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natta225 [31]
4 years ago
13

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

Physics
2 answers:
ElenaW [278]4 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]4 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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Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnit
Snowcat [4.5K]

Answer:

894 electrons

Explanation:

The electrostatic force between the two charges is given by:

F=\frac{k q_1 q_2}{r^2}

where we have

F=4.57\cdot 10^{-21} N is the force

k is the Coulomb's constant

q1 = q2 =q is the magnitude of the charge on each sphere

r = 20.0 cm = 0.20 m is the distance between the two spheres

Substituting and solving for q, we find the charge on each sphere:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(4.57\cdot 10^{-21} N)(0.20 m)^2}{9\cdot 10^9 Nm^2C^{-2}}}=1.43\cdot 10^{-16} C

And since each electron has a charge of

e=1.6\cdot 10^{-19}C

the net charge on each sphere will be given by

q=Ne

where N is the number of excess electrons; solving for N,

N=\frac{q}{e}=\frac{1.43\cdot 10^{-16}C}{1.6\cdot 10^{-19}C}=894

7 0
4 years ago
an empty bottle has a mass of 24,25 grams and 86,55 grams completely full-filled with water. Then, water in the bottle is emptie
ladessa [460]
The density of a substance is the quotient we obtain when we divide its mass by the volume. The density is,
                          density = mass / volume
The mass of carbon tetrachloride is given to be 123.95 grams and that the volume is obtained by subtracting final weight of the completely full-filled with water bottle with the initial weight.
                              86.55 - 24.25 = 62.3 grams
Since the density of water is 1 grams/ cc. Then, the volume of the bottle is also 62.3 cc. The density is therefore,
                       density = 123.95 grams/ 62.3 cc = 1.99 g/cc
4 0
3 years ago
The melting point of a substance occurs at the same temperatures as it’s blank point
azamat

I think it's a.

Explanation:

melting point is boiling

4 0
4 years ago
Read 2 more answers
A parallel-plate capacitor consists of two plates, each with an area of 27 cm^2 separated by 3.0 mm. The charge on the capacitor
aev [14]

Answer:

Explanation:

Capacity of a parallel plate capacitor  C = ε₀ A/ d

ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.

C =(  8.85 X10⁻¹² X  27 X 10⁻⁴ ) / 3 X 10⁻³

= 79.65 X 10⁻¹³ F.

potential diff between plate = Charge / capacity

= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³

= 601 V

Electric field = V/d

= 601 / 3 x 10⁻³

= 2 x 10⁵ N/C

Force on proton

= charge x electric field

1.6 x 10⁻¹⁹ x 2 x 10⁵

= 3.2 x 10⁻¹⁴

Acceleration a = force / mass

= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷

= 1.9 x 10¹³ m s⁻²

Distance travelled by proton = 3 x 10⁻³

3 x 10⁻³ = 1/2 a t²

t = \sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }

t = 1.77 x 10⁻⁸ s

6 0
3 years ago
Find the specific heat of a substance that requires 8000 J of energy to heat up 400g by 20 C?
monitta

Answer:

c=1\ J/g^\circ C

Explanation:

Given that,

Heat required, Q = 8000 J

Mass, m = 400 g

The change in temperature, \Delta T = 20^{\circ}

The heat required due to change in temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C

So, the specific heat of the substance is 1\ J/g^\circ C

7 0
3 years ago
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