Answer:
(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m
(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J
Explanation:
Given;
magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C
distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m
strength of electric field, E = 1390 N/C
(a) the magnitude of the electric dipole moment
p = qd
p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)
p = 1.68 x 10⁻¹⁴ C.m
(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E
ΔU = U(180) - U(0)
ΔU = 2pE
ΔU = 2(1.68 x 10⁻¹⁴ )(1390)
ΔU = 4.6704 x 10⁻¹¹ J
The correct answer is: True
Explanation:
Indeed the chemical elements in the periodic table are arranged in terms of their atomic number. They are usually written at the top of the elements in the boxes of the periodic table. Furthermore, in the periodic table, the rows are called the periods and columns are called the groups.
1) In any collision the momentum is conserved
(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')
candel all the m factors (because they appear in all the terms on both sides of the equation)
2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')
2) Elastic collision => conservation of energy
=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2
cancel all the 1/2 and m factors =>
2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>
4(vo)^2 = 2(v1')^2 + (v2')^2
now replace (v2') = -2(v1')
=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>
(v1')^2 = [4/6] (vo)^2 =>
(v1')^2 = [2/3] (vo)^2 =>
(v1') = [√(2/3)]*(vo)
Answer: (v1') = [√(2/3)]*(vo)
ANSWER:
unknown
Explanation:
do jojo poses daily because yes.
:troll:
Cases A and B are all correct.
In Case C, the GPE is incorrect.
(But then, the power and HP have been correctly calculated
from the incorrect value of GPE, so we can't gig you too bad.)