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storchak [24]
3 years ago
12

A body is found in the woods and all you have left to analyze are bones and teeth. Explain what isotopes would be helpful to you

as you seek to identify these remains and match them to a missing person's report. How would these isotopes help in the identification of this person?
Physics
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

The isotopes that would be helpful to identify a missing person and complete its report are the stable isotopes.

The stable isotope forensic science analysis using Sr, O and S isotope variations has been developed recently using hair strands to identify recent geographic hints as temperature in climate causes the isotopes ratio to vary indicating if the person changed its geographic area or he/she  belongs to the area it was found, for what it drank.

Explanation:

The forensic analytical science uses different isotope´s methods like gas or liquid chromatography, HPLC, mass spectrometry, immunoassays, enzymatic assays, DNA profiling techniques, PCR, biosensing, metabolomics and LC-MS which can stablish the chemical nature of most of the samples provided.

As a basic principle consider isotopes are atoms of the same element that have different quantity of neutrons but the same number of protons and electrons.

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Work and Power Practice Calculations
Elza [17]

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

5 0
2 years ago
Identical spheres are dropped from a height of 100 m above the surfaces of Planet X and Planet Y. The speed of the spheres as a
Anna007 [38]

Answer:B

Explanation:

8 0
3 years ago
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}

4 0
3 years ago
What is Gravitational force?​
mote1985 [20]

Answer:

the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface

Explanation:

7 0
3 years ago
Read 2 more answers
A speed versus time graph is shown:
erma4kov [3.2K]

In the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

In the given graph we can deduce the following;

  • at the time interval, 0 s to 3.5 s, the speed of the object = 1 cm/s
  • when the time, t= 4 s, the <em>speed</em> of the object = 0 cm/s
  • at the time interval, 4.0 s to 8.0 s, the<em> speed </em>of the object = 0 cm/s

When the <em>speed</em> of an object is zero (0), the object is simply at rest.

Thus, we can conclude that in the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

Learn more here:brainly.com/question/10454047

6 0
2 years ago
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