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3 years ago

13

If you were to separate all of the electrons and protons in 1.00 g (0.001 kg) of matter, you’d have about 96,000 C of positive c

harge and the same amount of negative charge. If you placed these charges 8.00 m apart, how strong would the attractive forces between them be?

1 answer:

natima [27]3 years ago

3 0

Answer:

The attractive force between them is $1.296 \times 10^{18}$ N

Explanation:

Given:

Charge $q = 96000$ C

Distance between two charges $r = 8$ m

According to the coulomb's law,

$F = \frac{kq^{2} }{r^{2} }$

Where $k = 9 \times 10^{9}$ = force constant.

$F = \frac{9 \times 10^{9} \times (96000)^{2} }{8^{2} }$

$F = 1.296 \times 10^{18}$ N

Therefore, the attractive force between them is $1.296 \times 10^{18}$ N

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VMariaS [17]

A. the light bulb goes out once the circuit is open since it causes the flow of electricity to cut off. the light bulb dosent get the energy it needs to light up

Explanation:

B. a simple example of this in our every day life is a light switch. when you switch the light on then the circuit is closed and the energy transfers to the light bulb, when u switch the light off then you cut off the lights source of energy which causes the light to turn off.

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1 year ago

Is a spring stores 5 J of energy when its conpresses by 0.5 m what is the spring constant of the spring?

d1i1m1o1n [39]

Answer:

k = 40 N/m

Explanation:

A spring's energy is given:

$U = 0.5kx^2$

U is the energy in the spring, k is the spring constant and x is the spring displacement.

We are told that the spring stores 5J of energy, therefore, U = 5J. We are also told that the spring is compressed by 0.5m, so the spring x = 0.5m

$5 J = 0.5k(0.5m)^2\\5J = 0.5k(0.25m^2)\\5J = k*(0.125)m^2\\\\k = 5J/(0.125)m^2\\k = 40 N/m$

k = 40 N/m

Hope this helps!

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3 years ago

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

3 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

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Answer:

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2 years ago