Physics, 06.05.2020 05:02 elf55

Two 3 kg spherical masses are connected by a 20 m long thread of negligible mass that does not stretch. The masses are charged so they have the same amount of negative charge and the system is hung from the ceiling by the center of the thread.

Determine the charge of each sphere if the angle between the suspended sides of the thread is 17 degree.

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Answer from: ddddre4460

The charge of each sphere is $q = - 1.84 *10^{-4}C$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The mass of the two sphere is $m = 3 \ kg$

The length of the connection is $l = 20\ m$

The angle between the suspended side is $\theta = 17 ^o$

At Equilibrium the force acting at the horizontal is = 0N and the net force acting at the vertical is zero

This can be represented mathematically as

$\sum F_y = 0N , \ \ and \ \ \sum F_x = 0N$

For $\sum F_y = 0N$

$T cos \theta = mg$

=> $cos \theta = \frac{mg}{T}$

For $\sum F_x = 0N$

$T sin \theta = F$

=> $sin \theta = \frac{F}{T}$

Now $tan \theta = \frac{sin \theta }{cos \theta }$

$= \frac{\frac{F}{T} }{\frac{mg}{T} }$

$= \frac{F}{mg}$

Where F is the electrical force which is mathematically represented as

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

Therefore

$tan \theta = \frac{kq^2}{mgr^2}$

Where is the distance between the two masses and from the diagram it is

$r = 2l sin \theta$

$tan \theta = \frac{kq^2 }{(2l sin \theta )^2 * mg}$

making q the subject of the formula

$q = \sqrt{\frac{mg}{k} * tan \theta (2l sin \theta )^2 }$

Where k is the Coulomb's constant with a value of $k = 9*10^{9} kg \ cdot m^3 s^{-4} A^{-2}$

Substituting values

$q = \sqrt{\frac{3 * 9.8 }{9*10^9} * tan (17) (20 sin 17)^2 }$

$q = - 1.84 *10^{-4}C$

The negative sign is because we are told from the question that they are negatively charged

Two 3 kg spherical masses are connected by a 20 m long thread of negligible mass that does not stret

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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