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3 years ago

PLS PLS PLS PLS PLS PLS PLS PLS PLS PLS PLS HELP

Physics

2 answers:

PIT_PIT [208]3 years ago

7 0

Answer:

So, instead of fossil fuels, the magnetic field thats creatd by electrified coils in the guideway walls, and on the track combine to propel the train. Just think of it as playing with magnets. The opposite attract while if you put norh to north or south to south, they will push away from each ther. This is just the basic principle of electromagnetic propulsion

Explanation:

rodikova [14]3 years ago

7 0

Answer:

C motors

Electric motors use the forces produced by magnetic fields to produce a turning motion. If you put a length of wire in a magnetic field and pass a DC current through it (such as from a battery), the wire will move. This is called the motor effect.

To make a simple DC motor, you need:

two bar magnets

a coil of wire wrapped around something to support it

an axle for the coil of wire to spin around

two half rings (‘split rings’)

hope this helps

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Which would fall with greater acceleration in a vacuum a leaf or a stone?

Sergeu [11.5K]

Because its a vacuum, there's no air resistance, they will fall at same time
Applying gravity acceleration rule g=9.8m/s which is taken as 10m/s sometimes.

3 0

3 years ago

Read 2 more answers

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

3 years ago

For the equilibrium reaction N2O4(g) ⇀↽ 2 NO2(g) taking place inside a sealed container fitted with a piston, predict the effect

Butoxors [25]

Answer:

4. The equilibrium will shift to favor formation of NO2(g)

Explanation:

According to La Chatalier's Principle which states that when an equilibrium system undergoes changes either in temperature, volume or concentration; there will be in a change in the system in order to reach equilibrium.

From the above equation,

N2O4(g) ⇀↽ 2 NO2(g)

From the above reaction, there are 2 moles of gaseous product on the left and 1 mole of gaseous reactant.

Therefore, there are more moles of gases in the left hand side than the right hand side.

Because a decrease in volume favors the direction that produces fewer moles, an increase in volume will therefore shift this system towards the side with more moles of gases that is, more products are formed hence, this system will shift to right and produce more moles of products i.e more NO2(g) formed.

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3 years ago

Which statement best describes the atoms of the gas neon?

Leya [2.2K]

Answer:

They:

-Are far from each others

-Move constantly

-Move freely (all directions)

-Move at high speed

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3 years ago

If you were in charge of designing a wire to carry electricity across your city, state or province, which of

Anon25 [30]

Answer:

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.

7 0

2 years ago