Answer:d
Explanation:
Drift velocity is given by

where
=drift velocity
I=Current
n=no of electron
Q=charge of Electron
A=cross-section
If area of cross-section decreases gradually then drift velocity will increase because drift velocity is inversely proportional to Area of cross-section
Answer:
2240.92365 m/s
Explanation:
= Mass of electron = 
= Speed of electron = 
= Neutrino has a momentum = 
M = total mass = 
In the x axis as the momentum is conserved

In the y axis

The resultant velocity is

The recoil speed of the nucleus is 2240.92365 m/s
For this case we have that by definition, the momentum is given by:

Where,
- <em>m: mass
</em>
- <em>v: speed
</em>
Therefore, replacing values we have:

From here, we clear the value of the speed:

Answer:
The magnitude of velocity is:

Answer:
0.47 m
Explanation:
= Number of vibrations = 37
= total time taken = 33 s
= time period of each vibration
frequency of vibration is given as
Hz
= distance traveled along the rope = 421 cm = 4.21 m
= time taken to travel the distance = 8 s
= speed of the wave
Speed of the wave is given as

= wavelength of the harmonic wave
wavelength of the harmonic wave is given as
