<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:
Hey there
Your answer should be Joules, Watts, and J/s.
The joule
The joule (pronounced jool) is a derived unit of energy in the International System of Units. It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre (1 newton-metre or N⋅m). It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second. It is named after the English physicist James Prescott Joule (1818–1889).
The Watt
The watt (symbol: W) is a unit of power or radiant flux. In the International System of Units (SI), it is defined as a derived unit of (in SI base units)[1][2] 1 kg⋅m2⋅s−3 or, equivalently,[3] 1 joule per second (J/s). It is used to quantify the rate of energy transfer. The watt is named after James Watt (1736-1819), an 18th-century Scottish inventor.
I beg your parden
If I am correct may I please have the brainly
I do not know if it is correct but I think it is between B or C
Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N
