All categories

3 years ago

Where was forensic science first used to solve a crime?

Physics

2 answers:

LuckyWell [14K]3 years ago

8 0

Answer:

In 1836, Scottish chemist, James Marsh, did the first application of this forensic science technique. This test was actually used successfully in a murder trial at that time.

Explanation:

I hope that is what you were asking

Ludmilka [50]3 years ago

5 0

Answer: London , United Kingdom

Explanation:

In 1836, Scottish chemist, James Marsh, did the first application of this forensic science technique.

You might be interested in

John's grandfather clock has a pendulum that keeps the seconds. What does this graph say about the observed pattern of motion?

Radda [10]

D. The osculations show a variable rate of motion. Hope this helps:)

7 0

3 years ago

Read 2 more answers

An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1

Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0

3 years ago

A double-slit experiment yields an interference pattern due to the path length difference from light traveling through one slit

Lapatulllka [165]

Answer:

Therefore the correct statement is B.

Explanation:

In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.

The sum of all these waves results in the diffraction curve of the slit that has the shape

I = Io sin² θ /θ²

Where the angle is a function of the wavelength and the width of the slit.

From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)

Therefore the correct statement is B

6 0

3 years ago

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$