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Elena-2011 [213]
3 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]3 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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1 poir
djverab [1.8K]

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

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Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

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p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm

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