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makvit [3.9K]
1 year ago
8

Do you think there could be another period of time when so many brilliant discoveries are made in such a span of time? Why or Wh

y not?
Physics
1 answer:
Arada [10]1 year ago
8 0

There won't be another period where many brilliant discoveries are made in such a span of time because of the high degree of advancement already.

<h3>What is Technology?</h3>

Tgis is defined as the application of scientific knowledge for practical purposes.

We have advanced technological advancement which is why there won't be many discoveries in such time span.

Read more about Technology hee brainly.com/question/7788080

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Which of the following is the best definition of a closed system?
lisov135 [29]

The answer is either

b A system in which Newton's Laws are valid

or

c A system in which there are no external forces.

Explanation:

not a, and not d

There are energy changes in a closed system.

A closed system obeys the conservation laws in its physical description.

4 0
3 years ago
Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w
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3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
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SVETLANKA909090 [29]

Answer:

So what's the answer???

Explanation:

8 0
3 years ago
Alright eksqijakojqnlqozjzbw.wlisj
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Answer:

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