Atoms have no electric charge because they

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. True or False

Vadim26 [7]

Answer:

True.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Furthermore, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = G.P.E + K.E

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. This is ultimately in accordance with the law of conservation of energy, which states that energy cannot be destroyed but can only be converted or transformed from one form to another.

Hence, Mechanical energy that has been ‘lost' to friction isn't really lost but converted into heat energy.

4 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$