A turntable rotates counterclockwise at 78 rpmrpm . A speck of dust on the turntable is at 0.46 radrad at t=0t=0. What is the angle of the speck at ttt = 8.0 ss ? Your answer should be between 00 and 2π2π radrad. Express your answer in radians to two significant figures.

The answer is 2.97 rad.

Explanation:

the angular velocity will be equal to:

Angular speed = 78 rpm = 78 rotation/60 s = (78 * 2pi rad)/60 s = (13/5)*pi rad/s

in 8 seconds the angle will move = 8 * (13/5) = (104/5) * pi = 20.8 * pi

θinitial = 0.46 rad = 0.146 * pi rad

θfinal = (20.8 * pi) + (0.146 * pi) = 20.946 * pi = 10 * 2* pi + 0.946 * pi = 0.946* pi = 2.97 rad

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

(1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

is the antineutrino

We observe that:

In this problem, the original nucles that we are considering is iodine-131, which is

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be: