The velocity of the target and arrow after collision is 6.67m/s
<u>Explanation:</u>
Given:
Mass of arrow, mₐ = 415g
Speed of arrow, vₐ = 68.5m/s
Mass of the target, mₓ = 3.3kg = 3300g
speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction
Velocity of the target and arrow after collision, vₙ = ?
Applying the conservation of momentum,
mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ
415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ
28427.5 - 3630 = 3715 X vₙ
24797.5 = 3715 X vₙ
vₙ = 6.67m/s
Therefore, the velocity of the target and arrow after collision is 6.67m/s
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Your right. It's C. This question is difficult
Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
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