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tatuchka [14]
3 years ago
9

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

GPE) at various points on the slide At each point, her GPE can be found by using an equation. GPE = weight * height above ground Graph her GPE versus height every 2 m as she goes down the slide. Use height in meters above ground on the y-axis. Use GPE in Joules on the x-axis.​
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

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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:

A

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

B

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
The amount of pressure exerted by a solid is based on
Veronika [31]

Answer:

pressure= force/area

A solid resting on a horizontal surfaceexerts a normal contactforce equals to its weight. The pressure of the solid on the surface depends on the area of contact. (b) the area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.

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That answer is spiral galaxies
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The distance between two charges a and b is r, and the force between them is F. What is the force between them if the distancbet
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See coulomb's law. Force is inversely proportional to the distance squared. So if you multiply r by 2, the force is multiplied by (½)² = ¼.

a. F/4
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Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
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Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

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