This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.
<span>The key equation is going to come from Mr Planck: E=h \nu
Where h is Plancks constant; and ν is the frequency. This equation gives you the energy per photon at a given frequency. Alas, you're given wavelength, but that's easy enough to convert to frequency given the following equation:
c= lambda / nu
where c is the speed of light; λ (lambda) is the wavelength; and ν is again frequency. As soon as you know the energy of a photon with a wavelength of 550nm, you should know how many photons you would require to accumulate 10^-18J. Be careful with your units.</span>
The average current density in the wire is given by:

where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:

where r is the radius of the wire. In this problem,
, so the cross-sectional area is

and the average current density is

Answer:
Given that
V= 0.06 m³
Cv= 2.5 R= 5/2 R
T₁=500 K
P₁=1 bar
Heat addition = 15000 J
We know that heat addition at constant volume process ( rigid vessel ) given as
Q = n Cv ΔT
We know that
P V = n R T
n=PV/RT
n= (100 x 0.06)(500 x 8.314)
n=1.443 mol
So
Q = n Cv ΔT
15000 = 1.433 x 2.5 x 8.314 ( T₂-500)
T₂=1000.12 K
We know that at constant volume process
P₂/P₁=T₂/T₁
P₂/1 = 1000.21/500
P₂= 2 bar
Entropy change given as

Cp-Cv= R
Cp=7/2 R
Now by putting the values


a)ΔS= 20.79 J/K
b)
If the process is adiabatic it means that heat transfer is zero.
So
ΔS= 20.79 J/K
We know that

Process is adiabatic




Answer:
Therefore,
Current through Nichrome wire is 0.3879 Ampere.
Explanation:
Given:
Length = l = 10 meter


V = 12 Volt
To Find:
Current, I =?
Solution:
Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

Where,
R = Resistance
l = length
A = Area of cross section = πr²

Substituting the values we get




Now by Ohm's Law,

Substituting the values we get

Therefore,
Current through Nichrome wire is 0.3879 Ampere.