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## Rank the objects based on the maximum height they reach along the curved incline.

## Find current and potential difference across each resistor

## The capacitor in the figure (figure 1)is initially uncharged. the switch is closed at t=0.

## Hat is the magnitude of the electric force on charge a in the figure?

The electrostatic force may be created through F = K*Q1*Q2/R2 The force charges 1 exerts on the charge will result in F13 = k*q1*q3/d1^2 in which d1 represents the distance that exists between q1 and q3 Charge 2 forcefully on charge 3 now. F23 = k*q2*q3/d2^2 n which d2 represents the distance that eRead more

The electrostatic force may be created through

F = K*Q1*Q2/R2

The force charges 1 exerts on the charge will result in

F13 = k*q1*q3/d1^2

in which d1 represents the distance that exists between q1 and q3

Charge 2 forcefully on charge 3 now.

F23 = k*q2*q3/d2^2

n which d2 represents the distance that exists between q2 and q3

In light of that

|F13| = |F23|

k*q1*q3/d1^2 = k*q2*q3/d2^2

Based on that

q1 = q & q2 = 4q & q3 = q, So

k*q*q/d1^2 = k*4q*q/d2^2

(d2/d1)^2 = 4

2/d1 = sqrt 4, which is 2

d2 = 2*d1

distance between q2 and equals twice of the distance between them.

Q1 and q3

In another way in other words, d2 > d1

In the configurations above, only A and C is able to follow the previous relation.

See less=> The Correct Solution: A, C## A parallel-plate vacuum capacitor is connected to a batteryand charged until the stored electric energy is…

Concepts and Reason To resolve the issue that energy storage in the Parallel Plate capacitor can be the main idea. First, select the first instance. After that, utilize the formula to determine the amount of energy that is stored in the parallel plate capacitor. Find the energy stored within the capRead more

Concepts and Reason

To resolve the issue that energy storage in the Parallel Plate capacitor can be the main idea.

First, select the first instance. After that, utilize the formula to determine the amount of energy that is stored in the parallel plate capacitor. Find the energy stored within the capacitor that is parallel to the plate without a dielectric. Then, determine the amount of energy stored inside the capacitor in the event that a dielectric is present. The equation will be transformed into the energy stored in the capacitor without the necessity of dielectric. In this case the equation must not be written in the form of voltage.

Part B employs the equation to calculate the energy lost by the resistor. The equation must be written in the format of an equation for the energy stored in the capacitor in the parallel plate. The equation should be written in the format of voltage, which tells you what the length of time that the battery will remain connected.

Fundamentals

The dielectric as well as the parallel plate capacitors are the storage of energy in the capacitor.

In the case of capacitors, their energy is equivalent to the energy that is lost to the resistor. This equation indicates that Uk is the energy that is lost in resistors. Q is the charge, and k is the dielectric constant. C is the capacitance.

The equation for energy storage in the capacitor with a parallel-plate without dielectric is no longer valid.

In this equation, in this equation, the term “U” is the term used to describe that energy contained in the capacitor that is parallel to the plate without dielectric. The term Q is charge , and C is capacitance.

Part B has the same configuration as part A. However, the battery remains connected. The equation Q = CVq = CV is, therefore, completed.

The energy stored by capacitors made of dielectric in parallel plates is therefore absorbed by

This equation illustrates that Uk is the energy that is lost to resistors. V is the voltage, and k is the dielectric constant. C is the capacitance.

The equation that describes the energy stored by the capacitor with a parallel-plate design without dielectric is

In this equation, in this equation is the energy that is dissipated by the resistance, the term V is the voltage, k the dielectric constant, and the C is the capacity.

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## Draw structural formulas for organic products a and b in the window below.

This answer was edited.1) Lithium reacts to butyl bromide and froms butyl lithium , which is transformed into Gilmann reagent after treamenting with Cul. The mechanism of eactions is as follows 2) Lithium reacts with isopentyl Iodide. From this reaction is lithium that is transformed into Gilmann chemical reagent aRead more

1) Lithium reacts to butyl bromide and froms butyl lithium , which is transformed into Gilmann reagent after treamenting with Cul. The mechanism of eactions is as follows

2) Lithium reacts with isopentyl Iodide. From this reaction is lithium that is transformed into Gilmann chemical reagent after treating with Cul. The eaction pattern is as follows

Butyl bromide reacts with lithium from butyl. It is converted into Gilmann reagent after treatment by Cul. The sequence of eactions is as follows: 2L Cul H2CBr H3C-H3C H3C non-commodo cu . CH3 pentane 130 (A) -LIB (B. Gilmann reagent) ii. Lithium reacts with isopentyl iodide and also from itopentyl-lithium. It is converted to Gilmann-reagent upon treatment by Cul. The reaction is as like this Ch3 Chi3 Li Cul H2C1 LA *HAC the HOC pentane. -Lil (B. Gilmann reagent)

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