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sveta [45]
3 years ago
8

A volleyball is spiked so that it has an initial velocity of 17.6 m/s directed downward at an angle of 51.3 ° below the horizont

al. What is the horizontal component of the ball's velocity when the opposing player fields the ball?
Physics
1 answer:
nata0808 [166]3 years ago
4 0

<u>Answer:</u>

 Horizontal component of the velocity when the opposing player fields the ball = 11.00 m/s

<u>Explanation:</u>

   The velocity of a body in 2 dimension can be resolved in to 2 parts, horizontal and vertical component. In the case of free falling or projectile body the horizontal component remains the same but vertical component is affected by acceleration due to gravity.

   In this case Initial velocity = 17.6 m/s

   Angle between horizontal axis and direction of velocity = 51.3 °

   We know that horizontal component = v cos θ

                               Vertical component = v sin θ

    Since the horizontal component remains the same, it is unchanged when the opposing player fields the ball.

    So horizontal component of the velocity = v cos θ = 17.6 * cos 51.3 °

                                                                         = 11.00 m/s

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One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

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N = Number of loops

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A = Cross-sectional Area

\omega = Angular velocity

Re-arrange to find N,

N = \frac{\epsilon}{BA\omega}

Our values are given as,

\epsilon = 19V

B = 0.434T

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N = \frac{\epsilon}{( 0.434)A\omega}

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Explanation:

<em>mirrors : concave mirror gives upside down images everytime except when the object is located in front of F (focal point)</em>

  1. The object is located beyond C
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   2. The object is located between C and F  or on C

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<u>The properties of images</u>

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Characteristics of the images of each mirror and lens are mentioned in tables for further reference

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