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qwelly [4]
4 years ago
11

Which of the following equations can be used to directly calculate an object's impulse?

Physics
1 answer:
Iteru [2.4K]4 years ago
7 0
I believe the answer is B
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How are systems different from industries? Use an example to support your answer.
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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
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Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

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3 years ago
To model time-variant data, one must create a new entity in an m:n relationship with the original entity.
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To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.

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To know more about  entity AND relationship  visit : brainly.com/question/28232864

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