Consider an electrophysiology experiment in which a 0.200 ~mm0.200 mm diameter round silver wire is used as a stimulation electr
ode. In this experiment, a current of 4.69~ mA4.69 mA is applied through the wire. The density of free electrons in silver is 5.86 \times 10^{28}~ \frac{\text{electrons}}{m^3}5.86×10 28 m 3 electrons . Calculate the magnitude of the drift velocity of the electrons in the silver wire under these conditions. Enter your answer in units of \frac{\text{meters}}{\text{second}} second meters using calculator scientific notation (e.g., 1.23e-41.23e−4 for 0.0001230.000123).
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Answer:8 m/s
Explanation:
Given
kinetic Energy of
initially is at rest and let say
is moving with velocity u
kinetic Energy of
In Completely inelastic collision both mass stick together and move with common velocity
Suppose v is the common velocity
therefore Final velocity with which both blocks moves is 1 m/s