Quiz: Galvanic Electrolysis QUESTIONS
AND ANSWERS
1. A probe is inserted to anagen intensity and the follicle is treated with a galvanic epilator
having the settings at 0.1 ma and four seconds. The probe is then pulled 1/2 manner
and the follicle is treated again the usage of the equal setting. The modern-day density
around the probe for the half of intensity treatment is - ANSWER-Greater than the
whole anagen intensity remedy
(The more the density the greater lye is produced. A shallow insertion creates
more contemporary density. Things that could affect current density is: the scale
of probe, depth of insertion, time the modern flows, and the current depth)
2. During galvanic electrolysis with the terrible electrode inserted right into a hair follicle, the
subsequent chemicals are produced on the terrible electrode. - ANSWER-Sodium
hydroxide (lye) and hydrogen gasoline
3. Faraday's law states that the amount of lye produced at some point of galvanic
electrolysis is at once proportional to: - ANSWER-The amount of
modern-day(milliamps) and time (seconds) it flows
4. For the "a laugh of it" an electrologist comes to a decision to attempt pulsing along with
her galvanic epilator. She treats on follicle the usage of Three 1 second pulses with a
half 2d relaxation between every and any other follicle with 1 1/2 2d pulses with a half
second relaxation between each. The full quantity of lye produced is - ANSWER-The
same for each
(The overall time for every treatment = three seconds (3 (1 seconds) and a pair of
(1 half of seconds), so the amt of lye produced could stay the same. The amt of
time among pulses doesn't have an effect on amt of lye produced)
5. If an electrologist increases the timing of her galvanic epilator from 2 seconds to five
seconds at the same time as keeping the current setting at zero.2 ma, what happens to
the modern-day density across the probe - ANSWER-There isn't any alternate is
contemporary density
(cutting-edge density is the the quantity of electrons/modern coming from the
probe into the encircling tissues)
6. If the electrologist will increase the intensity of her galvanic system from 0.1 ma to 0.5
ma leaving the timing the identical the amount of lye produced will - ANSWER-Increase
by way of 5 times
(Faraday's Law: UL = ma x seconds zero.1 ma x 5 2nd= 0.Five UL ; zero.5ma x 5
seconds= zero.25 UL; an increase of five UL)
7. If the electrologist will increase the depth of her galvanic device from zero.1 ma to
zero.Five ma leaving the timing the same, the cutting-edge density will -
ANSWER-Increase by five instances
(Current density is the amount of electrons flowing via the probe to the tissue. 0.1
to zero.Five ma is an increase of five times)
AND ANSWERS
1. A probe is inserted to anagen intensity and the follicle is treated with a galvanic epilator
having the settings at 0.1 ma and four seconds. The probe is then pulled 1/2 manner
and the follicle is treated again the usage of the equal setting. The modern-day density
around the probe for the half of intensity treatment is - ANSWER-Greater than the
whole anagen intensity remedy
(The more the density the greater lye is produced. A shallow insertion creates
more contemporary density. Things that could affect current density is: the scale
of probe, depth of insertion, time the modern flows, and the current depth)
2. During galvanic electrolysis with the terrible electrode inserted right into a hair follicle, the
subsequent chemicals are produced on the terrible electrode. - ANSWER-Sodium
hydroxide (lye) and hydrogen gasoline
3. Faraday's law states that the amount of lye produced at some point of galvanic
electrolysis is at once proportional to: - ANSWER-The amount of
modern-day(milliamps) and time (seconds) it flows
4. For the "a laugh of it" an electrologist comes to a decision to attempt pulsing along with
her galvanic epilator. She treats on follicle the usage of Three 1 second pulses with a
half 2d relaxation between every and any other follicle with 1 1/2 2d pulses with a half
second relaxation between each. The full quantity of lye produced is - ANSWER-The
same for each
(The overall time for every treatment = three seconds (3 (1 seconds) and a pair of
(1 half of seconds), so the amt of lye produced could stay the same. The amt of
time among pulses doesn't have an effect on amt of lye produced)
5. If an electrologist increases the timing of her galvanic epilator from 2 seconds to five
seconds at the same time as keeping the current setting at zero.2 ma, what happens to
the modern-day density across the probe - ANSWER-There isn't any alternate is
contemporary density
(cutting-edge density is the the quantity of electrons/modern coming from the
probe into the encircling tissues)
6. If the electrologist will increase the intensity of her galvanic system from 0.1 ma to 0.5
ma leaving the timing the identical the amount of lye produced will - ANSWER-Increase
by way of 5 times
(Faraday's Law: UL = ma x seconds zero.1 ma x 5 2nd= 0.Five UL ; zero.5ma x 5
seconds= zero.25 UL; an increase of five UL)
7. If the electrologist will increase the depth of her galvanic device from zero.1 ma to
zero.Five ma leaving the timing the same, the cutting-edge density will -
ANSWER-Increase by five instances
(Current density is the amount of electrons flowing via the probe to the tissue. 0.1
to zero.Five ma is an increase of five times)
