So, the first question is: how many meters are 10 nm?
1nm =<span>0.000000001 m.
So 10 nanometers are </span><span>0.00000001 m!
Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!
now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
</span>
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
Radius of nuclei is given by formula

now we can say volume of the nuclei is given as

now the density is given as
density = mass / volume
mass of nuclei = mass of neutron + mass of protons



Now density is given as

here we know that
= 1.2 fm


So from above we can say that density of all nuclei is almost same.
Answer:
Yes, if the system has friction, the final result is affected by the loss of energy.
Explanation:
The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.
In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces
= ΔEm
=
-Em₀
As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.
=
+ Em₀
Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer
½ m v² = -fr d + mgh
v = √(-fr d 2/m + 2 gh)
v = √ (2gh - 2fr d/m)
Now it is clear that there is a decrease in the final body speed.
Consequently, if the system has friction, the final result is affected by the loss of energy.