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2 years ago

What is the kinetic energy of a 6 kg and 2 m long thin rod that rotates with angular velocity of 10 rad/s

Physics

1 answer:

Leni [432]2 years ago

6 0

Answer:

1200 J

Explanation:

From the question

K.E = 1/2mv²...................... Equation 1

Where K.E = Kinetic Energy, m = mass of the rod, v = velocity of the rod

But

V = ωr..................... Equation 2

Where ω = angular velocity of the rod, r = radius of the rod.

Substitute equation 2 into equation 1

K.E = 1/2m(ωr)²..................... Equation 3

Given: m = 60 kg, ω = 10 rad/s, r = 2 m

Substitute these values into equation 3

K.E = 1/2(6)(10×2)²

K.E = 3×(20)²

K.E = 3×400

K.E = 1200 J

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<u>Electric Potential Due To Point Charges </u>

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$\displaystyle V=k\frac{Q}{r}$

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Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.