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4 years ago

Name the 9 types of energy

Physics

1 answer:

Ugo [173]4 years ago

6 0

Hello There!

-Energy Is The Ability To Do Work-

Energy comes into different forms but they can all be placed in two different categories. These categories are "potential" which means stored energy and energy of position and "kinetic" which is the energy of motion.

POTENTIAL ENERGY Stored energy and the energy of position

CHEMICAL ENERGY Energy in the bonds of atoms and molecules

MECHANICAL ENERGY Energy stored in objects by tension

NUCLEAR ENERGY Energy stored in the nucleus of an atom

GRAVITATIONAL Energy that is stored in an object's height

ELECTRICAL ENERGY Delivered by tiny charged particles called electrons

KINETIC ENERY The energy of motion

THERMAL ENERGY The vibration and movement of atoms and molecules in substances.

RADIANT ENERGY Electromagnetic energy that travels in waves

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Sort the items based on whether they are simple machines or compound machines.

Tom [10]

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Simple machine: pair of tongs, seesaw and wheelbarrow

Compound machine: sewing machine, fishing rod and reel and crane.

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Simple machine is the simplest device to use mechanical advantages. It has simplest mechanism to multiply the magnitude of force.

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Compound machine is a device that is made up of more than one simple machine.

so, sewing machine, fishing rod and reel and crane are example of compound machine because these devices shows more than one machine in it.

Thus, classification is as follows:

Simple machine: pair of tongs, seesaw and wheelbarrow

Compound machine: sewing machine, fishing rod and reel and crane.

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4 years ago

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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

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

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago