but not to 30°C but to an equilibrium temperature say 40°C
But in this case, the temperature of the chocolate will drop almost to the room temperature eventually,
The heat in the chocolate will even out into the room temperature and the room will get slightly warmer until the are both in equilibrium temperature. But you won't notice this little change but if you go out of the room, you might notice that change...
So this little fraction is not always notice, so we will still considered it as the room temperature
