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N76 [4]
2 years ago
11

Explain diffraction at a single slit (light)

Physics
1 answer:
Leto [7]2 years ago
4 0

Answer:

At some point on say, the receiving screen, light emanating from the left side of the slit will be out of phase (a difference of 1/2 wavelengths) from   light coming from the center of the slit.

Thus for every point that is left of the center of the slit, there will be a point on the right side of the slit that is out of phase,

There will be no light on the screen at that particular point and thus there will be a dark fringe there.

That is the basic explanation for the appearance of dark and bright fringes on the receiving screen.

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A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead, which is much denser than aluminum.
zalisa [80]

Answer:

Aluminium

Explanation:

When a body is immersed in a liquid partly or wholly it experiences an upward force which is called buoyant force.

The amount of buoyant force depends on the volume of body immersed, density of liquid and the value of acceleration due to gravity.

Here, the density of liquid is same in both the cases and g be the same. So, here the amount of buoyant force depends on the volume of body immersed.

As the density of lead is more than the density of aluminium, so the volume of aluminium is more than lead, as volume is equal to mass divided by density. So, the buoyant force acting on the aluminium is more than lead.

7 0
3 years ago
What would happen to the amount of matter on earth if mass were not conserved during changes of state?
attashe74 [19]
<span>earth would be thrown off its balance and nature would be in danger of too many resources and not enough resources </span>
4 0
3 years ago
Read 2 more answers
Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t
Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field.  </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy U_{p} will be:  

U_{p}=mgh  

Where m is the mass of the object, g=9.8m/s^{2} the acceleration due gravity and h=500m the height of the object.  

Knowing this, let's begin with the calculaations:

For m=3kg

U_{p}=(3kg)(9.8m/s^{2})(500m)  

U_{p}=14700J=14.7kJ  

For m=6kg

U_{p}=(6kg)(9.8m/s^{2})(500m)  

U_{p}=29400J=29.4kJ  

For m=9kg

U_{p}=(9kg)(9.8m/s^{2})(500m)  

U_{p}=44100J=44.1kJ  

6 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
4 years ago
All of following are types of kinetic energy EXCEPT for: *
Valentin [98]
Light energy is not kinetic energy
7 0
3 years ago
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