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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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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v
-BARSIC- [3]

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, \omega=1.16\ rev/s=7.28\ rad/s

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N

So, the tension in the chain is 692.42 N.

5 0
2 years ago
How can an animal regeneration produce two results?
GenaCL600 [577]
In biology, regeneration<span> refers to the process by which plants and </span>animals<span> replace lost or damaged parts by growing them anew. Some </span>animals can regenerate<span> their limbs, tails, or even parts of internal organs, such as the liver. In plant </span>regeneration<span>, neighboring cells replace missing tissue.</span>
8 0
3 years ago
Read 2 more answers
The measure of the force with which air molecules push on a surface is called
alex41 [277]
It is called air pressure
7 0
3 years ago
If the phase angle for a block–spring system in SHM is ϕ and the block's position is given by x = xm cos(ωt + ϕ), what is the ra
matrenka [14]

<h2>K.E/P.E = m/k  tan²φ x ω²</h2>

Explanation:

The given position of block x = x₀ cos(ωt + φ)

The velocity of block  v = dx/dt = - x₀ sin(ωt + φ) x ω

The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²

The potential energy of spring = 1/2 k x² , where k is the spring constant

Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)

When t = 0

K.E = 1/2 m x₀²sin²φ x ω²

P.E = 1/2 k x₀² cos²φ

Dividing these , we have

K.E/P.E = m/k  tan²φ x ω²

7 0
2 years ago
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