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Hunter-Best [27]
3 years ago
10

Pls help with this question i’ll give brainliest

Physics
1 answer:
bearhunter [10]3 years ago
6 0

Answer:

5m/s to the right

Explanation:

Momentum = mass * velocity

Momentum before = momentum after

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

3000*10 + 1000*0 = 3000*v₁ + 1000*15

30000-15000=3000v₁

15000=3000v₁

v₁=5m/s to the right (to the right because answer is positive)

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A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

<h3>General wave equation for simple harmonic motion</h3>

y = A sinωt

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<h3>Amplitude of the oscillation</h3>

A = 6 units

<h3>Angular frequency of the wave</h3>

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A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

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Complete Question

The diagram of with this question is shown on the first uploaded image

Answer:

The value is  v = -6.543  \^  i + 9.47 \^ j + 0 \^ k

Explanation:

From the question we are told that

   The mass of the rock is  m = 0.250 \ kg

    The length of the string is  L = 0.75 \  m

    The angle the string makes horizontal is  \theta  = 11.9^o

     The angle which the projection of the string onto  the xy -plane makes with the positive x-axis is  \phi = 34.6^o

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      v_x  = -v__{R}} *  sin(\phi)

=>   v_x  =- 11.5 *  sin(34.6)

=>   v_x  = -6.543 \ m/s

The negative sign show that the velocity is directed toward the negative x-axis

Generally the tangential velocity along the y-axis is  

      v_y = v__{R}} *  cos(\phi)

=>   v_y  = 11.5 *  cos(34.6)

=>   v_y  = 9.47 \ m/s

Generally the tangential velocity along the y-axis is  

      v_z = v__{R}} *  cos(90)

=>   v_z = 0 \ m/s

Generally the tangential velocity at that instant is mathematically represented as

       v = -6.543  \^  i + 9.47 \^ j + 0 \^ k

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