Democritus was the one who did not have experimental evidence to support his theory of the atom.
Answer: Option 4
<u>Explanation:
</u>
The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.
After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.
This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.
The inner planets are not colder or larger than the outer ones,
and they're not comprised of gas.
The inner planets are the ones that are made of rock. ( D ).
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
Gauss law states that the electric flux through any closed
surface is proportional to the net electric charge inside the surface. This is
expressed mathematically in the form of:
Φ = Q / εo
Where,
Φ = the electric flux = unknown (which we have to find for)
Q = the net electric charge = 5.0 µC = 5 E-6 C
εo = the permittivity of free space = a constant value =
8.85 E-12 C^2 / N m^2
Plugging in the values
into the equation will result in:
Φ = 5 E-6
C / (8.85 E-12 C^2 / N m^2)
Φ = 564,971.75 Wb = <span>5.6 x
10^5 Wb </span>
