Answer:
a). 53.78 m/s
b) 52.38 m/s
c) -75.58 m
Explanation:
See attachment for calculation
In the c part, The negative distance is telling us that the project went below the lunch point.
Answer:
C.<u>ten</u><u> </u><u>times</u><u> </u><u>the</u><u> </u><u>intensity</u><u>.</u>
The De Broglie wavelength of the electron is

And we can use De Broglie's relationship to find its momentum:

Given

, with m being the electron mass and v its velocity, we can find the electron's velocity:

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
Answer:
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