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Elena-2011 [213]
2 years ago
12

The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the

ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.. . Part C. . What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.. . A) 0, 0, 0, 0. B) 0, 0, 0, -9.80. C) 15.0, 0, 0, 0. D) 15.0, 0, 0, -9.80. E) 0, 26.0, 0, 0. F) 0, 26.0, 0, -9.80. G) 15.0, 26.0, 0, 0. H) 15.0, 26.0, 0, -9.80
Physics
1 answer:
Digiron [165]2 years ago
7 0
<span>Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80

</span><span>Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.
 15, 26, 0, -9.81</span><span>

</span>
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nasty-shy [4]

Answer:

\theta = 25.3^\circ

Explanation:

The acceleration of the block can be found by the kinematics equations:

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Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ

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vazorg [7]
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3 years ago
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Answer:

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Explanation:

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