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mr_godi [17]
2 years ago
12

9 the only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 n. the canister initially has

a velocity of 4.0 m/s in the positive x direction and some time later has a velocity of 6.0 m/s in the positive y direction. how much work is done on the canister by the 5.0 n force during this time?
Physics
1 answer:
-Dominant- [34]2 years ago
6 0
We don't need to know the magnitude of the force.

The canister's initial kinetic energy is (1/2) M V₀² = 16 joules.

Its final kinetic energy is  (1/2) M V₁² = 36 joules.

Its kinetic energy has grown by 20 joules.  The additional energy
must have come from the 5.0 N force, since that's the only force
acting on the canister.  
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natita [175]

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

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<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

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CQ

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a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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