Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement
limits are x = 0 to x = 0.15 m
work done
W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =
% cubic term =
9. Why do atoms have relatively the same size even though they may have widely different masses?
1 answer:
You might be interested in
In this case, you need the formula below where:
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges
pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...