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Cerrena [4.2K]
3 years ago
11

Who pioneered the use of galvanoplastic compounds for preserving footprints and ballistics?

Physics
2 answers:
GarryVolchara [31]3 years ago
7 0

Answer:

D. Alphonse Bertillon

(confirmed for Edmentum courses)

Explanation:

"Alphonse Bertillon (1853-1914): He was a French policeman and biometrics researcher. Bertillon created forensic techniques like forensic document examination. <u><em>He also pioneered the use of galvanoplastic compounds as molds for footprints and ballistics in order to preserve evidence.</em></u> Bertillon also created a system based on photographs of the same person over time to study physical differences with age."

(edited to add image)

alukav5142 [94]3 years ago
5 0

Answer:

Option D (Alphonse Bertillon) is the correct response.

Explanation:

  • He seems to have been a policeman turned biometrics expert from France. Forensic techniques such as forensic record analysis were developed by Bertillon.
  • To retain proof, he always pioneered or developed the use of such galvanoplastic compounds as molds for footsteps as well as ballistics. To research physical changes with age, Bertillon has developed a method focused on images of almost the same person’s performance.

All those other choices weren’t connected to the instance offered. So, the best one is the one described.

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3 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
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Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

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3 years ago
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