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3 years ago

Which best describes technology's use in science?

Physics

2 answers:

Setler79 [48]3 years ago

7 0

Answer:

Technology influences scientific progress.

Studentka2010 [4]3 years ago

4 0

The best description os technology's use in science is that technology influences scientific progress.

Explanation:

I'll provide an example;

Advancement in technology has enabled the development of more powerful microscopes. Traditional microscopes have low-resolution power and therefore were not able to view tiny organelles in cells such as the ribosomes and mitochondria. However, today, due to the development of very powerful microscopes like the atomic force microscope, cells can be viewed almost to an atomic level. This has improved scientists' understanding of cells hence positively influencing scientific research.

Learn More:

For more on technology in science check out;

brainly.com/question/13743322

brainly.com/question/13118765

#LearnWithBrainly

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Explanation:

First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.

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A paper airplane with mass 0.1 kg is flying 1.5 m above the ground with a speed of 2 m/s. What is the total mechanical energy of

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Positive feedback interactions in earth’s systems are always a result of human action. T/F

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a snail takes 16 minutes 40 seconds to cover a distance of 1 m calculate the average speed of the swimmer

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Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that

elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p = 9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

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4 years ago