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Firlakuza [10]
2 years ago
9

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

ork done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^
Physics
1 answer:
tatyana61 [14]2 years ago
7 0
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
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In the first case:

when we heat any gas, the Kinetic Energy of the molecules increases, making it collide more frequently with the surface, increasing the pressure

more collisions with the surface means more force applied on it, which would push the piston harder than before, moving it outwards.

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since the molecules inside the beaker have no way to escape, they would keep compressing the more you push the beaker downwards.

since there is the same number of molecules and lesser volume to cover, the molecules will start colliding with the surfaces more frequently, which would resist the downward force.

<em>another way to think about it is to imagine yourself where the trapped air is. you would be happy when the room is spacious but if the wall starts moving towards you, you would resist the change by your body because you need space to exist. making it harder for the wall to move.</em>

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just like in the first case, the heated air will apply force on the surface, including the cork. which would pop off when enough force is applied.

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