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nika2105 [10]
3 years ago
11

What happens to light when it enters a lens? A) It speeds up B) It is reflected C) It is refracted D) It becomes brighter.

Physics
2 answers:
choli [55]3 years ago
6 0
When light enters a lens, C) it is refracted.
Since it is bent when it enters a lens, it has to be refracted.
dsp733 years ago
3 0
The answer would be C(refracted ),  it would be refracted because when light enters a lense the light gets bent and the angle changes so meaning this the only option would be refracted.


hope this helps :3

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A box is placed on a table. If an upward pulling force on the box is slightly less than the weight of the box, what will happen
aniked [119]

Answer:

<u><em>A for certain.</em></u>

Explanation:

I got it right on the test, thanks to the other brainly answerer teresecaway. But knowing the answer doesnt help much if you dont know WHY. The reason WHY is as follows. At first you might think it would be B, because the downwards force is greater than the upwards force. I thought that maybe upwards force would also count as the table supporting it, but no, thats just structural inanimate solid table. The answer is A. gravity would have to be HIGHLY raised or somebody pressing down on it, to make it fall through the table.

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3 years ago
Which graph shows an object that is dropped?
Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
4 years ago
During which two moon phases would the amount of light reflected from the moon appear to be equal from Earth? Why does this occu
shusha [124]

The moon phase shows that D. light reflected during the first quarter and third quarter moon phases appears equal because the sun, moon and Earth are at right angles.

<h3>What is a moon phase?</h3>

It should be noted that the <em>moon</em> phase simply means the portion of the moon that we can see from the Earth.

In this case, the light reflected during the first quarter and third quarter moon phases appears equal because the sun, moon and Earth are at right angles.

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6 0
3 years ago
The brightest, hottest, an
dangina [55]

The speed of the star is 255 km/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically:

KE=\frac{1}{2}mv^2

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m is the mass of the object

v is its speed

For the star in this problem, we have:

m=3.38\cdot 10^{31} kg is the mass

KE=1.10\cdot 10^{42} J is the kinetic energy

Therefore, the speed of the star is:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.10\cdot 10^{42})}{3.38\cdot 10^{31}}}=2.55\cdot 10^5 m/s

Which corresponds to 255 km/s

Learn more about kinetic energy:

brainly.com/question/6536722

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6 0
3 years ago
A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh
kolezko [41]

Answer:

The angular velocity is  w = 53.35 \ rounds /minute

Explanation:

From the question we are told that

    The mass of the block is  m = 61.2kg

     The of the pulley is  M = 14.2 kg

      The radius of the pulley is  R = 1.5m

       The radius  of the cord around the pulley is  r = 1.5 m

       The distance of the block to the floor is  d = 8.0 m

         

From the question we are told that the moment of inertia of the pulley is

          I  = \frac{1}{2} MR^2 kg \cdot m^2

Substituting value  

         I = \frac{1}{2}  * 14.2 * (1.5)^2

         I = 15.975 kg \cdot m^2

Using the Newtons law we can express the force acting on the vertical axis as

              ma = mg -T

         =>  T = mg -ma

Now when the pulley is rotated that  torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

                  \tau = I \alpha

     Here \alpha is the angular acceleration

           Here \tau is the torque which can be equivalent to

              \tau = T r

  Substituting this above

            Tr = I \alpha      

Substituting for T

         (mg - ma ) r =  I\  r \alpha

Here a is the  linear acceleration which is mathematically represented as

           a = r\alpha

    (mg - m(r\alpha ) ) r =  I\  r \alpha

     mgr = I\alpha  + m(r\alpha ) r

    mgr = \alpha  [ I + mr^2]

   making \alpha the subject

          \alpha  = \frac{mgr}{I -mr ^2}          

   Substituting values

            \alpha  = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}

             \alpha =5.854 rad /s^2

Now substituting into the equation above to obtain the acceleration

             a = 5.854 * 1.5

                a=8.78 m/s^2

This acceleration is a = \frac{v}{t}

and v is the linear velocity with is mathematically represented as

         v = \frac{d}{t}

Substituting this into the formula acceleration

        a = \frac{d}{t^2}

making t the subject

         t = \sqrt{\frac{d}{a} }

substituting value

      t = \sqrt{\frac{8}{8.78}}

     t = 0.9545 \ s

Now the linear velocity is

       v = \frac{8}{0.9545}

       v = 8.38 m/s

The angular velocity is  

       w = \frac{v}{r}

So

       w = \frac{8.38}{1.5}

        w = 5.59 rad/s

Generally 1 radian is equal to  0.159155 rounds or turns

        So  5.59 radian is  equal to x

Now x is mathematically obtained as

         x = \frac{5.59 * 0.159155}{1}

            = 0.8892 \ rounds

 Also

      60  second =  1 minute

So   1 second  = z      

Now z is mathematically obtained as

         z = \frac{ 1}{60}

            z = 0.01667 \ minute

Therefore

              w = \frac{0.8892}{0.01667}

              w = 53.35 \ rounds /minute

           

8 0
3 years ago
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