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2 years ago

Metals are used in many products because of the characteristic properties that most metals have.

Physics

1 answer:

Kaylis [27]2 years ago

8 0

a) Mirrors

Mirror requires the high luster of metals in order to work.

<h3>What causes the highest luster of a metal?</h3>

When light shine on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals.

The electrons immediately fall back down to lower energy levels and emit light.

This process is responsible for the high luster of metals.

Here,

Mirrors have metal coating on its back side, because of high reflective property , high lustrous mirror is preferred . When metal is more lustrous, the less light reflect into it and bounces back.

Coating one side of a piece of glass with shiny metals can turn it into a mirror, reflecting light coming toward it.

Window glass can reflect only eight percent of light hitting it, while mirrors can reflect 95 percent of light hitting them.

The glass in a mirror is usually coated with a layer of silver or aluminum.

Mirrors are shiny because when photons (rays of light) coming from an object , it strike the smooth surface of a mirror, then they bounce back at the same angle. Our eyes see these reflected photons as a mirror image.

Therefore,

Mirror requires the high luster of metals in order to work.

Learn more about properties of metal here:

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sergejj [24]

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3 years ago

Particle A of charge 3.30 10-4 C is at the origin, particle B of charge -6.24 10-4 C is at (4.00 m, 0), and particle C of charge

gulaghasi [49]

Answer:

a) $E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) zero

Explanation:

a) To find the electric field at point C, you sum the contribution of the electric fields generated by the other two charges. The total electric field at C is given by:

$E_T=E_1+E_2$

E1: electric field of charge 1

E2: electric field of charge 2

It is necessary to calculate the x and y components of both E1 and E2. You take into account the direction of the fields based on the charge q1 and q2:

$E_1=k\frac{q_1}{r_{1,3}}[cos\theta\hat{i}+sin\theta \hat{j}]\\\\E_2=k\frac{q_2}{r_{2,3}}[cos\phi\hat{i}-sin\phi \hat{j}]\\\\$

r13: distance between charges 1 and 3

r12: charge between charges 2 and 3

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

Thus, you first calculate the distance r13 and r23, and also the angles:

$r_{1,3}=3.00m\\\\r_{2,3}=\sqrt{(3.00m)^2+(4.00m)^2}=5.00m\\\\\theta=90\°\\\\\phi=tan^{-1}(\frac{4.00}{3.00})=53.13\°$

Next, you replace the values of all parameters in order to calculate E1 and E2:

$E_1=(8.98*10^9Nm^2/C^2)(\frac{3.30*10^{-4}C}{(3.00m)^2})\hat{j}\\\\E_1=329266.66\frac{N}{C}\\\\E_2=(8.98*10^9Nm^2/C^2)(\frac{6.24*10^{-4}C}{(5.00m)^2})[cos53.13\°\hat{i}-sin(53.13\°)\hat{j}]\\\\E_2=224140.8[0.6\hat{i}-0.8\hat{j}]=134484\hat{i}-179312\hat{j}$

finally, you obtain for ET:

$E_T=134,484\frac{N}{C}\hat{i}+(329266.66-179312)\frac{N}{C}\hat{j}\\\\E_T=134,484\frac{N}{C}\hat{i}+149954.66\frac{N}{C}\hat{j}$

b) The x component of the force exerted by A on C is zero because there is only a vertial distance between them. Thus, there is only a y component force.

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_____ _____hello mate_______

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Explanation let me know if im wrong

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Answer:

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