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3 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

Physics

1 answer:

DIA [1.3K]3 years ago

7 0

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

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The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

chubhunter [2.5K]

Answer:

The angular separation equals $0.35^{o}$

Explanation:

We have according to Snell's law

$n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})$

$\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})$

Using this equation for both the colors separately we have

$\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}$

Similarly for violet light we have

$\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}$

Thus the angular separation becomes

$\Delta \theta =14.48-14.13=0.35^{o}$

6 0

3 years ago

State Schrodinger equation.

Elenna [48]

Hi!Schrodinger equation is written as HΨ = EΨ, where h is said to be a Hamiltonian operator.

3 0

2 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

An object located near the surface of Earth has a weight of a 245 N

gayaneshka [121]

Answer:

ثر أنواع التربة خصوبة التربحمراء .

ج- السوداء

Explanation:

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7 0

3 years ago

A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Î© resistor, and an open switch.A 8.2-V battery is connected

tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$

Put the value into the formula

$I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})$

$I(t)=0.01925\ A$

$I(t) = 19.25\ mA$

(B). We need to calculate the store energy in the inductor

Using formula of energy

$E=\dfrac{1}{2}LI^2$

Put the value into the formula

$E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2$

$E=7.04\times10^{-6}\ J$

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

8 0

3 years ago