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Klio2033 [76]
4 years ago
14

Why can scientists ignore the gravitational force when studying the physics of an atom?

Physics
2 answers:
Nataliya [291]4 years ago
8 0

Answer:

Scientists can ignore the gravitational force because the electrical force between protons and electrons in the atom is millions of times stronger than the gravitational force between the protons and electrons. The gravitational force is insignificant for masses as small as the proton and electron.

Explanation:

frozen [14]4 years ago
7 0
The short answer is that gravitational force is so small compared to other forces (such as the attractions/repulsion between charged particles) that gravity is negligible.
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The answer your looking for is, A. alpha particle.
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How does black paint interact with a light wave?
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Light rays that come from a source such as the sun reflect off items and enter our eye. ... While black objects absorb the energy from all colors and become hot, the objects gradually release some of that energy back into the air around it

Explanation:

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2 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
Nezavi [6.7K]

Answer:

the volume that has 1.0m^2 of water from the surface of the lake is 4.3

Explanation:

The computation of the volume is as follows;

The Bulk modulus is

= K = 2.3 × 10^9 Pa

K = - V dP ÷ dV

dV = - V dP ÷ K

dV = - (1 × 101.3 × 1000) ÷  2.3 × 10^9

= - 4.3 × 10^-3 m^3

So the volume that has 1.0m^2 of water from the surface of the lake is 4.3

Hence the same is relevant and considered

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3 years ago
Forces can be added together only if they are?
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The answer is a) because otherwise the forces can't be added as the objects will be different.
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Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate s
Serggg [28]

Answer:

The POSITION of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the VELOCITY from A to B is/are the same as expressed in both coordinate systems.

Explanation:

Imagine that you got 2 coordinate systems; A and B. Now, this coordinate systems start at different points.

Measured from A, the starting point of the A coordinate system its, of course, at (0,0,0)_A. But B its somewhere else. Lets put the origin of the system B MEASURED from the system A, at (b_x,b_y,b_z)_A.

This position, in the system B will be at (0,0,0)_B, cause its it origin, be careful, we could write:

(b_x,b_y,b_z)_A = (0,0,0)_B

as this two vectors are the same, the starting position for the B system.

Lets imagine that in thei got something displaced from the starting point of the system B by (1,0,0). Of course, in the system B this is located at (1,0,0)_B, What is the position for the system A? Well, it will be at

(b_x,b_y,b_z)_A + (1,0,0)_A = (b_x +1,b_y,b_z)_A.

The distance is measured the same way in both system.

Any position

(x,y,z)_B

in the system B, will be at

(b_x,b_y,b_z)_A + (x,y,z)_A = (b_x + x,b_y + y,b_z + z)_A

at system A, cause this positions ARE THE SAME, even though they are represented with different vector in this systems.

Now, why the velocity its the same? Well, the velocity its the derivative of the position with respect time.

So, for a position R=(x,y,z)_B, in the system B we got:

\frac{d}{dt} R = \frac{d}{dt} (x,y,z)_B = ( \frac{dx}{dt} , \frac{dy}{dt} , \frac{dz}{dt} )_B

R in the system A takes the form:

R=(b_x,b_y,b_z)_A + (x,y,z)_A

and the derivative its:

\frac{d}{dt} R = \frac{d}{dt} [(b_x,b_y,b_z)+ (x,y,z)]_A =\frac{d}{dt} (b_x,b_y,b_z)_A + \frac{d}{dt}  (x,y,z)_A

now, the (b_x,b_y,b_z)_A its constant, so:

\frac{d}{dt} (b_x,b_y,b_z)_A  = 0

which leave us with:

\frac{d}{dt} R =\frac{d}{dt}  (x,y,z)_A

\frac{d}{dt} R = (\frac{d}{dt} x,\frac{d}{dt} y,\frac{d}{dt} z)_A

This is the same for both systems!

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