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hjlf
3 years ago
12

The umbra during a SOLAR OR LUNAR eclipse is smaller than during a SOLAR OR LUNAR eclipse.

Physics
1 answer:
skad [1K]3 years ago
3 0
The umbra during a solar eclipse is smaller than during a lunar eclipse.
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What is the index of refraction of a material in which the speed of light is 7.50% slower than the speed of light in vacuum?
Inessa [10]

Answer:

The value is n = 1.081

Explanation:

From the question we are told that

   The speed of in a vacuum is c = 3.0 *10^{8} \  m/s

    The speed of light in the material is  v  =  c -0.075c = 0.925 c =  0.925 * 3.0*10^{8}  = 2.775 *10^{8} \ m/s

   Generally the reflection of the material is mathematically represented as

             n = \frac{c}{v}

=>    n = \frac{3.0*10^{8}}{2.775 *10^{8}}

=>    n = 1.081

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Answer:

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What type of seismic waves travel through Earth?
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Answer:d. Ground waves

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A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas
Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is D = 1.38 m

Explanation:

From the question we are told that

      The length of the string is  L = 1.6 \ m

      The mass of the ball is  m = 200 g = \frac{200}{1000} = 0.2 \ kg

       The height of ball is  h = 1.5 \ m

Generally the work energy theorem can be mathematically represented as

               PE = KE

   Where PE is the loss in potential energy which is mathematically represented as

                   PE =mgh

Where h is the difference height of ball at A and at B  which is mathematically represented as

                 h = y_A - y_B

So        PE =mg(y_A - y_B)              

While KE is the gain in kinetic energy which is mathematically represented as

               KE   = \frac{1}{2 } (v_b ^2 - 0)

Where v_b is the velocity of the of the ball

  Therefore we have from above that

                    PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)

               Making v_b the  subject we have

      v_b = \sqrt{2g (y_A - y_B)}

substituting values

      v_b = \sqrt{2g (1.5 - 0.40)}

     v_b = 4.6 \ m/s

Considering velocity of the ball when it hits the  floor in terms of its vertical and horizontal component we have

         v_x = 4.6 m/s \ while \ v_y = 0 m/s

The time taken to travel  vertically from the point the ball broke loose  can be obtained using the equation of motion

            s = v_y t - \frac{1}{2} g t^2

Where s is distance traveled vertically which given in the diagram as s = -0.4

The negative sign is because it is moving downward

     Substituting values

              -0.4 = 0 -\frac{1}{2}  * 9.8 * t^2

         solving for t we have  

               t = 0.3 \ sec

Now the distance traveled on the horizontal is mathematically evaluated as

           D =  v_b * t

           D =  4.6 * 0.3

           D = 1.38 m

8 0
3 years ago
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