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a 3.6 kg object is accelerated from rest to a speed of 34.4 m/s in 42 s. What average force was exerted on the object during this period of acceleration. Answer in units of N.
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I just did that, scroll down.
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a 3.6 kg object is accelerated from rest to a speed of 34.4 m/s in 42 s. What average force was exerted on the object during
Top answer:
F = m*a a = V-Vo)/t = (34.4-0)/42 = 0.819 m/s^2. F = 3.6*0.819 = 2.95 N.
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a 3.6 kg object is accelerated from rest to a speed of 34.4 m/s in 42 s. What average force was exerted on the object during
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F=ma=mv/t
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A 3.6 kg object is accelerated from rest to a
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average acceleration = change in speed / change it time = (34.4 - 0) / (42 - 0 ) = .819 m/s^2 Force
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A 2.0 kilogram object that is intially at rest is accelerated to a speed of 5.0 meters per second
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whatever your question is, I suspect you are missing some information.
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acceleration = (8.0 m/s - 0 m/s) / 2.0 s = 4.0 m/s^2 The correct answer is C. 4.0 m/s^2
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An object originally at rest is uniformly accelerated to a speed of 8.0 meters per second in 2.0 seconds. What is the
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The cruising speed in this case is the speed attained after the 1 min of acceleration. Call it V.
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Using the formula for acceleration: acceleration = (vfinal - vinitial)/time Given: vinitial = 0 m/s
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The acceleration of the object can be calculated using the formula: acceleration = (final velocity -
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To find the acceleration of the object, we can use the formula: acceleration = (final velocity -
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