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Elza [17]
3 years ago
8

Present the ways to observe interference patterns on thin films. Why the thickness of thin films should be in scale of wavelengt

h?
Physics
2 answers:
frez [133]3 years ago
6 0

the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs

antoniya [11.8K]3 years ago
5 0

Answer:

A colorful interference pattern is observed when light is reflected from the top and bottom boundaries of a thin oil film. The different bands form as the film's thickness diminishes from a central run-off-poin

Explanation:

<h3>hope it help brainliest pls</h3>
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A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t
diamong [38]

Answer:

11.98 N

Explanation:

Normal force =   mg =  2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

       .602 = F / (2.03 * 9.81)   = 11.98  N

5 0
2 years ago
The weight of an object is the same on Earth and the moon.
madreJ [45]

Answer:

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Explanation:

3 0
3 years ago
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Which equation represents the law of conservation of energy in a closed system?
Irina-Kira [14]

Answer:

KE + PE = KE + PE

Explanation:

In a closed system, the mechanical energy of the system is constant.

Mechanical energy is given by the sum of kinetic energy and potential energy; mathematically:

U = KE + PE

where

KE is the kinetic energy

PE is the potential energy

This means that if we consider two situations, one at the beginning and one at the end, the value of U will not change if the system is closed; this means that the sum KE + PE will remain the same, so we can write:

KE + PE = KE + PE

7 0
3 years ago
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Pls help me on question 4 it is true of false and it is super easy sixth grade stuff
koban [17]
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3 0
3 years ago
Find the deBroglie wavelength of an electron with 4.0 eV.
pantera1 [17]

Explanation:

It is given that,

Voltage, V=4 eV=4\times 1.6\times 10^{-19}\ V= 6.4\times 10^{-19}\ V

De broglie wavelength in terms of voltage is given by :

\lambda=\dfrac{h}{\sqrt{2meV} }

m and e are the mass and charge on electron. So,

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

\lambda=\dfrac{12.27}{\sqrt{6.4\times 10^{-19}} }\ A  

\lambda=1.53\times 10^{10}\ A

\lambda=1.53\ m

So, the De broglie wavelength of an electron is 1.53 meters. Hence, this is the required solution.      

3 0
3 years ago
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