You are using a geiger counter to measure the activity of a radioactive substance over the course of several minutes. if the reading of 400. counts has diminished to 100. counts after 66.7 minutes , what is the half-life of this substance? express your answer with the appropriate units.

33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

twist boundary

explanation:

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.