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a 10-kg bos slides down a frictionless ramp with acceleration 3.6 m/s calculate the weight
anlge? f ?
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Whay did you post the same question four times?
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a 10-kg bos slides down a frictionless ramp with acceleration 3.6 m/s calculate the weight....the angle and normal force
Top answer:
Weight W = M g = 98 N Acceleration (a) = F/M where F = W sin A a = M g sin A/M = g sin A sinA =
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a 10-kg bos slides down a frictionless ramp with acceleration 3.6 m/s calculate the weight....the angle and normal force
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enough, already
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a 10-kg bos slides down a frictionless ramp with acceleration 3.6 m/s calculate the weight....the angle and normal force
Top answer:
Duplicate post, including the misspelling of "box" See my other answer.
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I am having great difficulty with this question.
A block is placed on a frictionless ramp at a height of 12.5 m above the ground.
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To answer these questions, we need to apply principles of conservation of energy and motion. Let's
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A block is placed on a frictionless ramp at a height of 14.5 m above the ground. Starting from rest, the block slides down the
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.022
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(a) Suppose a hockey puck slides down a frictionless ramp with an acceleration of 5.60 m/s2. What angle does the ramp make with
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To determine the angle of the ramp with respect to the horizontal, we need to use trigonometry. The
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EXERCISE 4.9 (a) Suppose a hockey puck slides down a frictionless ramp with an acceleration of 5.00 m/s2. What angle
does the
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(a) Well, the angle of the ramp must be really "acute" to have such a high acceleration! Badum tss!
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Suppose a hockey puck slides down a frictionless ramp with an acceleration of 4.70 m/s2.
(a)What angle does the ramp make with
Top answer:
weight component down ramp = m g sin T so m g sin T = m a 4.7 = 9.81 sin T sin T = .479 T = 28.6
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Suppose a hockey puck slides down a frictionless ramp with an acceleration of 4.70 m/s2. What angle does the ramp make with
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To find the angle that the ramp makes with respect to the horizontal, we can use trigonometry.
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A 1.80-kg block slides down a frictionless ramp.The top of the ramp is h1 = 1.21 m above the ground; the bottom of the ramp is
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(1/2) m u^2 = m g (Hup - Hdown) so u = sqrt(2*9.81*.954) u is horizontal speed, does not change How
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