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3 years ago

Which of the following would be an example of basic research?

Physics

2 answers:

ElenaW [278]3 years ago

8 0

I think it's Edison research and use the other Sciences research light bulb, this is what I think I'm not totally sure

lara [203]3 years ago

6 0

Answer: Experimenting to determine the fundamental properties of x-rays.

Explanation:

Scientific basic research aims to find the fundaments or principles of the phenomena.

The basic research is intended to understand the matter, its nature, its properties, its behavior. It searches to make theory. To finding how the things work but not how to use those things for a determined purpose.

The study of the properties of the atom to understand what it is, how it interacts with other atoms, how it determines the properties of the matter are some examples of basic research, such as experimenting to determine the fundamental properties of x-rays is.

On the other hand, the other examples given, Spencer's research on WWII radar technology that led to the invention of the microwave oven, Edison's research and use of other scientists' work to invent the light bulb, and Morrison and Franscioni's research done to create the Frisbee, are examples of applied research science. This is research to find a valuable use of some scientific knowledge.

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3 years ago

A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito

Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

$c = \frac{c1 \times c2}{c1 + c2}$

$= \frac{2 \times 4}{2 + 4}$

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

$= 1.33 \times 10 {}^{ - 6} \times 8$

$= 10.64 \times 10 {}^{ - 6}$

charge on 2.0μf capacitor is

$\frac{Qeq}{2 \times 10 {}^{ - 6} }$

$= \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }$

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

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1 year ago

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3 years ago

A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes

Ivan

Answer:

f = 19,877 cm and P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

1 / f = 1/97 + 1/25

1 / f = 0.05

f = 19,877 cm

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P = 1 / f

P = 1 / 19,877 10-2

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2 years ago