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:
The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.
Explanation:
It is given that,
A high jumper jumps over a bar that is 2 m above the mat, h = 2 m
We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :
g is acceleration due to gravity
v = 6.26 m/s
So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.