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2 years ago

6

Two point charges are placed on the x axis as follows: charge q1=+3.75nc is located at x=0.205m and charge q2=−5.60 nc is at x=+

0.400 m. what is the magnitude of the net force exerted by these two charges on a third point charge q3=−0.610nc placed at the origin?

1 answer:

Mademuasel [1]2 years ago

6 0

I might have did mistake with calculations but this is how you should do.

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lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
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By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
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part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
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Explanation:

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Can you please stop pasting this question, just go to his profile and ask him.

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