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san4es73 [151]
3 years ago
13

At exactly 3:14PM, the velocity of a dog running in a park points toward a group of flowers. Which of the following best describ

e the forces acting on the dog at 3:14PM?
There may be many forces acting on the dog, but the forces pointing toward the flowers are larger than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are smaller than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are equal to the forces pointing away from the flowers.
Options A and B are both possible.
Options A and C are both possible.
Options A, B, and C are all possible.
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

Options A, B, and C are all possible.

Explanation:

We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?

If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.

If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.

If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers  nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.

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Explanation:

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10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
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Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

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The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

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F_f = ma

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