a steady horizontal force lasting for 2.10 s gives a 1.25 kg object an acceleration of 3.20
Homework Answers
We know that impulse
J = F * t
= m * a * t
= 1.25 * 3.20 * 2.10
= 8.4 N-s
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a steady horizontal force lasting for 2.10 s gives a 1.25 kg object an acceleration of 3.20
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Two forces act on a 3.20 kg object, the gravitational force and a second, constant force. The object starts from rest and in 1.20 s is displaced (4.251 - 3.301) m. Write the second force in unit vector notation. (Enter your answer in kg. m/s. Assume the gravitational force acts in the - direction.) = (18.888)i + (17.152); First, find the acceleration vector using kinematics, then substitute into Newton's second law to determine the unknown force vector. Recall that the...
A 1.30-kg object is attached to a spring and placed on
frictionless, horizontal surface. A horizontal force of 21.0 N is
required to hold the object at rest when it is pulled 0.200m from
its equilibrium position......
Would you write out the intermediate steps, too, please?
A 1.30-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 21.0 N is required to hold the object at rest when it is pulled 0.200 m...
A 2.31-kg object on a frictionless horizontal track is attached to the end of a horizontal spr equilibrium position and then released, initiating simple harmonic motion ng whose force constant is 5.00 N/m. The object is displaced 2.88 m to the right from its (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released? magnitude (b) How many times does the object oscillate in 3.50 s? (Do not round your answer to...
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