<span>Depends on the precision you're working to.
proton mass ~ 1.00728 amu
neutron mass ~ 1.00866 amu
electron mass ~ electron mass = 0.000549 amu
Binding mass is:
mass of constituents - mass of atom
Eg for nitrogen:
(7*1.00728)-(7*1.00866)-(7*0.000549)
-14.003074 = 0.11235amu
Binding energy is:
E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So:
0.11235 * 931.5 = 104.6MeV
Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV
This is probably about right; it sounds like the right size!
Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :)
1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules).
It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>
Answer:
Your answer would be
A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:
I belive this is the answer sorry if im wrong!
Answer:
halve the slit separation
Explanation:
As we know that
In YDS experiment, the equation of fringe width is as follows
where,
D denotes the separation in the middle of screen and slits
d denotes the distance in the middle of two slits
And to increase the Δx we have to decrease the d i.e, the distance between the two slits
Hence, the first option is correct
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
In order to clearly qualify this problem, we use relationships between linear displacement (x) and angular displacement (theta). The relationship would be, x= r*theta, where r is the radius. From here, you can deduce that the smaller the radius, the larger the angular displacement. Hence, the boy that is closer to the center would experience greater angular displacement.
