You are a researcher on the first manned expedition to Proxima Centauri b! One o
ID: 2039978 • Letter: Y
Question
You are a researcher on the first manned expedition to Proxima Centauri b! One of your tasks is to verify the local value of& You possess a long thin wire labeled 1.00 gm and a 1.25 kg mass. The chronometer in your lab is quite accurate, but Supply forgot to pack a meter stick. Undeterred, you go about your task by finding the midpoint of the wire by folding it in half. You attach one end to the laboratory wall, stretch it horizontally over a pulley at the midpoint of the wire, and then tie the mass to the hanging end. By vibrating the wire and measuring the time, you find that the wire's 2nd harmonic frequency is 100 Hz. Next, with the 1.25 kg mass still attached, you hang the other end from the ceiling to make a pendulum. You find that the pendulum requires 314 seconds to complete 100 oscillations. Using your trusty TI-800,000g you get to work. What value of g will you report back to your superiors? 1) A 5.00 mm-diameter proton beam carries a total current of 1.5 mA. The current density in the proton beam, which increases with distance from the center, is given by J radius of the beam and Jedge is the current density at the edge. 2) Jedge(r/R), where R is the a. b. How many protons per second are delivered by this proton beam? Determine the value of Jedge 3) What is the escape speed of an electron launched from the surface of a 1.0 cm-diameter glass sphere that has been charged to 10 nC? 4) You've been given a 1.0 g piece of aluminum and asked to make a wire, using all of the aluminum, which will dissipate 7.5W when connected to a 1.5V battery. What length and diameter will you choose for your wire?Explanation / Answer
3.
Q = charge on the sphere = 10 x 10-9 C
q = charge on electron = 1.6 x 10-19 C
r = distance = 0.5 cm = 0.005 m
v = speed
m = mass = 9.1 x 10-31 kg
Using conservation of energy
kinetic energy = electric potential energy
(0.5) m v2 = k Q q/r
(0.5) (9.1 x 10-31) v2 = (9 x 109) (10 x 10-9) (1.6 x 10-19)/0.005
v = 7.96 x 107 m/s
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