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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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A softball has a (negative) acceleration when it is thrown. A soft ball has a (positive) acceleration when it is caughtExplanation:

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Which of the following is not a an example of dissipated energy?
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When energy is changed from one form to another, ____.

b. all of the energy can be accounted for

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How much heat energy is needed to raise the temperature of 2.06 kg of this oil from 23 °c to 191 °c?
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Q=mc(t2-t1)
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3 years ago
How is light used to make other form of transportations safer
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3 years ago
A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
andrew-mc [135]

Answer:

23

21.7391304348 m

Explanation:

L = Initial length = 500 m

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Change in length is given by

\Delta L=L\alpha \Delta T\\\Rightarrow \Delta L=500\times 12\times 10^{-6}\times (40-(-35))\\\Rightarrow \Delta L=0.45\ m

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The number of joints would be

n=\dfrac{0.45}{0.02}\\\Rightarrow n=22.5\\\Rightarrow n\approx 23

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Each bridge section length would be

\dfrac{L}{n}=\dfrac{500}{23}=21.7391304348\ m

The length of each bridge section would be 21.7391304348 m

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