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Lena [83]
3 years ago
8

Please help me with the equations for this! Three uniform spheres are fixed at the positions shown in the diagram. ( there is a

1kg mass .5 m to the right of the origin, a 1kg mass .5m directly above the origen, and a 2kg mass at (.5,.5))Assume they are completely isolated and there are no other masses nearby. If the 0.20 kg particle is placed at (x,y) = (-500 m, 400 m) and released from rest, what will its speed be when it reaches the origin? (c) How much energy is required to separate the three masses so that they are very far apart?
Physics
1 answer:
lora16 [44]3 years ago
7 0
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!

<span>PE=−[G<span>M1</span><span>M2</span>]÷R

</span><span> Potential energy of a particle due to mass A is not affected by presence of any other mass B !</span>
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The magnetic force acting on the proton is 
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\theta is the angle between the direction of v and B; since the proton is moving perpendicular to the magnetic field, \theta=90^{\circ} and \sin \theta=1, so the force becomes
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this force provides the centripetal force that keeps the proton in circular motion:
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Re-arranging the previous equation, we can find the radius of the proton's orbit:
r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(6.5 m/s)}{(1.6 \cdot 10^{-19} C)(1.8 T)}=3.77 \cdot 10^{-8}m

And now we can calculate the centripetal acceleration of the proton, which is given by
a_c =  \frac{v^2}{r}= \frac{(6.5 m/s)^2}{3.77\cdot 10^{-8}m}=1.12 \cdot 10^9 m/s^2

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