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Ivenika [448]
3 years ago
5

Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom

entum p⃗ = ( -3.10 kg⋅m/s )i^+( 3.90 kg⋅m/s )j^ .
What is the momentum of the box at t = 1.90 s ?


Enter the x and y components of the momentum separated by a comma.
Physics
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

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Answer:

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Explanation:

As per the option A statement, the force acting towards left is greater than the force acting toward right side. So the net force will be towards the direction having maximum magnitude. Thus, the box will move toward left side in option A. The same situation arises for the object in option B. But here the difference in the forces is only 1 N, so the change in the position of the object will be very less. Thus it may look like there is no acceleration in the box of option B.

Similarly, the force acting on the objects given in option C and D have magnitude greater towards the right side than towards the left side.  So these two will be accelerated toward the right side.

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2 years ago
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kow [346]

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

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Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

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