. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and th
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You did not include the quesetion, but I can help you to understand the problem and how to find the relevant information.
1) The angle of 13° with which the shark ascends meets this:
Vertical ascending velocity = 0.85m/s * sin(13°)
Horizontal velocity = 0.85m/s * cos(13°)
2) The length swan by the shark ascending meets this
Vertical ascending length = 50 m
Horizontal length, y:
From that y = 50 * tan(13°)
=> y = 11.54 m.
3) Conclusions:
1) The shark run 50 m vertically upward and 11.54 m horizontally.
2) The length of the path run by the shark may be calculated using Pythagoras' theorem:
hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2
hypotenuse = 51.35m
So, the shark swan 51.35 m to reach the surface.
1) The angle of 13° with which the shark ascends meets this:
Vertical ascending velocity = 0.85m/s * sin(13°)
Horizontal velocity = 0.85m/s * cos(13°)
2) The length swan by the shark ascending meets this
Vertical ascending length = 50 m
Horizontal length, y:
From that y = 50 * tan(13°)
=> y = 11.54 m.
3) Conclusions:
1) The shark run 50 m vertically upward and 11.54 m horizontally.
2) The length of the path run by the shark may be calculated using Pythagoras' theorem:
hypotenuse^2 = (50m)^2 + (11.54m)^2 = 2633.25m^2
hypotenuse = 51.35m
So, the shark swan 51.35 m to reach the surface.