Ask question

Ask question

All categories

3 years ago

7

A young woman was found dead on the side of the road. Two witnesses said they saw and red cadillac hit her and drive off earlier

that evening. Forensic investigators collected glass samples found on the victims body. a few hours later, police found what they thought to be the car in question: a red cadillac with a broken headlamp. they collected glass samples from the headlight.
Forensic Experts would use all BUT ONE analysis technique to help match the glass samples. That would be

a) density
b) shatter pattern
c) refractive index
d. microscopic examinat

1 answer:

nignag [31]3 years ago

7 0

I think the answer would be A. Density because many headlights could have the same density, but the rest of the analysis techniques would be specific to that crime.

You might be interested in

Check cashing businesses do not require that an individual be an account holder; they will cash any valid check. True False

Romashka-Z-Leto [24]

True. Free trade means they can't refuse to give you cash for a check. If you want the check deposited to an account, that's another story.

3 0

2 years ago

Read 2 more answers

A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp

Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2$

b) Assume the grind stone is a solid disk, its moment of inertia is

$I = mR^2/2$

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

$I = 28*0.15^2/2 = 0.315 kgm^2$

So the friction torque is

$T_f = I\alpha = 0.315*0.21 = 0.06615 Nm$

The friction force is

$F_f = T_f/R = 0.06615 / 0.15 = 0.441 N$

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

$N = F_f/\mu = 0.441/0.2 = 2.205 N$

7 0

2 years ago

Imagine a negative test charge sitting at the coordinate origin (0,0). Two bunches of positive charges are located on the x-axis

Oxana [17]

Answer:

the total force vector, on test charge is points from origin to point C( 1, 1 )

Explanation:

Given the data in the question, as illustrated in the image below;

from the Image, OA = 1, OB = AC = 1

so using Pythagoras theorem

a² = b² + c²

a = √( b² + c² )

so

OC = √( OB² + AC² )

we substitute

OC = √( OA² + AC² )

OC = √( 1² + 1² )

OC = √( 1 + 1 )

OC = √2

Coordinate of C( 1, 1 )

Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )

5 0

2 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

2 years ago

Help asap please I will give you 5stars

boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....$

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

$\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega$

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

$\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega$

Hence, this is the required solution.

6 0

2 years ago