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bearhunter [10]
2 years ago
9

A woman drives a 45 gram golf ball off of a tee and gives the ball a velocity of 28 m/s to the right. What is the change in mome

ntum of the ball?
Physics
1 answer:
lidiya [134]2 years ago
7 0

Answer:

1.26

Explanation:

change 45 grams to kilograms which is 0.045 then just multiply it by 28 m/s

0.045x28=1.260

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Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".  

Therefore, torque equation about A will be as follows.

   60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)

      d = \frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}

       d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.

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Explanation:

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During the final stages of descent, asky diver with an
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Answer:

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Explanation:

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Answer:

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