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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:

brainly.com/question/18153051

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Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the accelerat

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The period of a pendulum is given by
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length and g is the gravitational acceleration.

We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
$\frac{T_m}{T_e} = \frac{2 \pi \sqrt{ \frac{L}{g_m} } }{2 \pi \sqrt{ \frac{L}{g_e} } }= \sqrt{ \frac{g_e}{g_m} }$
where the labels m and e refer to "Moon" and "Earth".

Since the gravitational acceleration on Earth is $g_e = 9.81 m/s^2$ while on the Moon is $g_m=1.63 m/s^2$ , the ratio between the period on the Moon and on Earth is
$\frac{T_m}{T_e}= \sqrt{ \frac{g_e}{g_m} }= \sqrt{ \frac{9.81 m/s^2}{1.63 m/s^2} }=2.45$

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2 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

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2 years ago

You have a small piece of iron at 75 °C and place it into a large container of water at 25 °C. Which of these best explains what

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D. Heat energy will be transferred within the system and if left long enough, there will be enough transferred energy to make both of them the same temperature.

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3 years ago

Read 2 more answers

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

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3 years ago

What happens if :<br> . The test charge is not tiny.

docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

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Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

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2 years ago