I think D x=vxt because it's equation finding change of x (displacement) and using time
Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called
cavitation.
So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at

is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.
It is by gaining! I just learned this in school. Hope this helps:)
Answer:
option D
Explanation:
given,
A conductor is carrying current = 2.0 A is 0.5 mm thick
Hall voltage = 4.5 x 10-6 V
uniform magnetic field = 1.2 T
density of the charge = n =?
hall voltage =


n = 6.67 × 10²⁷ charges/m
hence the correct answer is option D