Permanent magnet. An induced magnet would be created when a piece of iron (for example) is in contact with a magnet. Temporary magnets would be something like an electromagnet. Bar magnets are permanently magnetic unless we heat them or hammer them to cause their domains to loose alignment.
Answer
A. the work done on the refrigerant in each cycle is 105kJ
B the coefficient of performance of the refrigerator is 4.8
Explanation
Given data
Work done at high temperature T2 Qh=610kJ
Work done at low temperature T1 Ql=505kJ
We know that the net work done by the refrigerator is expressed as
Wnet= Qh-Ql
=610-505
=105kJ
Also we know that the coefficient of performance is expressed as
COP= Ql/Wnet
COP= 505/105
= 4.8
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow

Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
Answer:
The correct option is;
(b) The end A of the solenoid behaves like a north pole
Explanation:
According to Lenz's law we have that the induced emf direction in the solenoid due to the rapid introduction of the bar magnet will be such that the electric current induced will have a resultant magnet field that will oppose to the movement of the north pole of the bar magnet that resulted in the magnetic field
Therefore, the opposing magnetic pole to the north pole of a magnet is a north pole and the solenoid end A will act like the north pole.