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finlep [7]
3 years ago
14

Which test was conducted on the skin found under the victim’s nails?

Physics
2 answers:
BARSIC [14]3 years ago
5 0
Y-STR analysis. It is a DNA test used to compute and match with the possibility via lap.
N76 [4]3 years ago
3 0
The answer to that question should be and I am pretty sure is, DNA.
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In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gent
Dmitry_Shevchenko [17]

Answer:

Explanation:

A )

speed of swimming in still water is given by the expression

distance / time

= 50 / 25

= 2 m /s

In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of  2 + 1.2 = 3.2 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / 3.2 = 15.625 s

In lane 8 , 1.2 cm/s current is flowing against  the direction that the swimmers are going so swimmers will cover distance at the rate of  2 - 1.2 = .8 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / .8 = 62.5 s

8 0
3 years ago
A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?
liubo4ka [24]

Answer:

137.2 in pounds and in Newton's it's 588.399

3 0
3 years ago
How do you find the range of a data set?
Helga [31]

The answer is A.

The answer is A.

The answer is A.

6 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
There are free body diagrams in the picture, number (#). Which of the diagram(s) shows an object with a net force right?
ryzh [129]
4 only diagram net Force
6 0
3 years ago
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