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3 years ago

Which is necessary for visualizing latent stains of blood, semen, or urine at a crime scene

Physics

2 answers:

Ksivusya [100]3 years ago

6 0

Answer: B) ultraviolet light source

Explanation:

An ultraviolet light source is used to locate bodily fluids like semen, sweat, saliva and vaginal fluids at the scene of crime. These fluids emit their own natural fluorescence in the presence of ultraviolet light which are otherwise remain hidden and undetected at the scene of crime by naked eyes. This technique helps in their detection, collection and identification.

Slav-nsk [51]3 years ago

5 0

The correct answer is B) Ultraviolet light source. Hope this helps.

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Question 2 of 25

True [87]

Answer: The symbol representing the result of a gas expanding against a constant pressure is W

Explanation:

According to first law of thermodynamics, energy can neither be destroyed nor created but rather transferred to one place to another and converted to and from other energy forms.

The mathematical form of first law of thermodynamics is:

$\Delta U=q+W$

where $\Delta U$ = change in internal energy

q = heat absorbed or released

W = work $(W= P\Delta V)$

P = pressure

$\Delta V$ = volume change

Thus the changes in heat or work leads to change in internal energy.

5 0

3 years ago

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

Black_prince [1.1K]

Answer:

Part A:

$E_{midpoint}=0$

Part B:

$E_{center}=2711.7558 N/C$

Explanation:

Part A:

Formula of Electric Field Strength:

$E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}$

Where:

x is the distance from the ring

R is the radius of the ring

$\epsilon$ is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C$

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C$

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

$E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0$

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

$E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C$

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

$E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C$

Electric Field Strength at center of left ring is same as that of right ring.

$E_{center}=2711.7558 N/C$

5 0

3 years ago

A driver brings a car to a full stop in 2.0 s. If the car was

Dvinal [7]

Working displayed in the picture below, the answer is -11 m s^-2

7 0

3 years ago

What is the gravitational field strength of a 5 kg object that weighs 25 N?

solniwko [45]

Answer:

ExplanaThis is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.

Gravitational field strength = Weight/mass unit is N/kg

Weight = mass x gravitational field strength unit is N

On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.

Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.

Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.tion:

pls brainlieste

7 0

2 years ago

Read 2 more answers

I need help pls now plleeeeeeeeaaassseeeee

GrogVix [38]

Answer:

$r = \frac{v}{i} = v = ri \\ i = \frac{v}{r}$

4 0

3 years ago