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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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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

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3 years ago

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grandymaker [24]

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