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Fudgin [204]
4 years ago
6

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

the Coulomb force between the two is to be maximized?
Physics
1 answer:
Lisa [10]4 years ago
6 0

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

F = \frac{Kq_{1}q_{2}}{r^{2}}

So, the force between the charges q and Q - q is given by

F = \frac{K\left ( Q-q \right )q}}{r^{2}}

For maxima and minima, differentiate the force with respect to q.

\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now \frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}

the double derivate is negative, so the force is maxima when q = Q / 2 .

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Explanation:

Vi = 12 m/s

a = 3 m/s^2

t = 2 s

Vf = Vi + a × t = 12 + 3 ×2 = 18 m/s

3 0
3 years ago
Roseanne and Jackie were asked to design and create a paper airplane that would stay in the air for as long as possible. They be
jekas [21]

Answer:

the planes times in the air

Explanation:

When a paper airplane is thown the path it takes is not the same every time. So, measuring the distance between the points where the airplane took off and the point where is landed would not show the time the plane was in the air.

The weight also does not determine the time the plane was in the air.

So, the planes air time recorded with the stopwatch gives the time the plane was in the air and will indicate the best plane.

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3 years ago
Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between
Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

\Delta V=Ed

where

\Delta V is the potential difference

E is the magnitude of the electric field

d is the  distance between the plates

In this problem, we have:

\Delta V=750 V is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m

6 0
4 years ago
8. Ella wants to replace old bulbs of her home with new LED bulbs. She buys LED bulb that has inbuilt AC to DC converter of outp
tankabanditka [31]

The power dissipated by the LED is 20 Watts and the work done for 1 hour 15 minutes is 56.25 kJ.

<h3>What is electrical power?</h3>

Electrical power is the rate at which electrical work is done.

  • Electrical power = voltage × current

The LED is 75% efficient means that 75% of power dissipated by the LED is converted to light.

Total power dissipated = 5 × 2.5 = 12.5 Watts

  • Work done = power × time (in seconds)

Work done = 12.5 × (1 × 3600 + 15 × 60)

Work done = 56250 J = 56.25 kJ

Therefore, the power dissipated by the LED is 20 Watts and the work done for 1 hour 15 minutes is 56.25 kJ.

Learn more about electrical power and work done at: brainly.com/question/23901751

#SP1

3 0
2 years ago
Which<br> factors will increase the speed of a sound wave in the air?
Dafna11 [192]
A higher temperature, stiffer materials, and less dense materials increase the speed of sound.
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4 years ago
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