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Physics vector question
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(a) The west bound velocity component = 41.cos(60) km/h = 20.5 km/h The north bound velocity component = 41.sin(60) km/h = 35.5 km/h Hence, velocity of hurricane = (-20.5i + 35.5j) km/h where i and j are unit vectors pointing in the east and north directions respectively. (b) New velocity = 25j km/h (c) Displacement vector in the first 3 hours = (-20.5i + 35.5j) x 3 km = (-61.5i + 106.5j) km (d) Displacement vector in 1.5 hours = 25j x 1.5 km = 37.5j km (e) Total displacement in 4.5 hours = [(-61.5i + 106.5j) + 37.5j ] km = (-61.5i + 144j) km Hence, distance from Grand Bahama = square-root[(-61.5)^2 + 144^2] km = 157 km
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Physics vector question
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As it passes over Grand Bahama Island, the eye of the hurricane is moving in a direction 60.0 degrees north of west with a speed of 41.0 km/h.a) What is the unit-vector expression for the velocity of the hurricane?It maintains this velocity for 3.00 h, at which time the course of the hurricane suddenly shifts... 顯示更多 As it passes over Grand Bahama Island, the eye of the hurricane is moving in a direction 60.0 degrees north of west with a speed of 41.0 km/h. a) What is the unit-vector expression for the velocity of the hurricane? It maintains this velocity for 3.00 h, at which time the course of the hurricane suddenly shifts due north, and its speed slows to a constant 25.0 km/h. This new velocity is maintained for 1.50 h. b) What is the unit-vector expression for the new velocity of the hurricane? c) What is the unit-vector expression for the displacement of the hurricane during the first 3.00h? d) What is the unit-vector expression for the displacement of the hurricane during the latter 1.50 h? e) How far from Grand Bahama is the eye 4.50 h after it passes over the island? plz help thx!!最佳解答:
(a) The west bound velocity component = 41.cos(60) km/h = 20.5 km/h The north bound velocity component = 41.sin(60) km/h = 35.5 km/h Hence, velocity of hurricane = (-20.5i + 35.5j) km/h where i and j are unit vectors pointing in the east and north directions respectively. (b) New velocity = 25j km/h (c) Displacement vector in the first 3 hours = (-20.5i + 35.5j) x 3 km = (-61.5i + 106.5j) km (d) Displacement vector in 1.5 hours = 25j x 1.5 km = 37.5j km (e) Total displacement in 4.5 hours = [(-61.5i + 106.5j) + 37.5j ] km = (-61.5i + 144j) km Hence, distance from Grand Bahama = square-root[(-61.5)^2 + 144^2] km = 157 km
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