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olga2289 [7]
4 years ago
15

a scientist designs an experiment that requires two atomic particles whose only fundamental force exerted between them is the gr

avitational force. Which combination of particles and separation distance will meet this condition?
Physics
2 answers:
natka813 [3]4 years ago
5 0

Answer:

Infinite distance; all objects in the universe.

Explanation:

In physics, there are four fundamental interactions, and the fundamental interactions include strong force, electromagnetic force, weak force and gravitational force. These forces governs how objects or particles interact and how certain particles decay.

With respect to their relative strengths, the strong force is regarded as the most powerful, followed by the electromagnetic force, the weak force, and the gravitational force.

Gravitational force and electromagnetic force operates at infinite distance. Gravitational force acts between all objects of the universe, no matter the distance/how far the body are apart.

yaroslaw [1]4 years ago
3 0

Answer:

A proton and neutron located 1.0 mm apart

Explanation:

As we know, atoms are composed of fundamental particles, atomic particles: protons, neutrals and electrons.

Protons and neutrons are found in the nucleus of the atom and are located very close to each other. The electrons, however, are located in the electrosphere, a little further away from the nucleus and consequently, away from the protons and neutrals, by approximately 1.0 mm. In addition, the only force that acts between protons and neutrals is the gravitational force, which is very important, even, so that the electrodes are not attracted over the nucleus.

For this reason, we can confirm that the combination of particles and separation distance that will meet the condition shown in the question above are: a proton and neutron located 1.0 mm apart.

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What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
kipiarov [429]

Answer:

The density of the woman is 950.8 kg/m³

Explanation:

Given;

fraction of the woman's volume above the surface = 4.92%

then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%

the specific gravity of the woman = \frac{95.08}{100 } = 0.9508

The density of the woman is calculate as;

Specific \ gravity \ of \ the \ woman = \frac{Density \ of \ the \ woman }{Density \ of \ fresh \ water }\\\\ Density \ of \ the \ woman  = Specific \ gravity \ of \ the \ woman \ \times \ Density \ of \ fresh \ water

Density of fresh water = 1000 kg/m³

Density of the woman = 0.9508 x 1000 kg/m³

Density of the woman = 950.8 kg/m³

Therefore, the density of the woman is 950.8 kg/m³

4 0
3 years ago
1. Johnny wants to know where the water line is in a dark well. He drops a penny into the well and counts until he hears the pen
777dan777 [17]

Answer:

3 feet down

Explanation:

i think

8 0
3 years ago
A red car passes a blue car. Which is true?
Marrrta [24]
D. The red car is moving faster than the blue car
4 0
3 years ago
Describe an experiment to show that pressure increases with the decrease in the area of surface
cluponka [151]
Answer:
press a baloon against one pin it bursts
but now arrange lot of pins parallel closely to each other if u press a baloon against them it does not burst hope this helps u
7 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
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