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Mkey [24]
2 years ago
12

Microscopes are used to study the structures of rocks. Over time, microscopes have become more and more powerful.

Physics
1 answer:
natta225 [31]2 years ago
5 0

As a result this technological improvement, our structural information of rocks has improved greatly.

<h3>What is a petrographic microscope?</h3>

A petrographic microscope is a type of optical microscope that is used to identify rocks and minerals in very thin sections and also to investigate microscopic textures and features that are present in minerals and rocks.

This has improved greatly recent structural information about rocks.

Hence, microscopes used in the study of structures of rocks has overtime improved greatly our structural information about them

Learn more about structure of rocks here:

brainly.com/question/20151682

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What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both
Triss [41]

Answer:

Both Thomson and Rutherford used charged particles in their experiments.

Explanation:

6 0
3 years ago
Read 2 more answers
A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?
algol13
The snail will go <span>0.18193752 miles </span>
5 0
3 years ago
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 certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are squa
Evgesh-ka [11]

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

7 0
3 years ago
1 point
klio [65]

Answer:

0

10

20

30

40

50

60

70

80

90

100 g

0

100

200

300

400

500

8

9

108

Explanation:

ikkk

8 0
3 years ago
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