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2 years ago

What are challenges of securing a crime scene

1 answer:

7 0

At any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

<h3>What is the meaning of physical evidence?</h3>

In evidence law, physical evidence (also called real evidence or material evidence) is any material object that plays some role in the matter that gave rise to the litigation, introduced as evidence in a judicial proceeding (such as a trial) to prove a fact in issue based on the object's physical characteristics.

The two types of evidence at crime scenes:

Biological evidence (e.g., blood, body fluids, hair and other tissues)

Latent print evidence (e.g., fingerprints, palm prints, footprints)

The biggest impediment to an investigation is the removal or loss of a piece of evidence from the scene of a crime.

Hence, at any crime scene, the two greatest challenges to the physical evidence are contamination and loss of continuity.

Learn more about the physical evidence here:

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A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min

Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

$v_1=89km/h$

$t_1=30min$

$v_2=100km/h$

$t_2=12min$

$v_3=40km/h$

$t_3=45 min$

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute= $\frac{102}{60}=1.7 hour$

1 hour=60 minutes

Distance= $speed\times time$

$d_1=v_1\times t_1=80\times\frac{30}{60}=40km$

$d_2=100\times \frac{12}{60}=20 km$

$d_3=40\times \frac{45}{60}=30 km$

Total distance= $d_1+d_2+d_3=40+20+30=90km$

Average speed= $\frac{total\;speed}{total\;time}$

Using the formula

Average speed= $\frac{90}{1.7}=52.9Km/h$

b.Total distance between the initial and final city lies along the route=90 km

4 0

3 years ago

10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m

Mariulka [41]

Work formula:

$W = Fd\cos(\theta)$

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

$W = (50N)(1.0)(0) = 0$

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

$W = (50 N)(1.0m)(1)$

$W = 50$

total work = 50 joules

6 0

2 years ago

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega$

And now we can use Ohm's law to find the current in the circuit:

$V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A$

6 0

3 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$

8 0

3 years ago

Starburst galaxies:

alexgriva [62]

Answer:

are often associated with a galaxy that is colliding with another galaxy.

Explanation:

A starburst galaxy is a galaxy that undergoes very fast formation of stars. The rate at which stars are born is 100 times more than 3 solar masses per year of the Milky Way. The starburst is stage of the formation of a galaxy. After this stage is complete the stars will have used almost all the gas in it. As the star formation rate is very fast the difference between the age of the stars and the galaxy itself is very less. The star formation is triggered by mergers and tidal interactions between gas-rich galaxies.

6 0

3 years ago