Explanation:
- Mass(m)= 20kg
- Acceleration (a)= 5m/s²
- Force(F)= ?
We know that,
Hence, the needed force is 100N.
Wires or silver and copper
The atomic number is the simply the number of protons in the atom. So in the first row with atomic number 2, the number of protons is 2.
If the atom has no charge, which I think you can assume for all of these, the number of electrons is equal to the number of protons. So the number of electrons is also 2.
The number of neutrons (which are the particles with no charge in the nucleus) is simply the mass number minus the atomic number i.e. 4 - 2 = 2.
The isotopic symbol is the symbol which is found on the periodic table of elements. There are 2 numbers associated which each element on the table. The larger is the mass number and the smaller is the atomic number. The atomic number or number of protons is what identifies the element. Looking at the periodic table ( https://sciencenotes.org/wp-content/uploads/2015/01/PeriodicTableOfTheElementsBW.pdf or https://simple.wikipedia.org/wiki/Periodic_table_(big) ), it can be seen that the element on the first row above with an atomic number of 2 is Helium with a symbol He. The number that is included with the name is simply the mass number which is 4 in this case, which tells us that this type of helium has 2 neutrons.
Another type (or isotope) of helium is Helium-3 which has one neutron.
Try the next row and post back if you have trouble with it
Answer:
So lift will be 30.19632 N
Explanation:
We have given area of the wing 
We know that density of air 
Speed at top surface 
and speed at bottom surface 
According to Bernoulli's principle force is given by
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e (
)
with this expression we can find the closest approach distance (r)
